# Calculate The Magnitude Of The Gravitational Force Exerted On A 4.20 Kg Baby By A 100 Kg Father 0.200 (2023)

Physics College

The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N

What is gravitational potential energy?

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as,

• (b) The magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Put the values in the above formula again as,

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

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(a). The magnitude of the gravitational force is

(b). The magnitude of the gravitational force is

Explanation:

Given that,

Mass of baby = 4.20 kg

Mass of father = 100 kg

Distance = 0.200 m

We need to calculate the gravitational force

Using gravitational formula

Put the value in to the formula

(b). We need to calculate the gravitational force

Using gravitational formula

Put the value in to the formula

Hence, (a). The magnitude of the gravitational force is

(b). The magnitude of the gravitational force is

## Related Questions

In otherwise empty space is a system of 4 particles, each of the same mass. The accelerations of the particles are as follows: a1 = ˆi a a2 = ˆj a 2 a3 = −( ˆi +ˆj) (2 a) a4 = (ˆi −ˆj) (3 a), where a > 0 is a constant. What is the acceleration of the center of mass of the system of 4 particles?

(a i - 2a j) / 2

Explanation:

a1 = a i

a2 = a j

a3 = - (i + j) (2a) = - 2a i - 2a j

a4 = (i - j) (3a) = 3a i - 3a j

Let the mass of each particle is m.

Use the formula of acceleration of centre of mass

a cm = (m1 a1 + m2 a2 + m3 a3 + m4 a4) / (m1 + m2 + m3 + m4)

a cm = (a i + aj - 2a i - 2a j + 3a i - 3a j) / 4

a cm = ( 2a i - 4a j) / 4

a cm = (a i - 2a j) / 2

A force F S applied to an object of mass m1 produces an acceleration of 3.00 m/s2. The same force applied to a second object of mass m2 produces an acceleration of 1.00 m/s2. (a) What is the value of the ratio m1/m2? (b) If m1 and m2 are combined into one object, find its acceleration under the action of the force F S .

a)

b) Acceleration = 0.75 m/s²

Explanation:

a) We have force , F = mass x acceleration.

We have force value is same

b) We have

Combined mass

Force

Star A has a radius of 200 000 km and a surface temperature of 6 000 K. Star B has a radius of 400 000 km and a surface temperature of 3 000 K. The emissivity of both stars is the same. What is the ratio of the rate of energy radiated by Star A to that of Star B?

Explanation:

In order to solve this problem, the Stefan-Boltzmann law will be useful. This law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

(1)

Where:

is the energy radiated by a blackbody radiator per second, per unit area (in Watts).

is the Stefan-Boltzmann's constant.

is the Surface of the body

is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (although the radiation of stars like our Sun is quite close). Therefore, we will use the Stefan-Boltzmann law for real radiator bodies:

(2)

Where is the star's emissivity

We have two stars where the emissivity of both is the same:

Star A with a radius and a surface temperature .

Star B with a radius and a surface temperature .

And we are asked to find the ratio of the rate of energy radiated by both stars:

(3)

Where is the rate of energy radiated by Star A and is the rate of energy radiated by Star B.

On the other hand, with the radius of each star we can calculate their surface area, using the formula for tha area of a sphere (assuming both stars have spherical shape):

(4)

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(5)

Writting the Stefan-Boltzmann law for each star, taking into consideration their areas:

(6)

(7)

Substituting (6) and (7) in (3):

(8)

(9)

(10)

Finally:

An airplane traveling at half the speed of sound emits a sound of frequency 4.68 kHz. (a) At what frequency does a stationary listener hear the sound as the plane approaches? kHz (b) At what frequency does a stationary listener hear the sound after the plane passes? kHz

(a) 9.36 kHz

(b) 3.12 kHz

Explanation:

(a)

V = speed of sound

= speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

f' = 9360 Hz

f' = 9.36 kHz

(b)

V = speed of sound

= speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

f' = 3120 Hz

f' = 3.12 kHz

Each plate of a parallel-plate air-filled capacitor has an area of 2×10−3 m2, and the separation of the plates is 5×10−2 mm. An electric field of 8.5 ×106 V/m is present between the plates. What is the surface charge density on the plates? (ε 0 = 8.85 × 10-12 C2/N · m2)

The surface charge density on the plate,

Explanation:

It is given that,

Area of parallel plate capacitor,

Separation between the plates,

Electric field between the plates,

We need to find the surface charge density on the plates. The formula for electric field is given by :

Where

= surface charge density

Hence, this is the required solution.

You’re on a team performing a high-magnetic-field experiment. A conducting bar carrying 4.1 kA will pass through a 1.3-m-long region containing a 12-T magnetic field, making a 60° angle with the field. A colleague proposes resting the bar on wooden blocks. You argue that it will have to be clamped in place, and to back up your argument you claim that the magnetic force will exceed 10,000 pounds. Are you right?

Yes we are right as the force on wire is approx 12500 Lb

Explanation:

Magnetic force on a current carrying bar is given by the equation

here we know that

L = 1.3 m

B = 12 T

i = 4.1 kA

now from above formula we have

So this is equivalent to 12500 Lb force

An Earth satellite moves in a circular orbit 561 km above Earth's surface with a period of 95.68 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Explanation:

It is given that,

Radius of earth, r = 6371 km

An earth satellite moves in a circular orbit above the Earth's surface, d = 561 km

So, radius of satellite, R = 6371 km + 561 km = 6932 × 10³ m

Time taken, t = 95.68 min = 5740.8 sec

(a) Speed of the satellite is given by :

d = distance covered

For circular path, d = 2πR

v = 7586.92 m/s

(b) Centripetal acceleration is given by :

Hence, this is the required solution.

A roller coaster, loaded with passengers, has a mass of 2 000 kg; the radius of curvature of the track at the bottom point of the dip is 24 m. If the vehicle has a speed of 18 m/s at this point, what force is exerted on the vehicle by the track? (g = 9.8 m/s2)

The force exerted on the vehicle by the track of mass 2000kg and with a speed of 18m/s is 46600 N.

What is force?

Force is an external agent which is capable of changing an object's state of rest or motion. Force is a push or a pull on an object with some mass that causes it to change its velocity. Force is a vector quantity. This is because it has both magnitude and direction.

The vehicle experiences two forces which are acting on it. The forces include its weight which is pulling it down that is away from the center of the track and the normal force of the track which is pushing up towards the center of the track.

Therefore, the sum of the forces towards the center of the track will be:

(Video) 4.30 | Find the magnitudes of the forces F1 and F2 that add to give the total force Ftot shown in

∑F = ma

N − mg = m v² / r

where, N = normal force,

m = mass of the object,

g = acceleration due to gravity,

v = velocity of the object,

r = radius of the track

N = mg + m v² / r

N = m (g + v² / r)

Given that, m = 2000 kg, v = 18 m/s, and r = 24 m:

Therefore, N = 2000 (9.8 + (18)² / 24)

N = 46600 N

Therefore, the force exerted on the vehicle by the track is 46600N.

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#SPJ5

46600 N

Explanation:

The vehicle has two forces acting on it: weight pulling down (away from the center of the track) and normal force of the track pushing up (towards the center of the track).

Sum of the forces towards the center of the track:

∑F = ma

N − mg = m v² / r

N = mg + m v² / r

N = m (g + v² / r)

Given m = 2000 kg, v = 18 m/s, and r = 24 m:

N = 2000 (9.8 + 18² / 24)

N = 46600 N

The track pushes on the vehicle with a force of 46600 Newtons.

A 13.7 N force is applied to a cord wrapped around a disk of radius 0.43 m. The disk accelerates uniformly from rest to an angular speed of 30.3 rad/s in 3.43 s. Determine the angular acceleration of the disk. Hint: the average acceleration is the change in angular speed over time.

Explanation:

w₀ = initial angular speed of the disk = 0 rad/s

w = final angular speed of the disk = 30.3 rad/s

t = time period of rotation of the disk = 3.43 sec

= Angular acceleration of the disk

Angular acceleration of the disk is given as

inserting the values

= (30.3 - 0)/3.43

An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?

q = 1.815 \times 10^{-8} C

Charge on one plate is positive in nature and on the other plate it is negative in nature.

Explanation:

E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm

According to the Gauss's theorem in electrostatics

The electric field between the two plates

Charge, q = surface charge density x area

q = 1.815 \times 10^{-8} C

At a carnival, you can try to ring a bell by striking a target with a 8.91-kg hammer. In response, a 0.411-kg metal piece is sent upward toward the bell, which is 3.88 m above. Suppose that 21.9 percent of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

4 m/s

Explanation:

M = mass of the hammer = 8.91 kg

m = mass of the metal piece = 0.411 kg

h = height gained by the metal piece = 3.88 m

Potential energy gained by the metal piece is given as

PE = mgh

PE = (0.411) (9.8) (3.88)

PE = 15.6 J

KE = Kinetic energy of the hammer

Given that :

Potential energy of metal piece = (0.219) Kinetic energy of the hammer

PE = (0.219) KE

15.6 = (0.219) KE

KE = 71.2 J

v = speed of hammer

Kinetic energy of hammer is given as

KE = (0.5) M v²

71.2 = (0.5) (8.91) v²

v = 4 m/s

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 µT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming the electron has an instantaneous velocity of 5.90 ✕ 106 m/s directed to the east in this environment.

(Video) 6.23 | A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What

Explanation:

For gravitational force we know that

F = mg

now we have

Now electrostatic force

here we have

Now magnetic force on it is given by

An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equivalent spring constant of the bow? N/m (b) How much work does the archer do on the string in drawing the bow? J

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down?

(a) The velocity of the object will be zero at the maximum height

(b) The velocity of the object changes direction.

(c) The acceleration due to gravity will not have the same sign on the way up as on the way down.

(A) The velocity of the object decreases as the object ascends upwards and eventually becomes zero at the maximum height. As the object descends downwards, the velocity increases and becomes maximum before the object hits the ground.

(B) The velocity of the object is always directed in the direction of the motion of the object. As the object moves upward, the direction of the object is upward and as the object moves downwards the direction of the object is downwards. Thus, the velocity of the object changes direction.

(c) The upward motion of the object is opposite direction to acceleration due to gravity and the sign of acceleration due to gravity becomes negative.

As the object moves downwards, it will be moving in the same direction as the acceleration due to gravity. The sign of acceleration due to gravity is positive.

Thus, the acceleration due to gravity will not have the same sign on the way up as on the way down.

a) The velocity is zero when the object is at the highest vertical point (peak). b) Yes, at the peak, the velocity for an instant gets to zero and then the direction of motion and the velocity will be downwards.

C) No, the direction of ‘g’ will be opposite in direction when the object is on the way down when compared to what it was during the upward motion.

Explanation:

a) Since gravity is acting on the object and trying to pull it down to earth, the ball will slow down to a velocity of zero when the ball starts its free fall downward.

b) When the direction of the object changes, its velocity will also change direction.

c) When the ball has a positive velocity, gravity is pulling it in the opposite direction, so it has a negative acceleration trying to slow it down. When the ball's velocity is negative (downward free fall), the acceleration will be positive at the instant when the ball hits the ground. The only way to stop the ball from falling is a positive force upward, which results in a positive acceleration when the ball hits the ground.

Hope this helps :)

A 1.50-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction between the block and the wall is 0.54. What is the minimum compression in the spring to prevent the block from slipping down?

0.032 m

Explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

= Static frictional force

= Normal force on the block from the wall

= Spring force due to compression of spring

= Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N

k = spring constant = 860 N/m

μ = Coefficient of static friction between the block and wall = 0.54

x = compression of the spring

Spring force is given as

= kx

From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as

=

= kx eq-1

Static frictional force is given as

= μ

Using eq-1

= μ k x eq-2

From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as

=

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m

A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 ✕ 10−4°C−1). At room temperature (20.0°C), the frames have circular lens holes 2.34 cm in radius. To what temperature must the frames be heated if lenses 2.35 cm in radius are to be inserted into them? °C

Final temperature = 52.44 °C

Explanation:

We have equation for thermal expansion

ΔL = LαΔT

We have change in length = Circumference of 2.35 cm radius - Circumference of 2.34 cm radius = 2π x 2.35 - 2π x 2.34 = 0.062 cm

Length of eyeglass frame = 2π x 2.34 = 14.70 cm

(Video) 8.8 | A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat

Coefficient of linear expansion, α = 1.30 x 10⁻⁴ °C⁻¹

Substituting

0.062 = 14.70 x 1.30 x 10⁻⁴ x ΔT

ΔT = 32.44°C

Final temperature = 32.44 + 20 = 52.44 °C

Initially, a particle is moving at 5.24 m/s at an angle of 35.2° above the horizontal. Four seconds later, its velocity is 6.25 m/s at an angle of 57.5° below the horizontal. What was the particle's average acceleration during these 4.00 seconds in the x-direction (enter first) and the y-direction?

- 0.23

0.56

Explanation:

= initial velocity of the particle = =

= final velocity of the particle = =

t = time interval = 4.00 sec

= average acceleration = ?

Using the kinematics equation

= + t

= + (4)

= 4

=

hence

Average acceleration along x-direction = - 0.23

Average acceleration along y-direction = 0.56

True False Suppose I have a resistor of some resistance R. If I were to double the length and double the cross-sectional area of the resistor, what is the new resistance?

Explanation:

The resistance of a wire is given by :

Where

is the resistivity of the wire

l = initial length of the wire

A = initial area of cross section

If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A

So, new resistance of the wire is given by :

R' = R

So, the resistance of the wire remains the same on doubling the length and the area of wire.

A 0.18-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o’clock position, the stick's tension is 19 N. Find the tension in the stick when the ball is (a) at the twelve o’clock and (b) at the six o’clock positions.

a) 17 N

b) 21 N

Explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.

A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle?

6.6 x 10^5 Nm^2/C

Explanation:

E = 125000 N/C

Area, A = length x width = 2.5 x 5 = 12.5 m^2

θ = 65 degree

electric flux, φ = E A Cosθ

φ = 125000 x 12.5 x Cos 65

φ = 6.6 x 10^5 Nm^2/C

Here, we are required to evaluate the electric flux through the rectangle.

The electric flux through the rectangle is;

φ = 6.6 × 10^5 Nm²/C

The electric flux through a body is given as;

electric flux, φ = E A Cosθ

where;.

• φ = electric flux
• E = Electric field magnitude
• A = Area
• θ = angle between the electric field vector and the vector normal

The Area of the rectangle, A = 2.5 × 5

Therefore, A = 12.5m².

Therefore, the electric flux, φ = E A Cosθ

φ = 125000 × 12.5 × Cos65

φ = 660,341 Nm²/C.

φ = 6.6 × 10^5 Nm²/C

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## FAQs

### How to calculate the force of gravitation due to Earth on a child? ›

1. Step1: Given data. Mass of ball = 1 Kg. Mass of earth = 6 × 10 24 Kg, The radius of earth = 4 × 10 3 Km, ...
2. Step 2: Formula used. F = G m 1 m 2 d 2 . . . . ...
3. Step 3: Calculating the gravitational force between ball and earth. On, putting the given values in equation (a), we get. F = 6 . 67 × 10 - 11 × 1 × 6 × 10 24 6 .

What is the force of gravity between a newborn baby of mass 4 kg and the obstetrician of mass 75 kg who is 0.3 m from the baby? ›

The equation for gravitational force found on page 172. The mass of baby is 4 kg, mass of obstetrician is 75 kg, distance between two is 0.3 m, and G = 6.67 × 1 0 − 11 . G= 6.67 \times 10^{-11}.

How do you determine the magnitude of the gravitational force on the 10 kg mass? ›

Gravitational force on a body on the surface of the earth = mass x acceleration due to gravity = 10 kg x 9.8 ms-2 = 98 N.

What is the magnitude of the gravitational force between Earth and a 1 kg body? ›

This shows that Earth exerts a force of 9.8 N on a body of mass 1 kg.

What is the equation for gravitational force? ›

In symbols, the magnitude of the attractive force F is equal to G (the gravitational constant, a number the size of which depends on the system of units used and which is a universal constant) multiplied by the product of the masses (m1 and m2) and divided by the square of the distance R: F = G(m1m2)/R2.

What is the formula for the gravitational force on Earth? ›

Newton's law of gravitation is: F = GMm r2 where the Gravitational Constant G = 6.673 × 10−11Nm2kg−2 (kg−1m3s−2). gravitational force per unit mass = gravitational acceleration g. g is approximately 9.8m/s2 at the surface of the Earth.

What is the magnitude of the gravitational force? ›

From the universal law of gravitation, the magnitude of the gravitational force (F) between the earth and the body is given by, F=GM×mR2. where G=6.67×10−11 Nm2/kg2 is the universal gravitational constant.

What is the force of gravity between a 3 kg newborn baby? ›

The force of gravity between a 3 kg newborn baby and a 75 kg doctor standing 1 m away is1. 5×10-8 N. where G is the universal gravitation constant, m1 is the mass of the first object, m2 is the mass of second object and r is the distance between them.

What is the gravitational force of 1kg? ›

Gravitational force which acts on 1 kg is 9.8 N.

What is the gravitational force between a body of 2 kg and the earth? ›

So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body. Where 'G' is the gravitational constant. Hence 2kg mass will experience 19.6N of force.

### What is the magnitude of the gravitational force exerted on the person by the earth? ›

Near the surface of the earth its magnitude is g = 9.8 m/s2 and its direction is downward.

What is the magnitude of the gravitational force exerted by a 15kg mass? ›

a2 = Gm10.252 =6.673∗10−11∗150.252= 1.6*10−8 ms2.

What is the magnitude of net gravitational force on 2kg mass? ›

Net force on 2 kg mass = (46.69 - 33.5) × 10−11​​ N = 13.19 × 10−11 ​​N.

What is magnitude in math examples? ›

For numbers such as 1, 2, 3, and so on, the magnitude is simply the number itself. If the number is negative, the magnitude becomes the absolute value of the number. For example, the magnitude of 10 is 10. The magnitude of -10 becomes the absolute value of -10, which is 10.

How do you find the magnitude of an object in physics? ›

The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

What is the formula for magnitude in physics Newton's law? ›

Newton's second law states that the magnitude of the net external force on an object is Fnet=ma. Since weight g=9.80m/s2 on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: w=mg=(1.0kg)(9.8m/s2)=9.8N.

What is the gravitational force between a body of mass 100 kg and the earth? ›

Therefore we get that force of attraction between body and earth is 9.8 × 10^8N. Therefore acceleration of the body is 9.8 × 10^6.

What is the magnitude of the gravitational force between the earth and a 50 kg object on its surface? ›

Therefore, gravitational force between the earth and an object is 9.77 N.

What is the formula to find the magnitude of the gravitational force between the earth and object on the surface of the earth? ›

Let m1 be the mass of the Earth and m2 be the mass of an object on its surface. If r is the radius of the earth, then according to the universal law of gravitation, the gravitational force (F) acting between the earth and the object is given by the relation: F=Gm1m2r2.

What are the three formulas of gravitation? ›

Important Gravitational Formulas
• Gravitational Force. F = $\frac{Gm_{1}m_{2}}{r^{2}}$ ...
• Gravitational Potential. i) For Point Charge: ...
• Gravitational Acceleration. g = GM / R²
• Variation of g with Depth. ...
• Variation of g with Height. ...
• Effect of Non-Spherical Earth Shape on g. ...
• Effect of Earth Rotation on Apparent Weight.

### What is force of gravity is 9.8 N? ›

The force of gravitational attraction on a mass of 1 kg on the Earth is 9.8 N. Another way of putting that is that the gravitational field strength on the surface of the Earth is 9.8 N/kg.

What is 9.8 force of gravity? ›

"9.81 meters per second squared" means that objects on Earth will accelerate (or go faster) 9.81 meters every second, if they are in free fall, due to the pull of gravity. Throughout space, gravity actually is constant.

Why is the magnitude of the gravitational force? ›

The magnitude of the gravitational force will be equal to the product of mass and acceleration due to the gravitational force. The Universal Law of Gravitation says that – any two objects having mass will attract each other with a force which is directly proportional to the product of their masses.

Is weight the magnitude of the gravitational force? ›

The weight of an object is the net force on a falling object, or its gravitational force. The object experiences acceleration due to gravity. Some upward resistance force from the air acts on all falling objects on Earth, so they can never truly be in free fall.

What if a baby is 3.5 kg? ›

A child's normal weight is somewhere between 2.5 and 3.5 kg. If the weight is slightly more than 3.5 kg it is also considered to be normal. If your baby weighs less than 2.5 kg, he or she is said to have a low birth weight. This is something that can happen to premature babies.

What is an infant with 2.5 kg birth weight termed as? ›

Preterm infants are born at less than 37 weeks gestational age and low birth weight infants are born with a birth weight below 2.5kg regardless of gestational age. An estimated 15 million newborns are born preterm and more than 20 million are born low birth weight each year.

What is the average weight of a newborn baby is 3 kg? ›

Assessing a newborn's weight

The average weight for full-term babies (born between 37 and 41 weeks gestation) is about 7 pounds (3.2 kg). In general, small babies and very large babies are more likely to have problems. Newborn babies may lose as much as 10% of their birth weight.

What is the force of gravity of a 10 kg child? ›

Therefore, the force of gravity on a 10 kg mass is 24.42 N.

What is the force exerted by gravity on 5 kg? ›

Different masses are hung on a spring scale calibrated in Newtons. The force exerted by gravity on 1 kg = -10 N. The force exerted by gravity on 5 kg = 49.05 N.

What is 9.8 gravity in kg? ›

Explanation: Consider the mass of an object is 1 kg. So, the weight of one kg is 9.8 N or we can say that One kgw = 9.8 N.

### What is the magnitude of gravitational force between two 1 − kg − kg bodies that are 1 m apart? ›

Explanation of Solution

The gravitational force between two 1 − kilogram bodies that are 1 m apart is 6.674 × 10 − 11 N .

What is the gravitational force between two objects of mass 1kg each? ›

The force of gravitation between two bodies of mass 1 kg each kept at a distance of 1 m is: 6.67 N.

What is the gravitational force between Earth and a 1 kg object placed on its surface? ›

Hence, the magnitude of the gravitational force between the earth and a 1 kg object on its surface is 9.77 N.

What is the gravitational force of a 50 kg person on Earth? ›

Weight, W= m x g =50 x 9.8=490N.

What is the magnitude of the gravity force that the planet exerts on you when you are standing on its surface? ›

This results in Earth having a gravitational strength of 9.8 m/s² close to the surface (also known as 1 g), which naturally decreases the farther away one is from the surface. In addition, the force of gravity on Earth actually changes depending on where you're standing on it.

What is the magnitude of the gravitational force exerted by ball A on ball B? ›

What is the magnitude of the gravitational force exerted by ball A on ball B? F₂ = (6.67 x 10-¹¹ N-m2/s2)(7.00 kg)(7.00 kg) (2.00 m)² Solving, F=8.17 x 10-10 N. The force of gravity acting on an object causes it to accelerate when dropped in a gravitational field.

What is the magnitude of the gravitational force acting on a 60 kg man? ›

The gravitational force between man of mass 60 kg and the earth is 586.23 N.

How to find the magnitude of the gravitational force between two objects? ›

Gravitational Force = (Gravitational Constant × Mass of first object × Mass of the second object) / (Distance between the centre of two bodies)2.

What are the magnitude and direction of the net gravitational force on the 20 kg mass? ›

The magnitude and direction of the net gravitational force on 20kg mass are 6 × 10 - 7 N and respectively.

What is the magnitude of the gravitational force exerted by ball A? ›

What is the magnitude of the gravitational force exerted by ball A on ball B? F₂ = (6.67 x 10-¹¹ N-m2/s2)(7.00 kg)(7.00 kg) (2.00 m)² Solving, F=8.17 x 10-10 N. The force of gravity acting on an object causes it to accelerate when dropped in a gravitational field.

### What is the formula for magnitude of the gravitational acceleration? ›

The value of the acceleration due to the gravity on earth is 9.8 m/s2. g = GM/r2 is the equation used to calculate acceleration due to gravity.

How do you find the magnitude of the gravitational force between the earth and an object on the surface of the earth Class 9? ›

Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
1. Let the mass of earth = M.
2. Let the mass of the object = m.
3. Distance between the the earth's centre and object = Radius of the earth = R.
4. Therefore,
5. Gravitational Force = F = GMm/ R 2
May 16, 2023

Is gravitational force 9.8 or 10? ›

You should assume g value as 9.8 m/s2 unless mentioned in the question as 10 m/s2. Q. What is the value of acceleration due to gravity on the moon?

What is gravitational force exerted by? ›

The Mass of the Objects The more mass two objects have, the greater. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. The Distance Between the Objects As distance between the objects. increases, the force of gravity decreases.

What is the magnitude of the force of gravity exerted on that object and is given by the product of the mass and acceleration due to gravity? ›

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the newton.

What is the equation of magnitude of gravitational potential at a point? ›

Gravitational potential at a distance 'r' from a point mass 'm' is V = -(GM)/(r) Find gravitational field strength at that point. The gravitational potential energy at a body of mass m at a distance r from the centre of the earth is U.

What is the formula to find the magnitude of the gravitational force exerted by the earth on an object placed on its surface? ›

Let m1 be the mass of the Earth and m2 be the mass of an object on its surface. If r is the radius of the earth, then according to the universal law of gravitation, the gravitational force (F) acting between the earth and the object is given by the relation: F=Gm1m2r2.

What is the gravitational force between the earth and an object of 2 kg mass placed on its surface? ›

So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body. Where 'G' is the gravitational constant. Hence 2kg mass will experience 19.6N of force.

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