All categories

3 years ago

Why is the force of gravity on your body weaker on the Moon than on the Earth?

Physics

1 answer:

andrezito [222]3 years ago

4 0

Answer:

The moon’s gravity is weaker than the earth gravity due to the smaller in size as compared to the earth. As there is no atmosphere present on the moon gravity, so there are fewer chances of withstanding temperature.

Please mark as brainliest

You might be interested in

An electronic balance shows the mass of a sample of sodium chloride to be 29.732 g. what is the uncertainty of the measurement

OLEGan [10]

The uncertainty of the measurement is 0.001 gm.

The uncertainty in the measurement of a physical quantity is given as how precisely we can measure that, in this case as we can see that the mass of the sodium chloride is precisely given as 29.732 gm, this means the electronic scale is precise to 0.001 gm and round of the values after that which means there is a uncertainty of 0.001 gm.

8 0

3 years ago

When you start the simulation, there is no friction and no air resistance ("Air Drag"). Apply full force to the right, 200N. Wil

kakasveta [241]

Answer:Like friction, the drag force always opposes the motion of an object. ... The drag coefficient can depend upon velocity, but we will assume that it is a ... At highway speeds, over 50% of the power of a car is used to overcome air drag. ... A zero net force means that there is no acceleration, as given by Newton's second law.

3 0

3 years ago

a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin

Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0

4 years ago

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

$K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

3 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago