All categories

3 years ago

A 15kg boulder is moved from the Earth to the Moon. Explain what happens to the mass and weight of the boulder.

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

The mass of the boulder remains constant, while its weight decreases with respect to the value of gravitational force on the moon.

Explanation:

The mass of the boulder = 15 kg

On the earth, its mass remains 15 kg. But its weight is;

weight = m x g

= 15 x 9.8

= 147 N

The boulder's weight on the earth is 147 N.

When transferred to the moon, the mass remains constant i.e 15 kg. But its weight decreases due to a change in the value of acceleration due to gravity on the moon. Thus, the boulder becomes lighter in weight.

You might be interested in

Proposed Kinematic Exercise I

valentinak56 [21]

Answer:

(a) 20 m

(b) 6 m/s²

Between t=2 and t=4, the body moves to the right.

Explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.

3 0

3 years ago

a charge Q exerts a 2.4 N force on another charge q.If the distance between the charges is doubled,what is the magnitude of the

GrogVix [38]

Heyyyyyyyyyy how is your day goingggg luvvvv

6 0

3 years ago

Water is flowing through a channel that is 12m wide with a

Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 $d_1$ = 12 m

So radius $r_1=\frac{d_1}{2}=\frac{12}{2}=6m$

Speed through the channel 1 $v_1=0.75m/sec$

Width , that is diameter of the channel 2 $d_2=4m$

So $r_2=2m$

From continuity equation

$A_1v_1=A_2v_2$

$\pi \times 12^2\times 0.75=4\times \pi\times 4^2\times v_2$

$v_2=1.6875m/sec$

So velocity from smaller channel will be 1.6875 m /sec

6 0

3 years ago

A force of 203 N is used to move a television 2.3 m. How much work is done?

tresset_1 [31]

Answer:

the answer is 466.9 J

Explanation:

workdone= force×distance

force=203N

distance=2.3m

workdone=203×2.3=

466.9J

4 0

3 years ago

Can some one please help!!!!!!<br> ASAP pleaseeeeeeee❗️❗️❗️❗️❗️❗️❗️❗️

Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$