If the mass of the moon is ...

As your rockets went upwards how would you describe how the energies changed?

chubhunter [2.5K]

As a rocket increases height and slows down, it gains more and more potential energy and loses more and more kinetic energy. Potential energy is store energy (usually determined by height), and kinetic energy increases as speed increases.

4 0

3 years ago

Physical Science - 02.05 - Question #3

Reika [66]

Answer:

1. They should not trust this study. 2. Pseudoscience.

Explanation:

Because they drank the water and after 24 hours, they said that it was gone. Head aches go away sooner than that. Another reason is that some people could have lied about having a headache just so they could have the water or their headache could have gone away before they drank the water.

6 0

3 years ago

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi

miv72 [106K]

Answer:

The time taken is $\Delta t = 1.5 \ s$

Explanation:

From the question we are told that

The mass of the ball is $m = 1.25 \ kg$

The time taken to make the first complete revolution is t= 3.60 s

The displacement of the first complete revolution is $\theta = 1 rev = 2 \pi \ radian$

Generally the displacement for one complete revolution is mathematically represented as

$\theta = w_i t + \frac{1}{2} * \alpha * t^2$

Now given that the stone started from rest $w_i = 0 \ rad / s$

$2 \pi =0 + 0.5* \alpha *(3.60)^2$

$\alpha = 0.9698 \ s$

Now the displacement for two complete revolution is

$\theta_2 = 2 * 2\pi$

$\theta_2 = 4\pi$

Generally the displacement for two complete revolution is mathematically represented as

$4 \pi = 0 + 0.5 * 0.9698 * t^2$

=> $t^2 = 25.9187$

=> $t= 5.1 \ s$

So

The time taken to complete the next oscillation is mathematically evaluated as

$\Delta t = t_2 - t$

substituting values

$\Delta t = 5.1 - 3.60$

$\Delta t = 1.5 \ s$

7 0

3 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

3 years ago

The trajectory of a projectile always ________________.

sattari [20]

Answer:

c) curves downward, below the initial velocity vector

Explanation:

A projectile is usually launched from a height, where it is launched with an initial velocity. From that point the gravitational force begins to act on the projectile causing it to decay. As time passes, the projectile advances but its height decreases. So its trajectory is curved downward, below the initial velocity vector.

8 0

3 years ago