During the last glacial maximum, ice sheets could be found at _____. A. both poles B. the North Pole C. the South Pole

How long would it take for a rock falling 97.2 m/s to reach the ground from 100 meters

mote1985 [20]

Answer;

velocity(v) = 97.2 m/s ,

distance (S) = 100 m

determine time(t) = ?

We know that,

distance (S) = velocity(v) × time(t)

So, time (t) = distance ÷ velocity

= 100 ÷ 97.2

t = 1.02 sec.

I hope this will help you.

5 0

4 years ago

what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately 3.331 × 10³⁶ J

Reasons:

The parameters required for the question are;

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs.

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = 3.331 × 10³⁶ Joules

Learn more here:

brainly.com/question/13623190

The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere.

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs

3 0

3 years ago

A van slows down as it travels up a steep hill. Its speed goes from 20m/s at the bottom of the hill to 12m/s when it reaches the

Paladinen [302]

The van's AVERAGE speed on the way up the hill is

(1/2) (20 + 12) = (1/2) (32) = 16 m/s .

At an average speed of 16 m/s, it travels

(16 m/s) x (8 s) = 128 meters in 8 seconds.

5 0

4 years ago

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$ .

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$ , then $\frac{v_{2}}{v_{1}} > 1$ . That is to say, $v_{1} < v_{2}$ .

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$ , then $p_{1} > p_{2}$ . Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.

6 0

3 years ago

3. Using the F, m, a triangle, calculate the boy's mass. Use the "force

aliya0001 [1]

Answer: 30kg

Explanation:

F = 300N

m = ?

a = 10 m/s/s

m = f/a = 300N/10 m/s/s

m = 30kg

5 0

3 years ago