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3 years ago

9

Ross was taking a walk on a cold, winter day. A few minutes into his walk, Ross's hands got cold, so he rubbed them together to

warm them. Ross warmed his hands by using:

1 answer:

andrezito [222]3 years ago

6 0

He warmed his hands using FRICTION

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

4 years ago

The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s ju

zhenek [66]

Answer:

The overall velocity of the water when it hits the bottom is:

$v_f=63.61\ \frac{m}{s}$

Explanation:

Use the law of conservation of energy.

Call it instant [1] to the moment when the water is just before reaching the falls.

At this moment its height h is 206 meters and its velocity horizontally $v_i$ is $v_i = 2.90$ m/s.

At the instant [1] the water has gravitational power energy $E_g$

$E_g = mgh$

The water also has kinetic energy Ek.

$E_k = 0.5mv_i ^ 2$

Then the Total E1 energy is:

$E_1 = mgh + 0.5mv_i ^ 2$

In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity $v_f$

So:

$E_2 = 0.5mv_f ^ 2$

As the energy is conserved then $E_1 = E_2$

$mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2$

Now we solve for $v_f$ .

$gh + 0.5v_i ^ 2 = 0.5v_f ^ 2\\\\9.8(206) + 0.5(2.90) ^ 2 = 0.5v_f 2\\\\v_f^2 = \frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}\\\\v_f = \sqrt{\frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}}\\\\v_f=63.61\ \frac{m}{s}$

7 0

3 years ago

I need help asap !

Elodia [21]

Answer:

the answer is 100%!

your welcome

5 0

3 years ago

Check all of the following statements which are true. (You may need to consult Section 7.6 in the textbook.) You should check an

IrinaVladis [17]

Answer:

D

Explanation:

8 0

4 years ago

A ductile cast-iron bar is to support a load of 178 kN in a heat-treating furnace used to make malleable cast iron. The bar is l

Nitella [24]

Answer: L.M = 34.69

Area of the bar = 25835m^2

Explanation: To design a bar that can withstand temperature at 773K for 10years, we need to calculate and know:

1) the Larson Miller value

2) rupture time

3) creep rate

4) stress and load.

Please find the attached file for the solution

3 0

4 years ago