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2 years ago

8

Physics

1 answer:

aivan3 [116]2 years ago

5 0

The angle of the gun needs to be at 25.8 degrees from the ground..

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Convection occurs when thermal energy is transferred by the movement of

madam [21]

Answer:

Convection is the transfer of thermal energy from one place to another by the movement of gas or liquid particles. How does this happen? As a gas or liquid is heated, the substance expands. This is because the particles in liquids and gases gain kinetic energy when they are heated and start to move faster.

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2 years ago

ASTRONOMY !! PLS HELP!!

aleksandr82 [10.1K]

-- pick a planet from the table

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2 years ago

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nirvana33 [79]

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2 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

2 years ago

Very lost please help.

V125BC [204]

A) The acceleration is due to gravity at any given point if you look at it vertically, so $-10 m/s^2$ .

b) $sin(25) = V_y/V$ , so $V_y = V*sin(25)$ . We use $V = V_0 + a t$ and then the final speed must be 0 because it stops at the highest point. So $0 = V_y - 10t$ . Solve for $t$ and you get $t = 32sin(25)/10 = 16sin(25)/5$

c) $Y = Y_0 + V_0t + (1/2)at^2$ , and then we plug the values: $Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2$ and we already have the time from "b)", so $Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2$ ; then we just rearrange it $Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]$ and finally $Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20$

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2 years ago