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2 years ago

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A train composed of a small engine car and a massive cargo car are connected as they move along a track. The speed of the small

engine car is ____(blank 1)___ the speed of the massive cargo car. The magnitude of the acceleration of the small engine car is ____(blank 2)___ the magnitude of the acceleration of the massive cargo car because ____(blank 3)___ .

1 answer:

stepan [7]2 years ago

7 0

Answer:

The speed of the small engine car is <u>_equal to_</u> the speed of the massive cargo car. The magnitude of the acceleration of the small engine car is <u>_equal to_</u> the magnitude of the acceleration of the massive cargo car because <u>_the cars speeds cannot change by different amounts_</u> .

Explanation:

The two cars are connected, hence, they move with the same speed at all points in time.

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A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

Margarita [4]

Potential energy at any point is (M G H). On the way down, only H changes. So halfway down, half of the potential energy remains, and the other half has turned to kinetic energy. Half of the (M G H) it had at the tpp is (0.5 x 9.8 x 10) = 49 joules.

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2 years ago

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A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it

Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

v = v₀ + at

Substitute known values into the equation.

0 = 200 · sin(30) + (-9.8)t
-200 · sin(30) = -9.8t
t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

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1 year ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

7 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction

Vaselesa [24]

Answer:

H=1020.12m

Explanation:

From a balance of energy:

$\frac{m*Vo^2}{2} -mg*H=-Ff*d$ where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30) Replace this into our previous equation:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d$

From a sum of forces:

N -mg*cos(30) = 0 => N = mg*cos(30) Replacing this:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d$ Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

6 0

2 years ago

Read 2 more answers