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borishaifa [10]
3 years ago
12

A rail car moving at 10m/s collides with and connect to another stationary car. What is their final velocity?

Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

read this it might help some

When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change.

Explanation:

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Why do the graphs differ?​
melisa1 [442]

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

Take1/ f to be constant c

1/v = c - 1/u

it is of the form y = - x + k

Slope = -1 having intercept k as shown in fig 2

3 0
3 years ago
Pls help
belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these  value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0
2 years ago
How do resistors in parallel affect the total resistance?
4vir4ik [10]

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

4 0
2 years ago
An increase in a sound's pitch corresponds to an increase in what other property?
Wittaler [7]
An increase in a sound's pitch corresponds to an increase in the frequency!
8 0
3 years ago
If the car's speed decreases at a constant rate from 77 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
Lostsunrise [7]

The acceleration of the car,

a = \frac{v-u}{t}

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given u =77 \ mil/h ,v = 50 mi/h and t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h

Therefore, from above equation

a =  \frac{50 \ mi/h -77 \ mi/h}{8.33 \times 10^{-4} h} = - \frac{27 \ mi/h }{8.33 \times 10^{-4} h} = - 3.2 \times 10^{4} \ mi/h^2.

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is 3.2 \times 10^{4} \ mi/h^2.

3 0
3 years ago
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