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2 years ago

14

Which of the following is an elastic collision?

2 answers:

UNO [17]2 years ago

6 0

I think its B. Two cars colliding and sticking together.or C. A wad of gum sticking to a window.

fiasKO [112]2 years ago

3 0

D is also an example

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29. Jorge is conducting an investigation into perfectly inelastic collisions using equipment where two carts collide with

Readme [11.4K]

Answer:

He used the probability assumption

5 0

2 years ago

Running at maximum speed, it takes a boat times as long to go 10 miles upstream as it does to go 10 miles downstream. If the cur

Greeley [361]

Note: <em>The question states the time to go upstream is a number of times (not explicitly written) the time to go downstream. We'll assume a general number N</em>

Answer:

$\displaystyle v_b=\frac{N+1}{N-1}(4\ mph)$

Explanation:

<u>Relative Speed</u>

If a boat is going upstream against the water current, the true speed of motion is $v_b-v_w$ , being $v_b$ the speed of the boat and $v_w$ the speed of the water. If the boat is going downstream, the true speed becomes $v_b+v_w$ .

The question states the time to go upstream is a number of times N (not explicitly written) the time to go downstream. The speed of an object is computed as

$\displaystyle v=\frac{x}{t}$

Where x is the distance traveled and t the time taken for that. The time can be computed by

$\displaystyle t=\frac{x}{v}$

If $t_u$ is the time for the upstream travel and $t_d$ is the time for the downstream travel, then

$t_u=Nt_d$

Siince the same distance x= 10 miles is traveled in both cases:

$\displaystyle \frac{10}{v_b-v_w}=N\frac{10}{v_b+v_w}$

Simplifying and rearrangling

$v_b+v_w=N(v_b-v_w)$

Operating

$v_b+v_w=Nv_b-Nv_w$

Solving for $v_b$

$\displaystyle v_b=\frac{N+1}{N-1}v_w$

$If\ N=2,\ v_w=4\ mph$

$\displaystyle v_b=\frac{3}{1}(4)=12\ mph$

If N=3

$\displaystyle v_b=\frac{4}{2}v_w=2(4)=8\ mph$

We can use the required value of N to compute the speed of the boat as explained

7 0

3 years ago

A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction

Anna11 [10]

The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20

8 0

2 years ago

How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)

leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0

2 years ago

An incident ray of light strikes water at an angle of 30°. The index of refraction of air is 1.0003, and the index of refraction

defon

Answer: It’s A

Explanation:

22 degrees

6 0

2 years ago

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