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3 years ago

If two carts collide, what type of collision will conserve both momentum and kinetic energy?

2 answers:

8 0

The appropriate answer is d. perfectly elastic. When a perfectly elastic collision occurs both momentum and kinetic energy are preserved. Momentum is the mass of an object times its velocity while kinetic energy is the energy that a moving object has due to its motion. In an elastic collision kinetic energy would be converted to some other form of energy.

son4ous [18]3 years ago

7 0

I just took the test and got 100%

1. A

2. C

3. B

4. C

5. B

6. B

7. B

8. A

9. B

10. C

11. D

Good luck :)

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Compare and contrast instantaneous and average speed.

Harlamova29_29 [7]

Answer:

instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken

6 0

2 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

This is about the magnet fields. thanks in advance.

Wewaii [24]

The answer is A. The outer lines change as it moves

8 0

2 years ago

Daily life examples of pressure of solids on solids?

Ipatiy [6.2K]

Explanation:

There's a massive amount, just think of anything everyday. Like a table on the floor, or when your walking around and putting pressure on the floor. When you squeeze something which is solid. Anything like that will do.

6 0

3 years ago

A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a diff. One friend says to the other

olga55 [171]

Answer:

The statement is not correct.

Explanation:

To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.

Data obtained from the question include:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3 s

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 3)

v = 0 + 29.4

v = 29.4 m/s

Thus, the velocity of the car after 3 s is 29.4 m/s.

Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.

8 0

3 years ago