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Alex Ar [27]
3 years ago
12

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

astic, the ball bounces back with the same speed v going upward. Which of the following statements about the bounce are correct? (There could be more than one correct choice.) 1. The balrs mpmentum was conserved during the bounce. 2. The ball ha the same momentum just before and just after the bounce. 3. The magnitude of the ball's momentum was the same just before and just after the bounce. 4. All of the above 5. None of the above
Physics
1 answer:
artcher [175]3 years ago
7 0

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

                     p₀ = m v

The final moment

                     p_{f} = m (-v)

 

                  p₀ = - p_{f}

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

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svetlana [45]

Answer:

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Explanation:

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8 0
3 years ago
What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of?
madreJ [45]
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) /  m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.

7 0
3 years ago
a vector a has components a x equals -5.00 m in a y equals 9.00 meters find the magnitude and the direction of the vector
Murrr4er [49]

Answer:

The magnitude = 10.30 m

The direction of the vector proceeds at angle of 119.05°

Explanation:

Given that:

A vector \bar A has component A_x = -5 m and A_y = 9 m

The magnitude of vector  \bar A can be represented as:

\bar A  = \sqrt{A_x^2 + A_y^2}

\bar A  = \sqrt{(-5)^2 + (9)^2}

\bar A  = \sqrt{25 + 81}

\bar A  = \sqrt{106}

\bar A  = 10.30 m

If we make \bar A  an angle \theta with y- axis:

Then;   tan \theta  = \frac{A_x}{A_y}

tan \theta  = \frac{5}{9}

tan \theta  = 0.555

\theta  = tan⁻¹ (0.555)

\theta  = 29.05°

Angle with positive x-axis = 90 + \theta  

= 90° + 29.05°

= 119.05°

5 0
3 years ago
The astrometric technique of planet detection works best for
denpristay [2]

The astrometric technique of planet detection works best for massive planets around nearby stars.

6 0
4 years ago
g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie
polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
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