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In-s [12.5K]
4 years ago
15

You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the bal

l or deflect the ball back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to minimize your speed on the skateboard? Choose one of the following three answers.
A) Your final speed on the skateboard will be the same regardless whether you catch the ball or deflect the ball.

B) You should catch the hall.

B) You should deflect the ball

The answer is you should catch the ball to minimize speed. What I need is a math-based and THOROUGH explanation. Any formulas you use to prove, please show the ORIGNIAL equation with steps that you used to get to the final one. Please! will rate right away!
Physics
1 answer:
harina [27]4 years ago
7 0

To solve this problem we will apply the concept related to the moment, which describes the change in speed in proportion to the body mass. The momentum can be described under the general equation

p = mv

Where,

m = mass

v = Velocity

The change in momentum is the difference between the final moment and the initial moment

\Delta p = p_f - p_i

Where,

p_f = Final momentum

p_i = Initial momentum

After catching the ball, the ball comes to rest position and speed of the ball is zero.

p_f = 0

The change in momentum on catching the ball is

\Delta p = 0 - mv

\Delta p = -mv

The momentum during the catching of the ball is

p_i = mv

Now during deflection of the ball, we know that there is an initial momentum and a negative final momentum because it moves with the same speed but in the opposite direction, that is

p_f = -mv

The momentum is negative since during deflection the ball moves with same speed in opposite direction

\Delta p = -mv-mv

\Delta p = 2-mv

Therefore, the correct ansswer is B: catch the ball in order to minimize your speed on the skateboard

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sleet_krkn [62]

Answer:

a. 25000 J

b. 2500 J/s

Explanation:

Given,

Distance ( s ) = 50 m

Force ( f ) = 500 N

a.

To find : -

Work done ( W ) = ?

Formula : -

W = fs

W

= 500 x 50

= 25000 J

Therefore,

the work done by the force the horse exerts is

25000 J.

b.

To find : -

Power ( P ) = ?

Formula : -

W = Pt

P = W / t

P

= 25000 / 10

= 2500 J/s

Therefore,

the power produced if the movement took 10 s

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3 years ago
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Peter throws a snowball at his car parked in the driveway. The snowball disintegrates as it hits the car. By Newton’s third law,
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A plane electromagnetic wave varies sinusoidally at 90.0 MHz as it travels along the x direction. The peak value of the electric
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Answer:

Explanation:

frequency n = 90 x 10⁶ Hz .

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= 1 / 90 x 10⁶

= 1.11 x 10⁻⁸ S.

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= 3 x 10⁸ / 90 x 10⁶

= 3.33 m

maximum value of the magnetic field. ( B₀ )

E₀ / B₀ = c where E₀ and  B₀ are maximum electric and magnetic field .

E₀ / c=  B₀

200/ 3 x 10⁸

= 66.67 x 10⁻⁸ T .

expressions in SI units for the space and time variations of the electric field

E=E_{0y}sin(2\pi nt - \frac{2\pi x}{\lambda} )

E=200sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} ) N/C

B=B_{0z}sin(2\pi nt - \frac{2\pi x}{\lambda} )

B=66.67\times 10^{-8}sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} ) T

4 0
4 years ago
Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs
mafiozo [28]

Answer:

Magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

Explanation:

Magnitude of first vector = |A| = 13\ \text{units}

Angle = \theta_1=27^{\circ}

Magnitude of second vector = |B| = 11\ \text{units}

Angle = \theta_2=45^{\circ}

x component of first vector

A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}

y component of first vector

A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}

x component of second vector

B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}

y component of first vector

B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}

Adding the magnitudes

C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}

C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}

Magnitude of the sum of the vectors would be

|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}

The direction would be

\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}

The magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

4 0
3 years ago
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