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SVETLANKA909090 [29]
4 years ago
12

Find the net force it would take to accelerate an 800 Kg car at a rate of 5 m/s2

Physics
1 answer:
icang [17]4 years ago
3 0

Answer:

3,500 N.

Explanation:

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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
2. What kinds of statistics are used in sports? In a paragraph explain your answer with examples
Lostsunrise [7]

Answer: Sports Illustrated noted that analytics in a sport such as a football is widely used to manage injury prevention. Basketball coaches make use of statistics such as field-goal attempts, effective field goal percentage, free throw attempts, and free throw percentages.

Explanation: BRAINLEST?

4 0
3 years ago
When you compare the prices of two different pairs of shoes, money is a _____.
Vladimir [108]

unit of account is the answer

3 0
4 years ago
Read 2 more answers
An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol
iogann1982 [59]

Answer:

(a) B = 2.85 × 10^{-6} Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = \frac{mv}{qB}

⇒ B = \frac{mv}{qr}

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = \frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }

  = \frac{9.11*10^{-27} }{3.2*10^{-21} }

B = 2.85 × 10^{-6} Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×10^{-7}Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

 = \frac{2.85*10^{-6} }{4*10^{-7} *25}

I = 0.285 A

5 0
3 years ago
An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.
Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

4 0
3 years ago
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