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Elena L [17]
2 years ago
6

1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?

Physics
1 answer:
nignag [31]2 years ago
6 0

(1) The time taken for the drag racer to accelerate is 4.89 s

(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.

Formula:

  • t = (v-u)/a........... Equation 1

Where:

  • t = time
  • v = Final velocity
  • u = initial velocity
  • a = acceleration.

From the question,

Given:

  • a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²
  • v = 60 mph = (60×0.44704) = 26.82 m/s
  • u = 0 m/s ( from rest)

Substitute these values into equation 1

  • t = (26.82-0)/5.4864
  • t = 4.89 seconds

(2) To calculate the speed of the contestant, we use the formula below

Formula:

  • s = d/t............ Equation 1

Where:

  • s = speed of the contestant
  • d = distance
  • t = time.

From the question,

Given:

  • d = 100 m
  • t = 10.6 s

Substitute these values into equation 1

  • s = 100/10.6
  • s = 9.43 m/s

(a) In feet = (9.43/0.3048) = 30.94 ft/s

(b)  in miles per hr = (9.43×2.24) = 21.12 miles per hr

Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.

Learn more about speed here: brainly.com/question/4931057

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Newton's 2nd Law of Motion:  Force = (mass) · (acceleration)

Do I need to go any further ?

I get  575.25 Newtons .

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Sport is an attractive activity for young people, and is often used as a draw card to recruit children and .....................
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Explanation:

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1 year ago
A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

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Start from 0 m/s and accelerate at 2m/s? Calculate the speed in m/s after acceleration for 5 seconds.
valentinak56 [21]

Answer:

10m/s

Explanation:

2m/s  x  5s=10m/s

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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

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We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

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