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defon
3 years ago
10

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling

70 mph can stop in 270 ft, how many feet will it take the same car to stop whenit is travelling 60 mph?
Physics
1 answer:
Crank3 years ago
4 0
270/70^2  = x/80^2

Cross multiply

270 (6400) = 4900x   

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps


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What describes how fast an object is moving and in what direction
Mashcka [7]

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

6 0
3 years ago
Read 2 more answers
Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0
3 years ago
by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
3 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
3 years ago
Read 2 more answers
Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration
ollegr [7]

Answer:

Explanation:

The mass of the car doesn't matter because On a flat curve the mass of the car does not affect the speed at which it can stay on the curve. You would need the mass if you were solving the the centripetal force acting on the car, but not the acceleration.

a=\frac{v^2}{r} and filling in

a=\frac{(8.5)^2}{48.0} and we need 2 significant digits in our answer. That means that

a = 1.5 m/sec²

8 0
3 years ago
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