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Dvinal [7]
2 years ago
12

Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the

wheels. The most massive car is car F, the least massive is car H, and all three cars have identical tires.
Which car travels the longest distance to skid to a stop?
Physics
1 answer:
Nastasia [14]2 years ago
5 0

Answer:

Car H

Explanation:

Frictional force is a resistant force. It is given as:

F = u*m*g

Where u = coefficient of friction

m = mass

g = acceleration due to gravity

From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.

Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.

The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.

So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.

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Marissaâs car accelerates uniformly at a rate of +2.60 m/s². How long does it take for Marissaâs car to accelerate from a speed
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Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, a=+2.6\ m/s^2

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

t=\dfrac{v-u}{a}

t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.                                    

4 0
3 years ago
Rosa uses the formula (vi cos)t to do a calculation. Which value is Rosa most likely trying to find? the vertical acceleration o
blagie [28]
The horizontal displacement of a projectile launched at an angle
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3 years ago
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An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,
trasher [3.6K]

Answer:

The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

F=mg

mg=\dfrac{GmM}{R^2}....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

F'=mg'

mg'=\dfrac{GmM}{4R^2}....(II)

Dividing equation (II) by equation (I)

\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}

\dfrac{g'}{g}=\dfrac{1}{4}

Hence, The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

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Answer:

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3 years ago
A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
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Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0
2 years ago
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