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Dvinal [7]
3 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]3 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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Answer:

temperature

Explanation:

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I took the test, and the correct answer is not Hydrosphere and Geosphere, but instead, its actually Hydrosphere and Atmosphere

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4 0
3 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the earth (0.50×10−4t) at a distance of 12 cm
Studentka2010 [4]
The magnetic field generated by a wire carrying a current I is:
B(r) =  \frac{\mu_0 I}{2 \pi r}
where r is the distance at which the magnetic field is measured, and \mu_0 = 4 \pi \cdot 10^{-7} NA^{-2} is the magnetic permeability in vacuum.

The problem says that the magnetic field at a distance r=12 cm=0.12 m from the wire must be no larger than B=0.5 \cdot 10^{-4}T. Substituting these values, we can find the maximum value of the current I that the wire can carry:
I= \frac{2 \pi r B}{\mu _0} = \frac{2 \pi (0.12 m)(0.5 \cdot 10^{-4}T)}{ 4 \pi \cdot 10^{-7} NA^{-2}}= 30 A
4 0
3 years ago
If the reading of a linear scale is 4 mm and no of division of the circular scale is 50, then what will be the diameter of the w
Sliva [168]

Answer:

Diameter of Wire = 4.5 mm

Explanation:

First, we need to find the fractional part of the reading. The fractional part o the reading can be given by the following formula:

Fractional Part = Circular Scale Reading x Least Count

where,

Circular Scale Reading = 50

Least Count = 0.01 mm

Therefore,

Fractional Part = (50)(0.01 mm)

Fractional Part = 0.5 mm

Now, the diameter of the wire can be given by using the following formula:

Diameter of Wire = Linear Scale Reading + Fractional Part

Diameter of Wire = 4 mm + 0.5 mm

<u>Diameter of Wire = 4.5 mm</u>

8 0
3 years ago
Physics question, please answer this ASAP, thanks!
miv72 [106K]

(a) The gravitational force received by each 1 kg mass is 8.66 N.

(b) The magnitude of gravitational acceleration is 8.66 m/s².

(c) The orbital speed of the ISS is  7,663.6 m/s.

(d) The time take  for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

<h3>Gravitational force received by each 1 kg mass</h3>

The gravitational force received by each 1 kg mass is calculated as follows;

F = Gm₁m₂/r²

where;

  • m₁ is mass of Earth
  • m₂ is mass of ISS
  • r is the distance between the ISS and center of Earth

F = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1) / (6780,000)²

F = 8.66 N

<h3>Magnitude of gravitational acceleration</h3>

mg = GMm/r²

g = GM/r²

where;

  • M is mass of Earth

g = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ ) / (6780,000)²

g = 8.66 m/s²

<h3>Orbital Speed of the ISS</h3>

v = √GM/r

v = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴  / 6780,000)

v = 7,663.6 m/s

<h3>Time of motion of the ISS round the Earth</h3>

T = 2πr/v

T = (2π x 6780,000) / (7663.6)

T = 5,558.75 seconds

1 hour = 3600 seconds

=  5,558.75/3600

= 1.54 hours

Thus, the gravitational force received by each 1 kg mass is 8.66 N.

The magnitude of gravitational acceleration is 8.66 m/s².

The orbital speed of the ISS is  7,663.6 m/s.

The time take  for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

Learn more about orbital speed here: brainly.com/question/22247460

#SPJ1

4 0
2 years ago
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