In 7.0 s, a car accelerates uniformly from rest to a velocity at which its wheels are turning at 6.0 rev/s. If the tires of the car have a diameter of 42 cm, and they rolled on the ground without slipping, how far did the car go in those 7.0 s?

Answers

Answer 1

The car has traveled a distance of approximately 55.44 m.

In 7.0 s, a car accelerates uniformly from rest to a velocity at which its wheels are turning at 6.0 rev/s. If the tires of the car have a diameter of 42 cm, and they rolled on the ground without slipping, the car will travel a distance of approximately 50.9 meters in those 7.0 seconds.

To calculate the distance, we must first determine the car's final velocity. The angular velocity of the wheels is given by 6.0 rev/s. Since the diameter of the tires is 42 cm, the circumference is:πd = π(0.42 m) = 1.32 m. The velocity of the car can be calculated by multiplying the circumference by the angular velocity: v = 6.0 rev/s × 1.32 m/rev = 7.92 m/s.

Now that we know the final velocity of the car, we can use the formula:d = (vf + vi)t/2where vi = 0 m/s (since the car is initially at rest), vf = 7.92 m/s, and t = 7.0 sd = (7.92 m/s + 0 m/s)(7.0 s)/2 = 27.72 m. The car traveled approximately 27.72 m in the first half of the trip (from rest to the final velocity), and 27.72 m in the second half of the trip (from the final velocity to a complete stop).

Therefore, the total distance traveled by the car in those 7.0 s is approximately 27.72 m + 27.72 m = 55.44 m. However, this is the distance that the wheels have rolled, not the distance that the car has traveled. Since the wheels are not slipping, the distance that the car has traveled is equal to the distance that the wheels have rolled.

So, the car has traveled a distance of approximately 55.44 m. Rounding to the appropriate significant figures, the distance is approximately 50.9 meters.

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Related Questions

Question 3
A crate, with mass of 21 kg, needs to be lifted and
moved by a machine. The machine attaches a chain to
the crate and pulls on the crate at an angle of 60°
from the ground. What is the minimum pulling force
required by this machine to lift the crate off the
ground, when it pulls the crate at the given angle?
206.0 N
1.25 pts
O 237.9 N
O 305.5 N
O 412.0 N

Answers

The minimum pulling force required by the machine to lift the crate off the ground at a 60° angle is 237.9 N.

To find the minimum pulling force required to lift the crate off the ground, we need to consider the vertical component of the force exerted by the machine.

Given that the angle between the chain and the ground is 60°, we can determine the vertical component of the force as follows:

Vertical component = Force * sin(angle)

Let's calculate the vertical component of the force:

Vertical component = Force * sin(60°)

= Force * ([tex]\sqrt{3}[/tex] / 2)

We want to find the minimum pulling force required, so we need to consider the force that counteracts the weight of the crate.

Weight = mass * gravity

Weight = 21 kg * 9.8 m/[tex]s^2[/tex]

= 205.8 N

Since the vertical component of the force must be equal to the weight of the crate to lift it off the ground, we can set up the equation:

Force * ([tex]\sqrt{3}[/tex] / 2) = 205.8 N

Solving for Force:

Force = 205.8 N / ([tex]\sqrt{3}[/tex] / 2)

= 205.8 N * (2 / [tex]\sqrt{3}[/tex])

≈ 237.9 N

Therefore, the minimum pulling force required by the machine to lift the crate off the ground at a 60° angle is approximately 237.9 N.

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a platinum resistance thermometer has a resistance R=40ohm at T=30°c a for pt is 3.92×10^-3°C.the thermometer is immersed in a vessel containing melting tin,at which point R increases to 94.6ohm.what is the melting point of tin

Answers

The correct answer is 105.98 C

Given,

[tex]R_1 = 40 ohm\\R_2 = 94.6 ohm\\T_1 = 30 C\\[/tex]

Coefficient of resistance for Pt = 3.92×10^-3°C

[tex]R_1/R_2 = (1+\alpha T_1)/(1+\alpha T_2)\\[/tex]

[tex]40/94.6 = (1+(3.92×10^(-3) * 30)/(1+(3.92×10^(-3)* T_2)\\T_2= 105.98 C[/tex]

Therefore, the melting point of tin is 105.98 C

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3 Study the circuits in Figure 36.8. The switch S is open (there
is a break in the circuit at this point). In which circuit would
lamps Q and R light but not lamp P?

Answers

The circuit in which lamps Q and R light but not lamp P when switch S is open is circuit B.

What is an electric circuit?

An electric circuit is a path for transmitting electric current.

Given the circuits below, when switch S is open, we want to determine the circuit in which lamps Q and R light but not lamp P.

To determine the circuit, we proceed as follows.

To determine the circuit in which lamps Q and R light but not lamp P, it must satisfy this condition

The switch must be before lamp P and The circuit for lamp P must be different from that of lamps Q and RLamps Q and R must be in the same circuit

Looking at all the circuits, the circuit which satisfy these condition is circuit B

So, the circuit in which lamps Q and R light but not lamp P is circuit B.

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What happens to sound waves from an object as it moves toward you?(1 point) Responses

Answers

Answer:

Suppose the object were stationary and emitting waves that had a distance of 1 m between crests - the receiver would receive waves that had a distance of 1 between crests

Suppose the object were moving towards the receiver, then there would no longer be 1 m between the crests as measured in the laboratory frame because of movement of the object.

Then the receiver would receive waves that were less than 1 m apart and would report a higher frequency than if the object were stationary,

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