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adelina 88 [10]
3 years ago
14

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 11.7 revolutions/s. Find the linear speed v

of the car, assuming that the tires are not slipping against the ground. v
Physics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

The linear speed of the car, v, is 24.26 m/s

Explanation:

Given;

radius of the car's tire, r = 0.330 m

angular speed of the car, ω = 11.7 revolutions/s

The angular speed of the car in radian per second:

\omega = 11.7 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} \\\\\omega = 73.523 \ rad/s

The linear speed of the car, v, is calculated as;

v = ωr

v = 73.523 rad/s  x  0.33 m

v = 24.26 m/s

Therefore, the linear speed of the car, v, is 24.26 m/s

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True or False: If a point charge has electric field lines that point into it, the charge must be positive.
Licemer1 [7]

Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges

8 0
3 years ago
A car accidently roll of a cliff. As it leaves the cliff it has horzontal velocity of 13 m/s it hits the ground 60m from the sho
grandymaker [24]

Answer: The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

7 0
3 years ago
Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m
marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 m / s^{2}

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

             F_{n e t}=m_{t o t} \times a

Where,

F_{n e t} is net force acting on body

m_{\mathrm{tot}} is mass of body

a is acceleration of body

Given values  

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

                a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}

Put the value for m = hanging mass = 40 kg and g=9.8 \mathrm{m} / \mathrm{s}^{2}, we get

                  a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}

The tension in the ropes,  T=(m \times g)+(m \times a)

Here, m as hanging mass

T = tension, N or  k g m / s^{2}

m = mass, kg  

g = gravitational force, 9.8 \mathrm{m} / \mathrm{s}^{2}

a = acceleration, m / s^{2}

          T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N

3 0
3 years ago
A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart
Charra [1.4K]
11,000 m = 11 km

11 km/s over 380,000km

380,000 / 11 = 34545.4 seconds

34545.4 / 60 = 575.7 minutes
5 0
3 years ago
A train travels 250 km westward from Carthage to Johnson City. The train arrives 2.5 hours after it left. What was the average v
den301095 [7]
The answer is C 100 km/h west
5 0
3 years ago
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