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serious [3.7K]
3 years ago
13

A car is traveling at a constant speed on the highway. Its tires have a diameter of 68.0 cm and are rolling without sliding or s

lipping. If the angular speed of the tires is 55.0 rad/s , what is the speed of the car?
Physics
1 answer:
Bad White [126]3 years ago
5 0

Answer:

37.4m/s

Explanation:

since the car doesn't accelerate, we can use the formula v=ωr where v is linear speed, ω is angular speed (rads/second) and r is radius. Substitute values for equation:

v=55*0.68

v=37.40

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Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the
Naddik [55]

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

\Sigma F = F - f = m\cdot a (Eq. 1)

Where:

F - Force from the wind exerted on the sailboat, measured in newtons.

f - Drag force of the water, measured in newtons.

m - Mass of the sailboat, measured in kilograms.

a - Net acceleration of the sailboat, measured in meters per square second.

If we know that F = 3.90\times 10^{3}\,N, f = 4.35\times 10^{3}\,N and m = 1250\,kg, then the net acceleration of the sailboat is:

a = \frac{F-f}{m}

a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}

a = -\frac{9}{25}\,\frac{m}{s^{2}}

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

v = v_{o}+a\cdot t (Eq. 2)

Where:

v_{o}, v - Initial and final velocities of the sailboat, measured in meters per second.

t - Time, measured in seconds.

If we get that v_{o} = 11\,\frac{m}{s}, a = -\frac{9}{25}\,\frac{m}{s^{2}} and t = 15\,s, then the final velocity of the sailboat is:

v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}}  \right)\cdot (15\,s)

v = 5.6\,\frac{m}{s}

The velocity of the boat 15 seconds later is 5.6 meters per second.

3 0
3 years ago
The law of reflection states that if the angle of incidence is 38 degrees, the angle of reflection is ___ degrees.
Tasya [4]

Answer:

38

Explanation:

because the law of reflection states angle of incidence is equal to the angle of reflection

4 0
3 years ago
A rightward force is applied to a box in order to move it across the table at a constant velocity. (Ignore wind resistance).
anzhelika [568]
The forces would be gravity and friction
3 0
3 years ago
A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0
4 years ago
Read 2 more answers
What do you mean by velocity ratio of a wheel and axle​
IgorC [24]

Answer:

Explanation:

In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…

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3 years ago
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