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Volgvan
3 years ago
8

If we want to describe work, we must have

Physics
2 answers:
tresset_1 [31]3 years ago
8 0

The answer is C

Hope that helps!

Good luck :)

PtichkaEL [24]3 years ago
6 0

Answer:

The correct answer is option c.

Explanation:

Work is defined as product of the force applied on object and displacement of the object.the direction of the force and displacement of the object are same in direction.It is measures is  Joules (J).

W=F\times d

Joules(J)= Newton(N) × meter(m)

W = Work done in Joules

F = force applied in Newtons.

d = displace of the object due to applied force in miters

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Phase unbalance causes three-phase motors to operate at temperatures higher than nameplate ratings and, therefore, the motor cannot deliver its rated horsepower.

<h3>What is an electric vehicle (EV)?</h3>

An EV stands for an electric vehicle. Electric vehicles (EVs) are autos that are powered totally or partially by electricity.

Electric cars are extremely cost-effective to run because they have fewer moving parts to maintain and use little to no fossil fuels (petrol or diesel).

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1 year ago
State a Newtowns second law​
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How does the energy possessed by a ball bearing change as it travels along an incline ramp?
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3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

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