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3 years ago

If you exert a force of 10.0 n to lift a box a distance of 0.9 m how much work do you do

1 answer:

5 0

A bit vague since it's not clear whether the applied force is in the same direction as the box's displacement. But let's assume it is. Then the work done is:

W = Fd

W = work, F = force, d = displacement

Given values:

F = 10.0N, d = 0.9m

Plug in and solve for W:

W = 10.0(0.9)

W = 9J

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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.

3 0

3 years ago

Evaluate u+xy where U=3 X=4 Y=7

Reptile [31]

Answer:

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

6 0

3 years ago

CAN SOME ONE HELP PLEASE :))))

Ede4ka [16]

Line c is at rest . line a is going in a positive direction . line b is going in a negative direction . line d is negative too

5 0

3 years ago

A solid cube of side 5cm has a mass of 250g. What is its density in kgm ??

Mariulka [41]

Answer:

5kgm

Explanation:

convert cm to m and g to kg

250/1000=0.25kg

5/1000=0.05m

then find the density

density=mass/volume

=0.25kg/0.05m

=5kgm

7 0

3 years ago

What is velocity ratio ?

Rasek [7]

The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied, in the same time. In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage. Velocity ratio is sometimes called distance ratio.

5 0

3 years ago