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2 years ago

A force acting on an object does no work if

2 answers:

8 0

If the force is not in the direction of the object’s motion then a <span> force acting on an object does no work
work is done if force is applied on a body and body covers some distance in the direction of the force
W=FS
hope it helps</span>

KIM [24]2 years ago

3 0

Answer: the force is not in the direction of the object’s motion.

Explanation:

Work = Force. Displacement

⇒ W = F. s = F s cos θ

Thus, work is said to be done when a force displaces an object from its position in the direction of the force.

If the force is less the frictional force, the object would not move. An object accelerates when a force is applied. Sometimes, the direction of force and direction of motion is not same. In that case, no work is said to be done.

For example, when porter lifts the weight on head and moves forward. The weight is perpendicular to the direction of motion. cos 90° = 0 ⇒ W = 0

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A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

Lisa [10]

Answer:

Vertical distance= 3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time= 0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0

2 years ago

In the lab activity, you will examine sound waves as they are emitted from a moving source. Predict what will happen to the soun

irina1246 [14]

<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
Hope this helps</span>

6 0

2 years ago

Read 2 more answers

What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear

faust18 [17]

Answer:

1) The radius of the tire is approximately 0.28966 meters

2) The centripetal force is the force that keeps a body moving on a circular path

Explanation:

1) The linear speed of the automobile tire = 20.0 m/s

The period with which the tire turns = 0.091 s

The period = The time it takes to make a complete turn

Therefore;

The number of turns in 1 second = 1/0.091 ≈ 10.989 turns

The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire

∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire

From the speed of the car, 20.0 m/s, we have;

The distance covered in 1 second = 20.0 meters

Therefore;

10.989 × 2 × π × Radius of the tire = 20.0 meters

Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters

The radius of the tire ≈ 0.28966 meters

2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.

The centripetal force is given by the formula, $F = \dfrac{m \cdot v^2}{r}$

Where;

F = The centripetal force

m = The mass of the object

v = The linear velocity of the object

r = The radius of the rotational motion.

3 0

2 years ago

A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago

A 17 g audio compact disk has a diameter of 12 cm. The disk spins under a laser that reads encoded data. The first track to be r

sladkih [1.3K]

Answer:

0.00066518 Nm

Explanation:

v = Velocity = 1.2 m/s

r = Distance to head = 2.3 cm

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 2.4 s

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{1.2}{0.023}\\\Rightarrow \omega=52.17391\ rad/s$

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{52.17391-0}{2.4}\\\Rightarrow \alpha=21.73912\ rad/s^2$

Torque

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mR^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}0.017\times 0.06^2\times 21.73912\\\Rightarrow \tau=0.00066518\ Nm$

The torque of the motor is 0.00066518 Nm

6 0

2 years ago