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3 years ago

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?

Physics

2 answers:

Tema [17]3 years ago

5 0

Answer:

An applied force, drag force and gravitational force.

Explanation:

When Amy throws a softball through the air, there is some force that acts on it. The forces that are experienced by the ball in the air as follows :

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences a drag from the air it passes through. It also experiences a downward pull because of gravity.

The applied force is defined as the force applied by the person who is throwing the ball.

Drag is equivalent to the frictional force that is applied in the opposite direction.

The gravitational force always acts in downward direction.

aniked [119]3 years ago

4 0

An applied force is a force that is applied to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
Normal force is the component, perpendicular to the surface (surface being a plane) of contact.
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.

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As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes 2.91 back-and-forth cy

Lapatulllka [165]

Answer:

Wavelength, $\lambda=0.011\ m$

Explanation:

Given that,

Number of cycles in a spiral spring is 2.91 in every 3.67 s

The velocity of the pulse in the spring is 0.925 cm/s, v = 0.00925 m/s

To find,

Wavelength

Solution,

Number of cycles per unit time is called frequency of a wave. The frequency of the longitudinal pulse is,

$f=\dfrac{2.91}{3.67}=0.79\ Hz$

The wavelength of a wave is given by :

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{0.00925\ m/s}{0.79\ Hz}$

$\lambda=0.011\ m$

So, the wavelength of the longitudinal pulse is 0.011 meters. Hence, this is the required solution.

8 0

2 years ago

The ideal mechanical advantage of a pulley system is equal to the?

marishachu [46]

Answer: The correct answer is "Number of rope segments supporting the load".

Explanation:

Mechanical advantage: It is defined as the ratio of the force produced by a machine to the force applied on the machine. The ideal mechanical advantage of a machines is mechanical advantage in the absence of friction.

The ideal mechanical advantage of a pulley system is equal to the number of rope segments which is supporting the load. More the rope segments, It is more helpful to do the lifting the work.

It means that less force is needed for this task to complete.

Therefore, the correct option is (C).

7 0

3 years ago

Read 2 more answers

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$

4 0

3 years ago

Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?

Drupady [299]

The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d

We have:
W = ?
F = 20 N = 20 kg*m/s²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J

8 0

3 years ago

A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.

Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 $\frac{m}{s^{2} }$

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.

4 0

3 years ago