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Vitek1552 [10]
3 years ago
13

A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 437 N on the glove, so that the glove is d

isplaced 9.0 cm, how much work is done by the ball?
Physics
1 answer:
sergiy2304 [10]3 years ago
5 0
437x9 
is ur answer. I'm not sure tho hope it helps

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Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
Lapatulllka [165]

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

3 0
3 years ago
Hypothetically speaking, if an object were located at the center of the Earth, the gravitational force on that object due to the
Likurg_2 [28]

Answer:

D. ) The force would be zero newtons

Explanation:

Because

If you are at the center of the earth, gravity is zero because all the mass around you is pulling "up" (every direction there is up!)

So F=mg so if g is zero F is also zero

5 0
3 years ago
A slinky forms it’s third harmonic standing wave when the input frequency is 24 Hz. What is the fundamental frequency of the sli
jarptica [38.1K]

Answer: The fundamental frequency of the slinky = 8Hz

An input frequency of 28 Hz will not create a standing wave

Explanation:

Let Fo = fundamental frequency

At third harmonic,

F = 3Fo

If F = 24Hz

24 = 3Fo

Fo = 24/3 = 8Hz

If an input frequency = 28 Hz at 3rd harmonic

Let find the fundamental frequency

28 = 3Fo

Fo = 28/3

Fo = 9.33333Hz

Since Fo isn't a whole number, it can't create a standing wave

6 0
3 years ago
A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of
Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

h_i = Initial height

h_f = Final height

From conservation of Energy

\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}

h_i=2r

v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0
3 years ago
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