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Hunter-Best [27]
2 years ago
10

Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of ma

sses.​
Physics
1 answer:
tiny-mole [99]2 years ago
5 0

Explanation:

new non law neutron means neutral then it's important that baseball and softball features small respectable range of masses soft it means that when a ball hits anything hard it comes back by the Newton Law if the baseball is big and the small boy small and then if the contract with each other they ignore triple so when a ball hits the wall if the comeback because of the Mutants and when a big ball if we throw it to the wall it doesn't come that it comes back but in a very low way because it contains less neutrons in it if it is helpful please share with me

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If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the
agasfer [191]

Answer:

Output power of the circuit is 3 Watt.

Given:

loss in decibles = 3 dB

Input power = 6 Watt

To find:

Output power = ?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss = 0.5 ratio

Output power is given by,

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watt

Thus, output power of the circuit is 3 Watt.

4 0
2 years ago
Read 2 more answers
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

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Consider a ball rolling around in a circular path on the inner surface of a cone. the weight of the ball is shown by the vector
zimovet [89]

Answer:

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1 year ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

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if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

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1 year ago
A What CE describes the way energy is stored in a sandwich​
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