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Alona [7]
3 years ago
5

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

Physics
1 answer:
ioda3 years ago
3 0
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
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Find the force necessary to pull a 6 kg object 3 m/s2
mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

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We have the final answer as

<h3>18 N</h3>

Hope this helps you

6 0
3 years ago
Which of the following provides evidence that there must be at least two types of electrical charge, but that there is only one
polet [3.4K]

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= 6.67\times 10^{-11}

The Coulomb's law constant (K) will be:

= 9\times 10^9

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⇒ \frac{f_e}{f_g}\sim 10^{42}

  • By that same argument, the electrostatic force including its planet's atmosphere would have strongly influenced the effect, as well as maybe the planet's atmosphere, would have crashed, or perhaps the earth would have shifted at a much longer exposure from one another and.
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The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

5 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
3 years ago
A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}

\Rightarrow s=197.76\ m

The distance traveled by the jet is 197.76 m

8 0
3 years ago
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