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blondinia [14]
3 years ago
7

A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will

it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
Physics
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

Ball will drop by 0.82 meter.

Explanation:

Horizontal speed = 41 m/s

Horizontal displacement = 17 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    17 = 41 t + 0.5 x 0 x t²

     t = 0.41 s

Now we need to find how much vertical distance ball travels in 0.41 s.

Initial vertical speed = 0 m/s

Time = 0.41s

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 0 x 0.41 + 0.5 x 9.81 x 0.41²

    s = 0.82 m

So ball will drop by 0.82 meter.

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A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro
german

Answer:

20m

6.9s

Explanation:

The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.

E_k = E_p

0.5mv^2 = mgh

h = \frac{0.5v^2}{g}

we can subtitute v = 20m/s and g = 10m/s2

h = \frac{0.5*20^2}{10} = 20 m

So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.

The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

t = v / g = 20 / 10 = 2s

Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff

d = \frac{gt_2^2}{2}

t^2 = \frac{2d}{g} = \frac{2*120}{10} = 24

t = \sqrt{24} = 4.9s

So the total time it takes is 4.9 + 2 = 6.9s

3 0
3 years ago
a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy incre
tekilochka [14]

Answer:

A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential

B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential

Explanation:

3 0
3 years ago
A wave changing shape when passing through an opening is an example of what?
igomit [66]
It is diffraction !!
7 0
4 years ago
PLEASE HELP!!!!! Will give brainliest to best answer
PolarNik [594]

Answer:

the answer is d

Explanation:

I took the test

6 0
4 years ago
An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra
WARRIOR [948]

Answer:

The inventors  claim is not real

a)  No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

     The  power input is  P_i  = 0.25 kW  =  0.25 *10^{3} \ W

      The  rate of heat transfer J  =  3050 J/s

       The temperature of the freezer content is T = 270 \ K

       The  ambient temperature is  T_a  =  293 \ K

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

      COP  =  \frac{T }{Ta - T}

substituting values

     COP  =  \frac{270 }{293 - 270}

     COP  =11.7

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

       COP  =  \frac{J}{P_i}

substituting values

       COP  =  \frac{3050}{0.25 *10^{3}}

       COP  = 12.2

Now given that the COP  of an ideal refrigerator is  less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

4 0
3 years ago
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