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jekas [21]
3 years ago
8

An inexperienced catcher catches a 126 km/h fastball of mass 160 g within 1.36 ms, whereas an experienced catcher slightly retra

cts his hand during the catch, extending the stopping time to 13.6 ms. What is the average force imparted to the two gloved hands during the catches of the inexperienced catcher
Physics
1 answer:
Firdavs [7]3 years ago
6 0

Answer:

4117.65 N

Explanation:

Speed of ball, u = 126 km/h = 35 m/s

Mass of ball, m = 160 g = 0.16 kg

Time interval, t = 1.36 ms = 0.00136 s

We can calculate the force as a measure of the momentum of the ball:

F = P/t

Momentum, P, is given as:

P = mv

Therefore:

F = (mv) / t

F = (0.16 * 35) / (0.00136)

F = 4117.65 N

The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.

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A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?
Inga [223]

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

6 0
3 years ago
Otion
Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

     N- W = 0

     N = W

X axis (radial)

        fr = m a

the acceleration in the curve is centripetal

         a = \frac{v^2}{r}

the friction force has the expression

        fr = μ N

we substitute

       μ mg = m v²/r

       v = \sqrt{\mu g r}

we calculate

      v = \sqrt{0.1 \ 9.8 \ 3}

      v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

        v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0
3 years ago
Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
exis [7]

Answer:

it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
The diagram shows the Earth rotating on it's axis. The two stars show different locations on the surface... How long does it tak
amid [387]
My guess would be about 10 years because stars are hot balls of light that are reflections from years ago so it would most likely take awhile
6 0
3 years ago
Read 2 more answers
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