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Leona [35]
3 years ago
10

A baseball player applies a force of 200N onto a baseball. Determine the time that baseball player exerts on the baseball if an

impulse of 50 Ns is exerted on the baseball.
Physics
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

Time, t = 0.25 seconds  

Explanation:

Given that,

Force applied by a baseball player, F = 200 N

We need to find the time that baseball player exerts on the baseball if an impulse of 50 Ns is exerted on the baseball. The impulse exerted on the baseball is given by :

J=F\times t

t is time taken

t=\dfrac{J}{F}\\\\t=\dfrac{50\ N-s}{200\ N}\\\\t=0.25\ s

So, the time for which the baseball player exerts on the baseball is 0.25 seconds.

jolli1 [7]3 years ago
3 0

Answer:A football player kicks a ball with a force of 50N. Find the impulse on the ball if his foot stays in contact with the football for ... A hockey player applies an average force of 80N to a 0.25kg hockey puck for a ... She uses a hammer to drive the nail into the wall. ... A 0.5kg baseball experiences a 10N force for a duration of 0.1s.

Explanation:

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Alternative sources of energy include : geothermal
baherus [9]

Answer:

All of the above

Explanation:

The correct answer is option E (All of the above)

All the options are alternative source of energy.

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6 0
3 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many
Darya [45]

Answer:

t=45.7s

\alpha =116revolutions

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

w=w_{o}+at\\  0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s

To find revolutions we use below equation

w^{2}=w_{o}^{2}+2a\alpha

Substitute the given values to find revolutions α

So

0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha  \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad

To convert rad to rev:

\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions

7 0
3 years ago
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krok68 [10]

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

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V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V

4 0
3 years ago
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