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BabaBlast [244]
3 years ago
11

1. A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ba

ll, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?
Physics
1 answer:
MariettaO [177]3 years ago
8 0
Momentum = (mass) x (velocity)

Original momentum before the hit =

                   (0.16 kg) x (38 m/s) this way <==

               =             6.08 kg-m/s  this way <==

Momentum after the hit =

                   (0.16) x (44 m/s) that way  ==>

               =           7.04 kg-m/s  that way ==>

Change in momentum = (6.08 + 7.04) =  13.12 kg-m/s  that way ==> .
-----------------------------------------------

Change in momentum = impulse.

                                   Impulse = (force) x (time the force lasted)

                          13.12 kg-m/s  = (force) x (0.002 sec)

  (13.12 kg-m/s) / (0.002 sec)  =  Force

             6,560 kg-m/s² = 6,560 Newtons  =  Force   

                            ( about 1,475 pounds  ! ! ! )
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Anna35 [415]
(a) Period of the wave
The period of a wave is the time needed for a complete cycle of the wave to pass through a certain point.
So, if an entire cycle of the wave passes through the given location in 5.0 seconds, this means that the period is equal to 5.0 s: T=5.0 s.

(b) Frequency of the wave
The frequency of a wave is defined as
f= \frac{1}{T}
since in our problem the period is T=5.0 s, the frequency is
f= \frac{1}{5.0 s}=0.2 Hz

(c) Speed of the wave
The speed of a wave is given by the following relationship between frequency f and wavelength \lambda:
v=f \lambda = (0.2 Hz)(2.5 m)=0.5 m/s

8 0
3 years ago
Potential Energy Equation (GPE = mgh)
Shkiper50 [21]

Answer:

10

Explanation:

weight=25x9.2=230

25x9.2x10=2300

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6 0
3 years ago
A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs
adell [148]

Answer:

the magnitude of the charge Q on each plate is 3.053 *10^{-8} \ C

Explanation:

Given that :

mass (m) = 7.00 *10 ^{-3} \ kg

charge (q) = +0.155 µC = +0.155 *10^{-6}\ C

angle \theta = 30^0 \ C

Area A on each plate = 0.0135 m²

From the diagram below;

tan \ \theta = \frac{Eq}{mg}    ----- equation (1)

Also by using Gauss Law ;

Q = \epsilon_0 \phi

Q = \epsilon_0EA     ----- equation (2)

Combination equation 1 and 2 together ; we have

Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}

Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}

Q = 3.053 *10^{-8} \ C

8 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
A sheet of gold foil has a volume of 0.750 cm3. If the foil measures 50.0cm by 10.0cm, what is the thickness of the foil?
vlada-n [284]

To calculate the thickness of foil, we use formula of volume as

V= l w t

Here, l is the length of the foil, w is the width of the foil and t is the thickness of the foil.

Given,  V = 0.750 \ cm^3, l = 50 cm and w = 10 \ cm.

Substituting theses values in above equation, we get

0.750 \ cm^3 = 50 cm \times 10 \ cm \times t \\\\ t =  0.0015 cm

Thus, the thickness of the foil is 0.0015 cm.

6 0
3 years ago
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