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emmainna [20.7K]
2 years ago
4

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

ase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.
What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase?
Physics
1 answer:
kogti [31]2 years ago
5 0

Answer:

force = 116N

Explanation:

given,

mass of ball  = 0.145 kg

final velocity (v) = 40 m/s

initial velocity (u) = 0

time (t) =  50 ms

force = \dfrac{change\ in\ momentum}{time} \\force = \dfrac{m\times \Delta v}{t}\\force = \dfrac{0.145\times(40-0)}{50 \times 10^{-3}}

force= 116 N

hence, force exerted by the pitcher on the ball will be equal to 116N

You might be interested in
Differences between Pressure and upthrust​
Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0
2 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the
steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units,  5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters  ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )  

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom  ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter /  grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0
2 years ago
gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
2 years ago
A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the
likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

a\sin \theta_n=n\lambda

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

a\sin \theta_1=(1)\lambda

solving for θ1:

\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})

\theta_1=0.63 degrees

for the second minimum:

a\sin \theta_2=(2)\lambda

\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})

\theta_2=1.26 degrees

So, the angular separation between them is the rest:

\Delta \theta =1.26-0.63

\Delta \theta=0.63

4 0
2 years ago
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