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Oksanka [162]
3 years ago
11

lonnie pitches a baseball of mass 0.02kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to L

onnie with a return speed of 60 m/s. If the bat is in contact with the ball 0.050 s, what is the impulse experienced by the ball?
Physics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

I=2 kg.m/s

Explanation:

The impulse is defined as the change of momentum:

I=p_f-p_o\\I=m*v_f-m*v_o\\I=0.02kg*[(-60m/s)-40m/s]\\I=2kg.m/s

We took the final velocity as negative since it is going on the opposite direction of the intial motion of the ball.

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The diameter of Mars is 6794km, and its minimum distance from the earth is 5.58×10^7 km. When Mars is at this distance, find the
pav-90 [236]

Answer:

D = 0.2374 m

Explanation:

let D be the diameter of the image of mars, f be the focal length of the telescope mirror and β be the spatial resolution.

β = 6794×10^3/(5.58×10^7) = 0.1217562724

then:

The diameter of the telescope mirror is given by:

D = f×β

   = (1.95)×(0.1217562724)

   = 0.2374 m

Therefore, the diameter of the image of mars is 0.2374 m.

4 0
3 years ago
HELP PLS and HURRY
tangare [24]

Answer:

C - 50,000 * 77 * 3

Explanation:

At the top of the hill the potential energy is E= mgh= (160 kg)(9.81 m s^-2)(30 m)= 47088

hope it helps ,

<u>help me by marking as brainliest....</u>

5 0
2 years ago
एक वाक्य में उत्तर लिखिए : 1. आज किसको बचाने की मांग है? 2. जीव कब तक जगत में रह सकता है? १. कवि किसको शुद्ध रखने की बात करते है
lilavasa [31]

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5 0
2 years ago
Read 2 more answers
What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

5 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
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