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il63 [147K]
3 years ago
15

If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp

eed of stanton's bat decrease due to the collision? (give your response as a positive value in units of miles per hour. assume that the bat and ball are moving in a single dimension. a baseball has a weight of 5 ounces and the bat stanton uses has a weight of 32 ounces.)
Physics
1 answer:
Mice21 [21]3 years ago
3 0

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

m_1(v_f - v_i) = m_2(\Delta v)

given that

m_1 = 5 ounce

m_2 = 32 ounce

For ball the change in speed will be

v_f - v_i = (120 - 77)mph

now from above equation

5\times (120 - 77) = 32 \times \Delta v

\Delta v = 6.72 mph

so speed of bat will decrease by 6.72 mph

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Which two processes increase the motion of the molecules?
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5 0
3 years ago
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
3 years ago
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