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bonufazy [111]
3 years ago
10

HELP PLEASE!!!

Physics
1 answer:
irakobra [83]3 years ago
6 0

Answer:

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

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What is the difference between speed and velocity?
avanturin [10]

Answer:

speed is the rate of change in distance thus it is scalar physical quantity

while velocity is the rate of change in displacement thus it is a vector physical quantity

Explanation:

vector physical quantity: is a quantity that requires both magnitude and direction to identify

scalar quantity: requires only magnitude to identify.

6 0
3 years ago
1. What is the Formula's for Kinetic and Potential Energy?
Vilka [71]

Answer:

Total energy of the object = mgh. As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, the kinetic energy = 1/2mv2.

Explanation:

4 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 4.4 m/s2, what will its velocity be after 28s?
Norma-Jean [14]

Answer:

123.2 m/s after 28s

Explanation:

Vi= 0 m/s

a= 4.4 m/s^2

t=28s

Vf after 28s

To find Vf use your kinematics formula Vf=Vi+at

Vi is Zero so it gets removed and the equation becomes

Vf=at  

Simply Plug and Solve

Vf= 4.4(28)

Vf=123.2 m/s after 28s

6 0
3 years ago
What is the pressure at the very bottom of a lake, which is 432 ft deep? (assume an air pressure of 1.013 ✕ 105 pa.)?
aleksley [76]
Preasure at the bottom would be Air Pressure at the top added to the pressure due to the water height.
so pressure = air pressure + hdg
where,
h = depth of the lake, (432ft into metres)
d = density of water (1000kg/m^3)
g = 9.81 m/s^2 (approx.)
8 0
4 years ago
A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as a box car passes him, the fugitive
Dima020 [189]

Answer:Given:

Initial speed of fugitive, v0 = 0 m/s

Final speed, vf = 6.1 m/s

acceleration, a = 1.4 m/s^2

Speed of train, v = 5.0 m/s

Solution:

t = (vf-v0)/a

t = (6.1-0)/1.4

t =4.36 s

Distance traveled by train, x_T =v*t

x_T =5*4.36 = 21.8 m

Distance travelled by fugitive, x_f = v0*t+1/2at^2

x_f = 0*4.36+1/2*1.4*4.36^2

x_f =13.31 m

5*t = v(t-4.36)+x_f

5*t=6.1*(t-4.36)+13.31

solve for t, we get

t = 12.08 s

The fugitive takes 12.08 s to catch up to the empty box car.

Distance traveled to reach the box car is

X_T = v*t

X_T = 5*12.08 s

X_T = 60.4 m

Explanation:

5 0
3 years ago
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