Answer:
Explanation:
Given:
- initial gauge pressure in the container,
- atmospheric pressure at sea level,
- initial volume,
- maximum pressure difference bearable by the container,
- density of the air,
- density of sea water,
<u>The relation between the change in pressure with height is given as:</u>
where:
dz = height in the atmosphere
= standard value of gravity
<em>Now putting the respective values:</em>
Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)
Answer:
Final velocity of the block = 2.40 m/s east.
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum
Mass of bullet = 0.0140 kg
Consider east as positive.
Initial velocity of bullet = 205 m/s
Mass of Block = 1.8 kg
Initial velocity of block = 0 m/s
Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s
Final velocity of bullet = -103 m/s
We need to find final velocity of the block( u )
Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u
We have
2.87 = -1.442 + 1.8 u
u = 2.40 m/s
Final velocity of the block = 2.40 m/s east.
Answer:
a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw
Explanation:
τchild=τrock
Use the equation for torque in this equation.
(F)child(d)child)=(F)rock(d)rock)
The force of each object will be equal to the force of gravity.
(m)childg(d)child)=(m)rockg(d)rock)
Gravity can be canceled from each side of the equation. for simplicity.
(m)child(d)child)=(m)rock(d)rock)
Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.
(25kg)(1m)=(50kg)drock
Solve for the distance between the rock and the center of the seesaw.
drock=25kg⋅m50kg
drock=0.5m
Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
