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
Answer:
7.9 
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

Using algebra, we can rearrange our equation to find the final acceleration,
:

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8
= 8.33
Plug everything in:
7.9
= 
(1590kg the initial weight plus the weight of the added passenger)
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
Refer to the diagram shown below.
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s
Answer:
(a) 160000 kV/m
(b) 1336 keV
Explanation:
(a) magnetic filed, B = 10 T
energy of electron, E = 740 eV
mass of electron, m = 9.1 x 10^-31 kg
Let v be the velocity of electron.
E = 1/2 mv^2
740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 1.6 x 10^7 m/s
v = E / B
E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m
E = 160000 kV/m
(b) E = 16 x 10^7 V/m
B = 10 T
Let v be the velocity of protons.
v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s
Kinetic energy of proton, E = 1/2 mv^2
= 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14
= 2.14 x 10^-13 J = 1336000 eV = 1336 keV