A) Volume = displacement
Therefore the rock has a volume of 10mL
B) density = Mass/ Volume
Therefore density = 23/10 = 2.3 per unit volume
PEG = 0.8*KE.
M*g*h = 0.8 *(0.5*M*v^2) M, h, and v are the vaulter's mass, height, and speed, respectively.
g*h = 0.8*0.5*v^2 Divide both sides by M (the mass).
g*h = 0.4*v^2
(9.81 m/s^2)*h = 0.4*(7 m/s)^2. Note that m and s denote meter and second, respectively.
h = 0.4*(7 m/s)^2 /(9.81 m/s^2) Divide both sides by 9.81 m/s^2. Also, note that m is NOT mass here.
= 0.4*(49 m^2/s^2)/(9.81 m/s^2)
= 0.4*49 m/9.81
Answer = 1.99796 meters
Answer:
Right shoe
Explanation:
Let the mass and velocity of incoming puck be m and v respectively.
Momentum of the colliding puck will be mv
In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe
= mv - 0 = mv
If time duration of collision be t
rate of change of momentum
= mv / t
This is the force exerted by puck on the left shoe .
Now let us consider collision with right shoe
momentum after collision with right shoe
- mv
change in momentum
= mv - ( - mv ) = 2mv
If time duration of collision be t
rate of change of momentum
= 2mv / t
This is the force exerted by puck on the right shoe .
Since the force on the right shoe is more , this shoe will have greater speed
after collision.
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J