Answer:
Coefficient of dynamic friction= md= 0.09931
Explanation:
To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:
F= 90N × cos(30°) = 77.9423N
This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:
F= md × m × g= 77.9423N
m= mass of the block
md= coefficient of dynamic friction
g= gravity acceleration
F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)
md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931
Answer:
100 m/s
Explanation:
Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁
Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is completely stationary. Let v₁ is speed of bond's boat.
It is the concept of the conservation of momentum. It remains conserved. So,
Putting all the values, we get :
So, Bond's boat is moving with a speed of 100 m/s after the collision.
Answer:
The force of gravity does not affect the horizontal component of motion; a projectile maintains a constant horizontal velocity since there are no horizontal forces acting upon it.
The first car travels at 60km/h and skid at 30m away from the starting point while another car is also traveling at 180km/h. Now, we need to solve for the skidding distance.
We assigned variables such as:
V1=60km/h
V2=180km/h
Skid1=30m
Skid2=?
We solve this by ratio and proportion method such as shown below:
V1/V2=skid1/skid2
60/180=30/skid2
skid2=(30*180)/60
skid2=90meters
Th answer is 90 meters.
Answer:
Total kinetic energy of entire system is 3 mgl
Explanation:
Given two masses: m and 4m.
Since the pulley is frictionless and the thread is massless, the energy here is linked to the two masses.
Total kinetic energy of entire system = decrease in gravitational potential energy of the system.
Therefore, we have :
ΔKE = Δp
ΔKE = 4mgl - mgl
= 3 mgl
Total kinetic energy of entire system is 3 mgl
