Answer:
From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to
1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.
2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity
3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.
4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.
5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.
Answer:
the water level remains same
Explanation:
This can be explained by Archimedes's principle which says that the wood will sink if weight of wood is more than the weight of the water displaced with weight equal to the water displaced otherwise the wood will float.
Therefore, buoyancy or the buoyant force is the same as the weight of wood, the weight of the water displaced by wood is also the same as that of the weight of wood.
Thus, we can see that the weight of the wood remains same and so is the level of water.
Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
your welcome, and have a great day.
