Answer:
Explanation:
The bike's acceleration can be found by using the following suvat equation:
where
v is the final velocity of the bike
u is the initial velocity
a is the acceleration
s is the distance covered
For the bike in the problem,
u = 0
v = 7 m/s
d = 40 m
Solving the equation for a, we find the acceleration:
Mass: the amount of matter an object contains
Weight: mass•acceleration due to gravity(9.8m/s) / the force of gravity acting on an object
Answer:
d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy .
Explanation:
Given :
Initial Potential energy , 
.
Initial Kinetic energy , 
. ( because ball is in rest )
Now , we know , kinetic energy is maximum when an object reaches ground .
Also , potential energy is zero when an object is in ground .
We know , by conservation of energy :
Initial total energy = Final total energy
Therefore , option d) is correct .
Answer:
Force, 
Explanation:
It is given that,
Length of the room, l = 4 m
breadth of the room, b = 5 m
Height of the room, h = 3 m
Atmospheric pressure, 
We know that the force acting per unit area is called pressure exerted. Its formula is given by :
So, the total force on the floor due to the air above the surface is 
. Hence, this is the required solution.
Answer:
Somewhere between the two wires, but closer to the wire carrying λ₂
Explanation:
Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².
Electric Fied due to an electric charge is a vector and its direction is such that if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)
According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires are opposite
In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.
As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.
But for points closer to wire with λ₂ ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance to get equals E and then Ef = 0
