The de broglie wavelength of an electron is 8.7 Ã 10-11 m. the mass of an electron is 9.1 Ã 10-31 kg. the velocity of this elect
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Answer:
a = 0.1 m/s²
Explanation:
- If the maximum static frictional force is 90 N, this means that any applied force that will overcome this force, will cause the piano to slide, so kinetic frictional force applies.
- Under these conditions, the net force in the horizontal direction is just the difference between the applied force (which is larger that the static friction force) and the kinetic frictional force, as follows:
- By the same token, according Newton's 2nd Law, this force is just equal to the product of the mass of the piano, times the acceleration, as follows:
- Solving for a:
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
=
So,
Rearranging the above equation to get Electric field, we will get:
E =
Multiply and divide by
E = x
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x ) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
=
Rearranging the above equation for E:
E =
E = +
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E = +
Replacing the above equations to get E:
E = (σ1 x ) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x = - σ2 x
In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up
First we will find the mass of the stone
As it is given that stone is spherical in shape so first we will find its volume
Now it is given that it's specific gravity is 10.8
So density of rock is
mass of the stone will be
now change in potential energy is given as
here
g = gravity on planet = 0.278 m/s^2
H = height lifted upwards = 15 cm
Now energy supplied by internal circuit of robot is given by
V = voltage supplied = 10 V
i = current = 1.83 mA
t = time = 12 s
Now efficiency is defined as the ratio of output work with given amount of energy used
so efficiency will be 23 %