The number of wavelengths that pass a point each per second is:
1 answer:
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Explanation:
It is given that,
Mass of the object, m = 0.8 g = 0.0008 kg
Electric field, E = 534 N/C
Distance, s = 12 m
Time, t = 1.2 s
We need to find the acceleration of the object. It can be solved as :
m a = q E.......(1)
m = mass of electron
a = acceleration
q = charge on electron
"a" can be calculated using second equation of motion as :
a = 16.67 m/s²
Now put the value of a in equation (1) as :
q = 0.0000249 C
or
Hence, this is the required solution.
Explanation:
The given data is as follows.
Mass, m = 75 g
Velocity, v = 600 m/s
As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.
where, = mass of the projectile
= mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
=
v = 0.898 m/s
Now, equation for energy is as follows.
E =
=
= 13500 J
Now, energy after the impact will be as follows.
E' =
= 20.19 J
Therefore, energy lost will be calculated as follows.
= E E'
= (13500 - 20) J
= 13480 J
And, n =
=
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.
At a distance of 469.2 m from the original point below the airplane.
Explanation:
First of all, we have to calculate the time it takes for the sack to reach the ground.
To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:
where, taking downward as positive direction:
s = 300 m is the vertical displacement
u = 0 is the initial vertical velocity
t is the time
is the acceleration of gravity
Solving for t, we find the it takes for the sack to reach the ground:
Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:
We also know the total time of flight,
t = 7.82 s
Therefore, we can find the horizontal distance travelled by the sack:
So, the sack will land 469.2 m from the original point below the airplane.
Learn more about free fall and projectile motion:
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