Answer:
The magnitude of the tension in he string is equal to the magnitude of the weight of the object.
Explanation:
According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.
In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,
The tension in string(T) , The weight of the object(W) .
∴ F = 0
T - W = 0
So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of the object.
Answer: Last option
2.27 m/s2
Explanation:
As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.
If we call a_c to the centripetal acceleration then, by definition

in this case we know the speed of the runner

The radius "r" will be the distance from the runner to the center of the track



The answer is the last option
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
Answer:
Explanation:
There is electric field between the plates whose value is given by the following expression
electric field E = V /d where V is potential between the plates and d is distance between them
E = 300 / 5 x 10⁻³
= 60 x 10³ N/c
Force on electron = q E where q is charge on the electron
F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.
Acceleration a = force / mass
a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹
= 10.55 x 10¹⁵ m / s²
For midway , distance travelled
s = 2.5 x 10⁻³ m
s = 1\2 a t²
t = 
= 
t = .474 x 10⁻¹⁸ s
For striking the plate time is calculated as follows
t =
[/tex]
t = 0.67 x 10⁻¹⁸ s