Answer:
Electric field,
Explanation:
It is given that,
Magnitude of charge,
Force experienced,
We need to find the electric field at the origin. It is given by :
So, the electric field at the origin is . Hence, this is the required solution.
After an incandescent lamp is turned on, the temperature of its filament rapidly increases from room temperature to its operatin
2 answers:
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Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
The other angle is 30 degrees.
Explanation:
The range of projectile is given by :
Here,
u is the speed of launch of projectile
Here,
We need to find the other launch angle when the projectile have the same range, such that,
So, the other angle is 30 degrees. Hence, this is the required solution.
Answer:
The recoil velocity is 0.2687 m/s.
Explanation:
∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.
∴ There is no force acting acting on the person and book as a system in horizontal direction.
Hence , momentum is conserved for this system in horizontal direction of motion.
If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p) -
=
System initially is at rest
∴=0
∴ From the above 2 equations
=0
We know that ,
Momentum(p)=Mass of the body(m)×velocity of the body(v)
Let and
be the mass of the person and the book respectively and
and
be the final velocities of the person and book respectively.
∴=
+
=0
From the question ,
= 74.9 kg
= 2.44 kg
= 8.25 m/s
Substituting these values in the above equation we get ,
(74.9 × )+ (2.44×8.25) = 0
∴ = - 0.2687 m/s (Negative sign suggests that the motion of the person is opposite to that of the book)
∴ The recoil velocity is 0.2687 m/s.