If a 1.50 kg mass revolves at the end of a string 0.50 m long, and its tangential speed is 6.0 m/s, calculate the centripetal fo
2 answers:
Answer:
Explanation:
Mass of object,
Length of string,
Tangential speed,
Now, centripetal force F of a object moving in given radius r and mass m moving with velocity v.
Here , object moves along the ends of string. Therefore, radius is equal to length of string.
Putting values of m,v and r in above equation.
We get,
Hence, this is the required solution.
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Answer:
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
Initial speed of the projectile is v and final speed is 0.5 v.
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
So, the height of the projectile above the ground is . Hence, this is the required solution.
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
Force,
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :
Finally, it stops, v = 0
F = -22713.92 N
So, the magnitude of the force that stops the bullet is