Answer:
Bottom of the circle.
Explanation:
At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.
Answer:
a = -2.4 m/s²
Explanation:
Given,
The initial speed of the bus, u = 24 m/s
The final speed of bus, v = 12 m/s
Time taken to reach final speed is, t = 5.0 s
The acceleration of the body is given by the change in velocity by time
a = (v - u) / t
= (12 - 24) / 5
= -2.4 m/s²
The negative sign in the acceleration indicates that the bus is decelerating.
Therefore, the acceleration of the bus is, a = -2.4 m/s²
Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

B is magnetic field

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.
Answer:
Acceleration of the meteorite, 
Explanation:
It is given that,
A Meteorite after striking struck a car, v = 0
Initial speed of the Meteorite, u = 130 m/s
Distance covered by Meteorite, s = 22 cm = 0.22 m
We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :



So, the deceleration of the Meteorite is
. Hence, this is the required solution.