Answer:
6495.19 Joule
Explanation:
F = Weight of the crate = 250 N
d = Distance the cart is pushed = 30 m
θ = Angle of inclination = 60°
The weight of the crate will be resloved into two components
Fdsinθ and Fdcosθ
Work done by the force of gravity is
W = Fdsinθ
⇒W = 250×30×sin60
⇒W = 6495.19 Joule
∴ The work done by the force of gravity is 6495.19 Joule
Answer:
five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed
The answer is b i just did the test
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)