Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
It depends on their tension. Normally, the rubber band store more tension than a t shirt.
Answer:
The rate at which radar must rotate is 0.335 rad/s.
Explanation:
Given that,
Velocity = 65 m/h = 29.0576 m/s
Angle = 15°
Suppose, the radius given by
We need to calculate the rate at which radar must rotate
Using formula of linear velocity
Where, v = velocity
r = radius
Put the value into the formula
Hence, The rate at which radar must rotate is 0.335 rad/s.
Answer:17.44A
Explanation: Current density=I/Area
Area is given by 2.79mm^2=2.79×10^-6m^2
Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A
Answer:
distance = 21.56 m
Explanation:
given data
mass = 50 kg
initial velocity = 18.2 m/s
force = -200 N ( here force applied to opposite direction )
final velocity = 12.6 m/s
solution
we know here acceleration will be as
acceleration a = force ÷ mass
a = = -4 m/s²
we get here now required time that is
required time = ...............1
put here value
required time =
so distance will be
distance = ........2
distance =
distance = 21.56 m