Answer:
t = 39.60 s
Explanation:
Let's take a careful look at this interesting exercise.
In the first case the two motors apply the force in the same direction
F = m a₀
a₀ = F / m
with this acceleration it takes t = 28s to travel a distance, starting from rest
x = v₀ t + ½ a t²
x = ½ a₀ t²
t² = 2x / a₀
28² = 2x /a₀ (1)
in a second case the two motors apply perpendicular forces
we can analyze this situation as two independent movements, one in each direction
in the direction of axis a, there is a motor so its force is F/2
the acceleration on this axis is
a = F/2m
a = a₀ / 2
so if we use the distance equation
x = v₀ t + ½ a t²
as part of rest v₀ = 0
x = ½ (a₀ / 2) t²
let's clear the time
t² = (2x / a₀) 2
we substitute the let of equation 1
t² = 28² 2
t = 28 √2
t = 39.60 s
::Answer::
The answer you're looking for is Electrical Energy.
Answer:
The force applied on one wheel during braking = 6.8 lb
Explanation:
Area of the piston (A) = 0.4 
Force applied on the piston(F) = 6.4 lb
Pressure on the piston (P) = 
⇒ P = 
⇒ P = 16 
This is the pressure inside the cylinder.
Let force applied on the brake pad = 
Area of the brake pad (
)= 1.7
Thus the pressure on the brake pad (
) = 
When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.
⇒ P =
⇒ 16 =
⇒ = 16 ×
Put the value of we get
⇒ = 16 × 1.7
⇒ = 27.2 lb
This the total force applied during braking.
The force applied on one wheel = 
= 
= 6.8 lb
⇒ The force applied on one wheel during braking.
Answer:
λ = 0.4 x 10⁻⁶ m = 400 nm
Explanation:
The relationship between frequency, wavelength and speed of an electromagnetic wave is given as follows:
where,
c = speed of light = 3 x 10⁸ m/s
f = frequency of the light wave = 7.5 x 10¹⁴ Hz
λ = wavelength of the light = ?
Therefore,
<u>λ = 0.4 x 10⁻⁶ m = 400 nm</u>
Answer:
2,500 Joules (J) or Newton Meter (N M)
Explanation:
Work = Force x Distance
The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.
Work = 5m x 500n
Work = 2,500 Joules or Newton-Meters.
Therefore 2,500 Joules or Newton Meters of work is done on an object.
