Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
<u>Determine the maximum acceleration for these elevators</u>
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u>
</u>
<u>
</u>At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2
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
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is

where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:

where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,

So the closest answer among the option is -5020 N.
Learn more about acceleration and forces:
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