Answer:
c. an increase in the length of the rope.
Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
Answer:
Both Technician A and Technician B
Explanation:
In order to gain a better understanding of the solution above let define some terms
Break Accumulator
We can define a break accumulator as storage that that helps generate the required pressure in order for the breaking system to respond faster this accumulator is charged by turning the steering wheel slowly at once from lock to lock now this build the pressure in the accumulator and one way to depressurize is it is by turning the ignition switch ""off""
Now a scan tool is a device that can interface with a car it can also be used to diagnose a car an get the diagnostic information to help in the cars diagnoses and also be used to reprogram a car
Answer:
it's the distance between objects in space
Explanation: Light travels super fast; but it still takes a long time to travel between objects in space. This is because distances between objects in space are enormous.
And can i please receive a brainliest and have a good day
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash
