Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.
Then, in other words, we have 0.001 millimeters in a micrometer.
Answer:
In a micrometer there are 0.001 millimeters.
Answer:
F = 19.1 N
Explanation:
To find the force exerted by the string on the block you use the following formula:
(1)
k: spring constant = 95.5 N/m
x: displacement of the block from its equilibrium position = 0.200 m
you replace the values of k and x in the equation (1):
Hence, the force exterted on the block is 19.1 N
IF voltage remains constant, then current is
inversely proportional to resistance.
The correct response is "b).", signifying "false" as the choice.
