Answer:
On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket
How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period
When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned he uses more force and the ball also spends less time on the racket to produce the same momentum
Explanation:
The impulse of a force, ΔP is given by the following formula;
ΔP = F × Δt
Where ΔP is constant, we have;
F ∝ 1/Δt
Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.
Answer:
Projected area= Diameter of the bolt* thickness.
Explanation:
Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _Diameter of the bolt______ and the plate ___thickness____.
Projected area= Diameter of the bolt* thickness.
Projected area is a 2-dimensional area measurement of a 3-dimensional body by projecting its surface on an arbitrary plane
The answer to this question would be B (the battery is the electrical power supply)
Answer:
Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320.
Explanation:
The universal law of gravitation states that the force between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
We have to choose the satellite having greatest gravitational force with earth. In all options the distance from the earth is same i.e. 320 km. So, we have to select the satellite having maximum mass because the mass of the earth is constant.
Hence, the correct option is (D) " Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320 ".
<span>We can answer this using
the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²
-----> 1</span></span>
ω² = ω₀² + 2αθ
-----> 2
Where:
θ = final angular
displacement = 70.4 rad
θ₀ = initial
angular displacement = 0
ω₀ = initial angular
speed
ω = final angular speed
t = time = 3.80 s
α = angular acceleration
= -5.20 rad/s^2
Substituting the values
into equation 1:<span>
70.4 = 0 + ω₀(3.80)
+ ½(-5.20)(3.80)² </span><span>
ω₀ = (70.4
+ 37.544) / 3.80 </span><span>
ω₀ = 28.406
rad/s </span><span>
Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4
ω = 8.65 rad/s
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