7500J
Explanation:
Given parameters:
Weight of child = 150N
Height of lifting = 50m
Distance = 50m
Unknown:
Work done by the woman = ?
Solution:
Work done by a body is the force applied to move a body through a given distance.
Work done = Force x distance
Weight is a force impacted by the mass of a body and the gravity on it.
input the parameters:
Work done = 150 x 50 = 7500J
learn more:
Work done brainly.com/question/9100769
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Answer:
θ = 29.38°
Explanation:
The centripetal force is given by the formula;
F_c = F_n(sin θ) = mv²/r
Now, the vertical component of the normal force is; F_n(cos θ)
Now, this vertical component is also expressed as; F_n(cos θ) = mg
Thus, the slope is;
F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg
tan θ = v²/rg
v² = rg(tan θ)
The initial speed will be gotten from the relation;
(v_o)² = μ_s(gr)
Plugging rg(tan θ) for (v_o)², we have;
μ_s(gr) = rg(tan θ)
rg will cancel out to give;
μ_s = (tan θ)
Thus, θ = tan^(-1) μ_s
μ_s is coefficient of static friction given as 0.563
θ = tan^(-1) 0.563
θ = 29.38°
The Metric, and the US Standard systems. :)
First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
