<h2>Let us find the efficiency : Ans = 0.6</h2>
Explanation:
we know :
efficiency = output/input
We also know that :
output = m x g x h
where :
m = mass of body
g = acceleration due to gravity
h = height of body from floor
Thus, output = 0.6 x 10 x 1.2 = 7.2J
Similarly ,input = 0.6 x 10 x 2 = 12J
Thus efficiency = 7.2/12 = 0.6
Answer:A
Explanation:
Explanation:
Given that,
Gravitational force = 600 N
Frictional force = 25 N
Pulled by the Force = 250 N
We know that,
The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.
The balance equation is along y-axis
The box will not move in y-axis therefore, the net force in the y-axis will be zero.
Hence, The net force in the y-direction will be zero.
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Answer:
Yes cause he walks 6.7 miles
Given:
F = 4.719 N, the applied force
a = 33 m/s², the acceleration
Let m = the mass.
By definition,
F = m*a
Therefore,
m = F/a
= (4.719 N)/(33 m/s²)
= 0.143 kg
Answer: 0.143 kg