The equation to be used is for the rectilinear motion at constant acceleration:
x = v₀t + 0.5at²
a = (v-v₀)/t
where
x is distance
v and v₀ is the final and initial velocity
t is time
a is acceleration
Because the acceleration is decelerating, that would be reported as -7.5 m/s². Substituting,
-7.5 = (0 - v₀)/t
v₀ = 7.5 t --> eqn 1
x = v₀t + 0.5at²
60 = (7.5t)(t) + 0.5(-7.5)(t²)
Solving for t,
t = 4s
Thus,
v₀ = 7.5 m/s² * 4s
v₀ = 30 m/s
Answer:
the correct answer is c) 23 g
Explanation:
The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body
Q_lost = - Q_absorbed
The latent heat is
Q_absorbed = m L
The heat given by the body
Q_lost = M 
ΔT
where m is the mass of sweat and M is the mass of the body
m L = M c_{e} ΔT
m = M c_{e} ΔT / L
let's replace
m = 90 3.500 1.8 / 2.42 10⁶
m = 0.2343 kg
reduced to grams
m = 0.2342 kg (1000g / 1kg)
m = 23.42 g
the correct answer is c) 23 g
Answer:
3 * 10³J/s
Explanation:
Given :
Force applied, F = 300 N
Distance, d = 30 m
Time, t = 3 seconds
Power, P = Workdone / time
Recall :
Workdone = Force * distance
Workdone = 300 N * 30 m = 9000 Nm
Workdone = 9 * 10³ J
Power = (9 * 10³ J) / 3s
Power = 3 * 10³J/s
Answer:
E=252J
Explanation:
The total mechanical energy of an object or system is given by:
E mech=K+U
Where K is the kinetic energy of the object and U is the potential energy of the object. The carriage, sitting motionless at the top of the hill, has only potential energy in the form of gravitational potential energy.
Gravitational potential energy is given by:
Ug=mgh
Where m is the mass of the object, g is the gravitational acceleration constant, and h is the height of the object above some specific reference point, in this case the ground 21 m below.
The weight of a stationary object at the surface of the earth is equal to the force of gravity acting on the object.
W=→Fg=mg
We are given that the carriage weighs 12 N, therefore mg=12N.
Ug=12N⋅21m
⇒Ug=252Nm=252J
Hope it helped, God bless you!
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
