Answer:
-120000 W
Explanation:
Power = change in energy / time
P = ΔE / t
P = (½ mv₂² − ½ mv₁²) / t
P = m (v₂² − v₁²) / (2t)
Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:
P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)
P = -120000 W
Answer:
The answer is
<h2>4 hrs</h2>
Explanation:
To find the time taken we use the formula
where
d is the distance covered
v is the velocity
t is the time
From the question
d = 1200 miles
v = 300 mi/hr
We have
We have the final answer as
<h3>4 hrs</h3>
Hope this helps you
Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
...(I)
Using formula of friction force
....(II)
Put the value of F in the equation (II) from equation (I)
....(III)
Put the value in the equation (III)
We need to calculate the distance,
Using equation of motion
Hence, The distance is 11 m.
Answer:
1. it helps to change the direction.
2. it helps us to walk on ground.
3. it helps the vechils to break while moving.
4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.
5. it opposites the force.
6. it helps us to change shape of objects.eg we roll the dough to make it roti.
7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.
i all know is just 7..
Answer:
1) 3.07kgm/s
2) 5.56kgm/s
3) 76.16N
4) 4.33kgm/s
5) 0.57s
6) -8.66J
Explanation:
Given
m = 0.221kg
v = 13.9m/s
θ = 25°
t = 0.073s
1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,
P(i) = mv(i)
P(i) = 0.221 * 13.9
P(i) = 3.07kgm/s
2) Magnitude of the change in momentum of the ball,
P(i,x) = P(i) cos θ
P(i,x) = 3.07 * cos25
P(i,x) = 3.07 * 0.9063
P(i,x) = 2.78
ΔP = 2P(i,x)
ΔP = 2 * 2.78 = 5.56kgm/s
3) magnitude of the average force exerted by the wall,
F(ave) = ΔP/Δt
F(ave) = 5.56/0.073
F(ave) = 76.16N
4) ΔP(z) = mv(f) - mv(i)
ΔP(z) = 0.221*-7.8 - 0.221*11.8
ΔP(z) = -1.72 - 2.61
ΔP(z) = 4.33kgm/s
5) F(ave) = ΔP/Δt
Δt = ΔP/F(ave)
Δt = 4.33 / 76.16
Δt = 0.57s
6) KE(i) = 0.5mv(i)²
KE(f) = 0.5mv(f)²
ΔKE = 0.5m[v(f)² - v(i)²]
ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]
ΔKE = 0.1105 ( 60.84 - 139.24 )
ΔKE = 0.1105 * -78.4
ΔKE = -8.66J
