Which property is expressed in distance units?
1 answer:
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Given:
Mass (m) = 1200 kg
Distance (s) = 100 m
Time (t) = 10 seconds
Now,
=
= 10 m/s
<span><u>Note that this one is the final velocity.</u></span><u />
We also know that,
initial velocity (u) = 0 m/s .......<span> because the car starts from rest.
</span>Now,
=
=
= 1 m/s²
Now,
Force (F) = mass (m) * acceleration (a)
= 1200 kg * 1 m/s²
= 1200 kg.m/s²
= 1200 N
Now,
Work Done (W) = Force (F) * displacement (s) ....<span>note that displacement is same as distance.
</span><span> = 1200 N * 100 m
</span> = 120000 N.m
= 120000 J
Now,
Power (P) =
=
= 12000 J/s
= 12000 watt
SO,
A) The acceleration of the car is 1 m/s².
B) 1200 Newton (N) force must have acted on the car.
C) The velocity of the car after 10 seconds is 10 m/s.
D) 120000 Joule (J) work was done on the car.
E) The engine produced a minimum power of 12000 watt.
Mass (m) = 1200 kg
Distance (s) = 100 m
Time (t) = 10 seconds
Now,
=
= 10 m/s
<span><u>Note that this one is the final velocity.</u></span><u />
We also know that,
initial velocity (u) = 0 m/s .......<span> because the car starts from rest.
</span>Now,
=
=
= 1 m/s²
Now,
Force (F) = mass (m) * acceleration (a)
= 1200 kg * 1 m/s²
= 1200 kg.m/s²
= 1200 N
Now,
Work Done (W) = Force (F) * displacement (s) ....<span>note that displacement is same as distance.
</span><span> = 1200 N * 100 m
</span> = 120000 N.m
= 120000 J
Now,
Power (P) =
=
= 12000 J/s
= 12000 watt
SO,
A) The acceleration of the car is 1 m/s².
B) 1200 Newton (N) force must have acted on the car.
C) The velocity of the car after 10 seconds is 10 m/s.
D) 120000 Joule (J) work was done on the car.
E) The engine produced a minimum power of 12000 watt.
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º