Answer:
The magnitude of acceleration is reduced.
Explanation:
Force is defined as push or pull
The force is said to be<em> balance force </em>if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.
The force are to be<em> unbalanced force </em>if the force cause change in motion. ie the object has force greater than zero and has acceleration.
According to <em>Newton second law of motion </em>, acceleration depends on force acting on the object and mass of object.
F=ma
a=
When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
Answer:
<h2>400 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 100 × 4
We have the final answer as
<h3>400 J</h3>
Hope this helps you
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
<h3>Average power produced by the rocket</h3>
The average power produced by the rocket is calculated as follows;
P = FV
where;
- P is the average power
- F is the force exerted by the rocket
- V is the velocity of the rocket
Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
Learn more about average power here: brainly.com/question/19415290
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