Idk what is growing but if it’s a free than c
The acceleration due to gravity (g) on this planet is 39.44 m/s²
<h3>What is solar system?</h3>
Solar system consists of all the planets and the most importantly the center of the solar system is Sun.
Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.
The time period of the pendulum is
T = 2π √l/g
Squaring both sides, we get
l/g = T² / 4π²
g = 4π²l/ T²
Substitute Time period T = 2s and length l = 4m, we get
g = 4π²x 4/ 2²
g =39.44 m/s²
Thus, the acceleration due to gravity on this planet is 39.44 m/s²
Learn more about solar system.
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Because the people in the car are attached to the vehicle, the people inside the vehicle are going the same speed as the vehicle.
Hope this helps! :)
Answer:
puck decelerates due to the kinetic frictional force μk mg
Explanation:
given data
total distance = 12 m
coefficient of kinetic friction = 0.28
solution
we will apply equation of motion that is
v² - u² = 2 × a × s ................1
we know acceleration will be
a = 
Then we have
Force = mass × acceleration .................2
m × = -μk mg
The puck decelerates due to the kinetic frictional force μk mg
and frictional force is negative as it opposes the motion.
so we get initial velocity of the puck which is strike.
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
