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Aleks [24]
3 years ago
7

an object of mass 1kg was at rest before it fell through a height o 5m and hit the ground. Calculate the initial gravitational p

otential energy of the object
Physics
2 answers:
chubhunter [2.5K]3 years ago
7 0
\bold{Hello  \:\:  Friend}


Given :-

=> mass = 1 kg
=> height = 5 m


Potential energy = m*g*h

→ 1*10*5



Potential energy = 50 J




Hope It helps you. ^_^
viva [34]3 years ago
3 0

Hello!

Data:

h = 5 m

m = 1 Kg

PE = ? (Joule)

Adopting, gravity (g) ≈ 10 m/s²

Formula:

PE_{grav}  = mass * g * height

Solving:

PE_{grav} = mass * g * height

PE_{grav} = 1*10*5

\boxed{\boxed{PE_{grav} = 50\:J}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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Show your working please ​
saw5 [17]

Explanation:

There's not enough information in the problem to solve it.  We need to know either the initial speed of the lorry, or the time it takes to stop.

For example, if we assume the initial speed of the lorry is 25 m/s, then we can find the rate of deceleration:

v² = v₀² + 2aΔx

(0 m/s)² = (25 m/s)² + 2a (50 m)

a = -6.25 m/s²

We can then use Newton's second law to find the force:

F = ma

F = (7520 kg) (-6.25 m/s²)

F = -47000 N

3 0
3 years ago
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

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

8 0
3 years ago
Which of these is an example of physical change?
My name is Ann [436]
That will be A melting a substance that's because the rest are examples of chemical changes and also its the same thing but in a different form.
8 0
4 years ago
Read 2 more answers
I need help!<br> Pleeeeeeeeaaaaaaaaaaaaasssssssssssseeeeeeeeeeee?
serious [3.7K]

Answer:

np 500000 eeeeeeeeeeee

Explanation:

346 763 999 eeeeeeeeeeeee

4 0
2 years ago
4. The greater the speed of gas particles in a container, the:
GREYUIT [131]

Answer:

The answer to your question will be d) greater the pressure

Explanation:

5 0
3 years ago
Read 2 more answers
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