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madreJ [45]
3 years ago
10

suppose you were pushing on a heavy bucket of stones it 100 N of force, and it did not move. How many newtons of force would be

balancing the force you were applying?​
Physics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

100N

Explanation:

the limiting frictional force acts in the opposite direction and is equal to the force applied

frictional force=100N

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what type of energy appears when a gymnast jumps on to a spring board? Question is on energy stores!Thanks
Brrunno [24]
Hello There!

It is Spring potential energy. Also called Elastic potential energy.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
5 0
3 years ago
the shock absorbers in a car act as a big spring with k= 21900 N/m. when a 92.5 kg person gets in, how far does the spring stret
r-ruslan [8.4K]

Answer: 0.04139m

Explanation:

First, we need to calculate the weight of the man which will be:

Weight = mass × acceleration due to gravity

Weight = mg

Weight = 92.5 × 9.8

Weight = 906.5N

Then, we calculate the force which will be:

F = kx

mg = kx

x = mg/k

x = 906.5/21900

x = 0.04139m.

The spring stretched for 0.04139m.

4 0
2 years ago
a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?
Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is 3.13 m/s^2

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity 15 m/s. It then accelerates for 4.25 s and in the end its velocity is 28.3 m/s.

initial velocity u = 15 m/s

time t=4.25 s

final velocity v=28.3 m/s

The equation of acceleration is

a= \frac{(v-u)}{t}

= \frac{(28.3-15)}{4.25} =  \frac {13.3}{4.25} =3.13m/s^2

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

7 0
2 years ago
A rotating lens mechanism blocks the view of the object as the new lenses are rotated. Be sure to click each lense firmly into p
Jobisdone [24]

I think that mechanism is called a <em>"lens turret"</em>.

6 0
2 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
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