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liberstina [14]
3 years ago
6

Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)

Physics
1 answer:
rewona [7]3 years ago
3 0

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

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A 0.71 W point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a
sveticcg [70]

Answer:

(a) 2.88×10⁻² W/m²

(b) 8.36×10⁻³ W/m²

Explanation:

The intensity of sound from an isotropic point source, with distance L is given as

I = P/(4πL²) .................................... Equation 1

Where I = intensity of sound, P = Power from the source, L = length, π = pie.

(a)

1.4 m from the source.

I = P/(4πL²)

Given: P = 0.71 W, L = 1.4 m, π = 3.14.

Substitute into equation 1

I = 0.71/(4×3.14×1.4²)

I = 0.71/24.6176

I = 0.0288 W/m².

I = 2.88×10⁻² W/m²

(b) 2.6 m from the source.

Given: P = 0.71 W, L = 2.6 m, π = 3.14

Substitute into equation 1

I = 0.71/(4×3.14×2.6²)

I = 0.71/84.9056

I = 0.00836 W/m²

I = 8.36×10⁻³ W/m²

5 0
4 years ago
What is the result of seafloor spreading?
Over [174]

Answer:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

6 0
3 years ago
Two electrons are separated by 1.70 nm. What is the magnitude of the electric force each electron exerts on the other?
lidiya [134]

Answer:

F=7.96*10^{-11}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two equals charges (q) is given by:

F=\frac{kq^2}{d^2}

Here k is the coulomb constant and d is the distance between the charges. For two electrons we have:

F=\frac{ke^2}{d^2}\\F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)^2}{(1.7*10^{-9}m)^2}\\F=7.96*10^{-11}N

8 0
3 years ago
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
Vika [28.1K]

Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

P_{avg}=1000\times9.8\times100

P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

Hence, The hydro static force on the back of the dam is 1.96\times10^{11}\ N

7 0
3 years ago
What is the x component of (+3m)ι^ And (+3m/s) i^?
expeople1 [14]
The "i" component of a vector is in the x-direction. Therefore, the x-component is 3 m/s.
4 0
3 years ago
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