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Sliva [168]
2 years ago
6

A heat source could be considered a source of pollution Please select the best answer from the choices provided

Physics
1 answer:
jeyben [28]2 years ago
5 0

The answer is true, is called Thermal pollotion:).

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In diving to a depth of 248 m, an elephant seal also moves 296 m due east of his starting point. What is the magnitude of the se
Vinvika [58]

Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

R= sqrt(vector1+vector2)

Since this is a right angle triangle

R= sqrt(248^2 + 296^2)

R= sqrt(149120)

R= 386.16m

Displacement = 386.16m

4 0
3 years ago
If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement
alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

v(t)=3t^2-6t+24

In order to find the expression for the position of the car, we integrate this expression. We find:

x(t)=\int v(t) dt=t^3-3t^2+24t+C

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

x(0)=0^3-0+0+C=C

While the position after t = 3 s is

x(3)=3^3-3(3)^2+24(3)+C=72+C

Therefore, the displacement of the car in 3 seconds is

d=x(3)-x(0)=72+C-C=72

7 0
3 years ago
What is the formula for conservation of momentum
olga55 [171]

Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

Explanation:

6 0
3 years ago
What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

7 0
3 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
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