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GuDViN [60]
2 years ago
15

A 10-N force is needed to move an object with a constant velocity of 5.0 m/s. What power must be delivered to the object by the

force
Physics
1 answer:
vagabundo [1.1K]2 years ago
5 0

The power that must be delivered to the object by the force is 50 W

Power is defined as the rate of doing work. The power of an object in relation to the force and velocity is given by the following equation:

Power (P) = Force (F) × velocity (v)

P = F × v

From the question given above, the following data were obtained:

  • Force (F) = 10 N
  • Velocity (v) = 5 m/s
  • Power (P) =?

P = F × v

P = 10 × 5

<h3>P = 50 W </h3>

Thus, the power that must be delivered to the object by the force is 50 W

Learn more on power: brainly.com/question/19539420

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vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

V=E\times d

V=3000\ N/C\times 0.7\ m

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

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3 years ago
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of 34.9^{\circ}

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

d sin \theta = m \lambda

where

d is spacing between the lines in the grating

\theta is the angle of the maximum

m is the order of diffraction

\lambda is the wavelength of the light

For laser 1,

d sin \theta = m_1 \lambda_1

For laser 2,

d sin \theta = m_2 \lambda_2

where

\lambda_1 = 420 nm\\\lambda_2 = 630 nm

Since the position of the maxima in the two cases overlaps, then the term d sin \theta on the left is the same for the two cases, therefore we can write:

m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}

Therefore:

m_1 = 3

m_2 = 2

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

d sin \theta = m_1 \lambda_1

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m

m_1 = 3 is the order of the maximum

\lambda_1 = 420 nm = 420\cdot 10^{-9} m is the wavelength of the laser light

Solving for \theta, we find the angle of the maximum:

sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572

So the angle is

\theta=sin^{-1}(0.572)=34.9^{\circ}

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

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