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storchak [24]
3 years ago
7

Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is

a unit of energy. c. A watt (W) is a unit of energy. d. A kilowatt x hour per year (kWh/yr) is a unit of energy. e. A kilowatt (kW) is a unit of power.
Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

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Answer:

the atom cannot be divided into smaller particles

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3 years ago
Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s
Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

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7 0
3 years ago
A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

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In this graph, the displacement of the particle in the last two seconds is 2 meters.
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