3 years ago

How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?

Physics

1 answer:

ioda3 years ago

3 0

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

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From this diagram which of the following can you conclude

Volgvan

Answer: A

Explanation:

I say opposites attract

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2 years ago

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A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro

svlad2 [7]

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T

Angle between V and B = θ = 60

We know that,

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N

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2 years ago

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starting from rest a skateboarder accelerates at a rate of 2 [m/s 2] for 2 [s]. what is their speed afterwards?

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Answer:

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1 year ago

A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met

shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

$\epsilon=BvL$

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

$\epsilon=12 V$ (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving for B, we find the strength of the magnetic field:

$B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T$

Learn more about magnetic fields:

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3 years ago

A truck travels up a hill with a 5.4° incline. The truck has a constant speed of 25 m/s. What is the horizontal component of the

goldenfox [79]

Answer:

$v_{x}=24.88m/s$

$v_{y}=2.35m/s$

Explanation:

Knowing that the truck travels up a hill with a 5.4º incline

The horizontal component of the truck's velocity is:

$v_{x}=vcos\beta=(25m/s)cos(5.4)=24.88m/s$

The vertical component of the truck's velocity is:

$v_{y}=vsin\beta=(25m/s)sin(5.4)=2.35m/s$

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3 years ago