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PilotLPTM [1.2K]
2 years ago
9

Estimate the monthly cost of using a 700 W refrigerator that runs for 10 hours a day if the cost per kilowatt hours is cents $.2

0
Physics
1 answer:
Mrac [35]2 years ago
4 0

Answer:

$4.2

Explanation:

Given data

Power= 700W

time= 10 hours

Cost per kilowatt hours is cents $0.20

Let us find the number of hours in a month

=24*30

=720 hours

Energ= power*time

Energy= 700/1000*30

Energy= 7*3

Enery= 21 kwh

1 kwh= 0.2

21kwh= x

cross multiply

x=21*0.2

x= $4.2

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The exact location of the 'subtropical-jet stream' is located at the North phase of 30 Degrees and the reason behind this is due to the variation of air which lies on the region of mid-altitude and warmer equatorial air. the correct answer would be Cool and Warm air masses meeting near the equator.
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2 years ago
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A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam
Allushta [10]
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹ 
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
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3 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
GalinKa [24]

Answer:

R = 98304.75 m = 98.3 km

Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

Density = 1 x 10⁹ kg/m³

Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

Therefore,

1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]

4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

R = ∛[0.95 x 10¹⁵ m³]

<u>R = 98304.75 m = 98.3 km</u>

6 0
2 years ago
Match each term with the appropriate definition.
Ratling [72]

Answer:

mass- the amount of matter in an object

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scale- a tool used to measure weight

weight- the downward pull on an object due to gravity

8 0
2 years ago
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