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ANEK [815]
3 years ago
13

How much does it cost to leave a porch light on all night?

Physics
1 answer:
worty [1.4K]3 years ago
6 0
Cost = (0.001) x (the wattage of the light) x (the number of hours it's left on) x (the cost of each kilowatt-hour of electrical energy where you live).
You might be interested in
An elevator weighs 1500 Newtons. Calculate how much potential energy it has when it is lived 500 meters in the air.
avanturin [10]

An elevator has 750,000 J when it is lived 500 meters in the air

Answer: Option D

<u>Explanation:</u>

The expression to find the potential energy completely on the basis of forces that on two objects. It can be written as

potential\ energy (P . E)=m \times acceleration\ due\ to\ gravity\ (g) \times height (h)

Where,

m – mass of object

Acceleration due to gravity =  9.8 \mathrm{m} / \mathrm{s}^{2}

h = 500 meters

Force = 1500 Newton

As we know, according to Newton second law of motion, the force can be expressed in terms of mass and gravity (F= ma). So, the potential energy can be calculated as

P . E=F \times h=1500(500)=750000 J

Hence, an elevator has 750,000 J when it is lived 500 meters in the air.

4 0
3 years ago
Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0
3 years ago
Hurricane or typhoons occur when large areas of the open ocean soak up heat from the sun t or f​
11Alexandr11 [23.1K]

Answer: true

Explanation:

3 0
3 years ago
How long will it take a plane to fly 1256km<br> if it travels 500km/hr?
expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

d is the distance

v is the velocity

From the question we have

t =  \frac{1256}{500}  \\  = 2.512

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0
3 years ago
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
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