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quester [9]
4 years ago
8

the current flowing in an appliance connected to a 120V source is 2A. How many kilowatt-hours of electrical energy does the appl

iance use in 4h
Physics
1 answer:
skad [1K]4 years ago
8 0

Answer:

0.96 kWh

Explanation:

First of all, we can find the power of the appliance using the formula:

P=VI

where

V = 120 V is the voltage of the source

I = 2 A is the current

Substituting,

P(120 V)(2 A)=240 W = 0.24 kW

Now we want to find the amount of energy used by the applied in a time of

t = 4 h

The amount of energy is given by

E=Pt

where

P = 0.24 kW

t = 4 g

Substituting,

E=(0.24 kW)(4 h)=0.96 kWh

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A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between
eduard

Answer:

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ =+7.5 nC = +7.5*10⁻⁹C  

q₂ =  -2.0 nC = -2.0*10⁻⁹C

d₁ =d₂ = 1.5cm = 1.5 *10⁻²m  = 0.015 m

Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

8 0
4 years ago
HELP ASAP I WILL GIVE BRAINLIEST!!!!!!!!!!!
Mice21 [21]
Ok the answer for this question would be B. Hope this helps.

5 0
3 years ago
5. A student travels 11 m north and then turns around to travel 25 m south. If the total time of travel is 12 sec., find a) The
11111nata11111 [884]

Explanation:

Given that,

A student travels 11 m north and then turns around to travel 25 m south.

Total time, t = 12 s

The total distance or the total path covered by the student is equal to 11 m + 25 m = 36 m

Displacement of the student or the shortest path covered is d = 25-11 = 14 m

(a) The student's average speed = total distance/total time

s=\dfrac{36\ m}{12\ s}\\\\s=3\ m/s

(b) The student's average velocity = total displacement/total time

v=\dfrac{14\ m}{12\ s}\\\\v=1.17\ m/s

4 0
3 years ago
If a hockey player starts from rest and accelerates at a rate of 2.1 m/s2 how long does it take him to skate 30m
OLga [1]
Hello!

<span>Let us apply the time function of space, in the uniform uniform motion (UUM)
</span>
Formula: S = S_{o} +  V_{o}* t + a* \frac{t^2}{2}

Data:
S (Final position) = 30 m
So (Initial Position) = 0 m
Vo (Initial velocity) = 0 m/s
t (time) = ? (in seconds)
a (acceleration) = 2.1 m/s²

Solving:
S = S_{o} + V_{o}* t + a* \frac{t^2}{2}
30 = 0 + 0*t + 2.1* \frac{t^2}{2}
30 =  \frac{2.1t^2}{2}
30*2 = 2.1t^2
60 = 2.1t^2
2.1t^2 = 60
t^2 =  \frac{60}{2.1}
t^2 \approx 28.5
t \approx  \sqrt{28.5}
\boxed{\boxed{t \approx 5.3\:s}}\end{array}}\qquad\quad\checkmark

Answer:
<span>C. 5.3s</span>
5 0
4 years ago
Solar cells convert what type of energy into electrical energy?
zysi [14]
I think the correct answer from the choices listed above is option D. Solar cells convert electromagnetic energy into electrical energy. This is because the photons from the rays of light are electromagnetic waves or particles. Hope this answers the question.
6 0
4 years ago
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