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tester [92]
3 years ago
14

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Physics
1 answer:
Lena [83]3 years ago
5 0

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

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6) A penny is dropped from the roof of a house. The penny is traveling at 35 m/s right before it hits
nadezda [96]
Wait what is m/s?
Can you tell me that so i could help
8 0
3 years ago
The equation below is for potassium oxide.
ValentinkaMS [17]

Answer: The ratio of atoms of potassium to ratio of atoms of oxygen is 4:2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed, and remains conserved. The mass of products must be same as that of the reactants.

Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.

K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.

5 0
3 years ago
Read 2 more answers
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Masja [62]
Question 18: a
question 19: b
question 20: c
6 0
3 years ago
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in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th
Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

R=\dfrac{u^2\sin2\theta}{g}

Plugging all the values we get :

R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m

So, the maximum range of the ball is 73.25 m

8 0
3 years ago
satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close
GrogVix [38]

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

8 0
3 years ago
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