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Alborosie
3 years ago
13

Please help.Just True or False​

Physics
1 answer:
True [87]3 years ago
6 0

Answer:

1. T

2. F

3.F

4.T

thank you

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Can anyone help me on question 13
Leona [35]

Given that in a parallel circuit:

R1 = 12 ohms

R2= 15 ohms

I = 12 A

I2 = 4 A

V=?

R=?

R3 =?

P=?

Since,

V= IR

or,

V2 = I2 * R2

V2= 4* 15

V2 = 60V

Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.

Therefore,

V= V1 = V2 = V3 = 60V

Since,

V= IR

R= V/I

R= 60/12

R= 5 ohm

That is total resistance is equal to 5 ohms.

Since for parallel circuit,

1/R= 1/R1 + 1/R2 + 1/R3

1/5= 1/12+ 1/15 + 1/R3

or

1/R3= 1/5- 1/12- 1/15

1/R3= 1/20

or

R3= 20 ohms

Since,

V=IR

I= V/R

I1= V1/ R1

I1= 60/12

I1= 5 A

I3= V3/R3

I3= 60/20

I3= 3A

Since,

P=VI

P= 60*12

P=  720 watt

P1= V1* I1

P1= 60* 5

P1= 300 watt

P2= V2* I2

P2= 60* 4

P2= 240watt

P3= V3*I3

P3= 60*3

P3= 180 watt

Hence we have,

R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms

I1= 5A, I2= 4A, I3= 3A, I= 12 A

V1= V2= V3= V= 60V

P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt



6 0
3 years ago
Fill in the blanks! Please, help me ASAP ✌
Goryan [66]

Answer:

I don't know

Explanation:

i don't know this question answer for here

3 0
3 years ago
A projectile is fired vertically upward from the surface of planet Moorun of mass 2 Ã 10^24 kg and radius 7 Ã 10^6 m. If this pr
topjm [15]

To solve this problem we will apply the principle of energy conservation. Here we have that the gravitational potential energy must be equal to the kinetic energy of the body. So,

PE = KE

\frac{GMm}{R} = \frac{1}{2} mv^2

Here,

m = mass of projectile

G = Gravitational Universal constant

M = Mass of the planet

R = Total height from center of mass of the planet

v = Velocity

Rearraning to find the velocity we have,

\frac{GM}{R} = \frac{1}{2} v^2

v = \sqrt{2\frac{GM}{R}}

Our values are given as,

M = 2*10^{24} kg

r = 7*10^6 m

h = 6*10^6 m

R = h+r = 13*10^6m

G = 6.67259*10^{-11} N\cdot m^2/kg^2

Replacing we have,

v = \sqrt{2\frac{(6.67259*10^{-11})(2*10^{24})}{13*10^6}}

v = 4531.12m/s

Therefore the initial speed of the projectile must be 4531.12m/s

4 0
3 years ago
Which statement describes why energy is released in a nuclear fusion reaction based on mass-energy equivalence? For large nuclei
Delicious77 [7]

Answer: Option (c) is the correct answer.

Explanation:

When two or more small nuclei combine together to form a larger nuclei then this process is known as nuclear reaction.

The smaller is an atom, the more energy it requires to release an electron. This energy is known as binding energy.

Thus, when two small nuclei fuse together then there will be more binding energy as compared to when two large nuclei fuse together.

For example, fusion of two hydrogen atoms release more energy then one helium atom, and upon binding excess energy is released into the space.

Hence, we can conclude that energy is released in a nuclear fusion reaction based on mass-energy equivalence because for small nuclei, the binding energy of the lighter nuclei is greater than the binding energy of the heavier nucleus.




4 0
3 years ago
Read 2 more answers
If fnet equals fg, what is the magnitude and direction of the acceleration of the object? Assume that the object is on Earth's s
suter [353]
You're describing free-fall. The acceleration is 9.8 meters per second-squared downward (toward the center of the Earth).
3 0
3 years ago
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