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11Alexandr11 [23.1K]
3 years ago
5

John performs an experiment on an electric circuit. He increases the voltage from 25 volts to 50 volts while keeping the resista

nce constant. What will be the effect of John’s changes on the current?
The current will increase.

The current will decrease.

The current will stay constant.

The current will stop flowing.
Physics
2 answers:
RSB [31]3 years ago
7 0
According to the Ohm's law, We know, potential difference(v) is directly proportional to electric current (I):

So, as voltage is increased by double of it's initial value, Current will also increase to it's double amount.

In short, Your Correct Answer would be option A) The Current Will Increase

Hope this helps!
Anna71 [15]3 years ago
3 0

Answer:

The current will increase.

Explanation:

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Bonnie and Clyde are trying to steal the world's largest diamond from a 10 story
german

Answer:

The speed Clyde will be falling at is 33.72.

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3 years ago
A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of
Vikentia [17]

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)

mv^2 = mv_t^2 + 2mgR(1 + cos \theta)

we know that,

N - mgcos \theta = \dfrac{mv^2}{R}

N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}

N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)

N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)

N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)

N = 26.59 N

3 0
3 years ago
Someone got this paper?
Snezhnost [94]
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.

Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms

Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A

Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.

Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A

Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
6 0
3 years ago
An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (
nataly862011 [7]

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

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3 years ago
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Montano1993 [528]

Answer:

correct!

Explanation:

4 0
3 years ago
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