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BARSIC [14]
3 years ago
12

the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​

Physics
1 answer:
34kurt3 years ago
6 0

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

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a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball
MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

   Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .


8 0
3 years ago
Read 2 more answers
Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv
lys-0071 [83]

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)

(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0

Hope this Helps!!!

3 0
3 years ago
1.Calculate the energy transferred by a 12V hairdryer, running on a current of 0.50A, that is left on for 8.0 minutes.
CaHeK987 [17]

Answer:

1. Energy = 2880 Joules.

2. Energy = 60 Joules.

3. Quantity of charge = 120 Coulombs.

Explanation:

Given the following data;

1. Voltage = 12 Volts

Current = 0.5 Amps

Time, t = 8 mins to seconds = 8 * 60 = 480 seconds

To find the energy;

Power = current * voltage

Power = 12 * 0.5

Power = 6 Watts

Next, we find the energy transferred;

Energy = power * time

Energy = 6 * 480

Energy = 2880 Joules

2. Charge, Q = 4 coulombs

Potential difference, p.d = 15V

To find the total energy transferred;

Energy = Q * p.d

Energy = 4 * 15

Energy = 60 Joules

3. Voltage = 6 Volts

Current = 1 Amps

Time = 2 minutes to seconds = 2 * 60 = 120 seconds

To find the quantity of charge;

Quantity of charge = current * time

Quantity of charge = 1 * 120

Quantity of charge = 120 Coulombs

8 0
3 years ago
Object a travels in the +x-direction before hitting a stationary object
Leto [7]
The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:
 
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
 
Now, we know, tan</span>Ф = \frac{y}{x}

⇒tanФ = \frac{6.2 × 10^4 }{5.7 × 10^4}

⇒tanФ = 1.088

⇒ Ф = tan^{-1} 1.088 
         =  47.4 ≈ 47

8 0
3 years ago
Read 2 more answers
The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

7 0
3 years ago
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