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zloy xaker [14]
2 years ago
10

A 30 ohm resistor and a 20 ohm resistor are

Physics
1 answer:
Reptile [31]2 years ago
8 0

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

  • I = V/Rt........... Equation 1

Where:

  • I = Electric current passing through the 30 ohms resistor
  • V = Voltage
  • Rt = Total or effective resistance of the resistors.

From the question,

Given:

  • V = 100 volts
  • Rt = (30+20) ohms (since both resistors are connected in series)
  • Rt = 50 ohms

Substitute these values into equation 1

  • I = 100/50
  • I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

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AVprozaik [17]
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time 
The vo=0 and vf = 25 m/s and t=7 sec. This gives... 
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N. 
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again). 
5 0
3 years ago
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If your vehicle has a steering wheel lock, should you ever turn off the ignition while it is moving?
umka2103 [35]

It is not advisable to turn off the ignition while it is moving, so it is a no. Why? Even though the vehicle has steering wheel lock, turning off the ignition while it is moving is not advisable because it causes the vehicle to lose out of control, leading to complications and accidents.

5 0
3 years ago
If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?
jeka94

Answer:

v=60m/s

Explanation:

Use the Kinematic Equation:

v=v_o+at

Plug in what is given and solve

v=0+(12)(5)

v=60m/s

3 0
2 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
3 years ago
If a ball speeds up as it is rolling down a hill, the forces acting on it must be -
VMariaS [17]

Answer:

the forces acting on it must be strong because gravity is pushing the ball down

Explanation:

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