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2 years ago

When two 10-Ohm resistors are in parallel in a circuit, what is the net resistor value

Physics

1 answer:

Elza [17]2 years ago

6 0

Answer:

Net capacitance

let net capacitance be R

1/R = 1/10+1/10

1/R=2/10

1/R=1/5

cross mutiply

R=5

the net capacitance is 5ohms

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A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha

Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

$T = \frac{1}{f}$

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

$T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s$

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:

$A = \frac{20 cm}{2} = 10 cm$

Therefore, the amplitude is 10 cm.

I hope it helps you!

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2 years ago

A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has

Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

8 0

2 years ago

if a car is moving on a highway at 70 kilometers per hour going east, what happens to its speed and velocity

Andrej [43]

The speed is constant since there is no opposite force facing the car

That’s my best guest hope it helps !! :)

7 0

2 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

2 years ago

A baseball player throws a baseball at a speed of 40 meters per second at an angle of 30 degrees. What is the velocity of projec

Vesna [10]

Answer:

The horizontal component of the velocity is the cosine of 30 degrees multiplied by 40m/s. The cosine of 30 degrees is the 0.8660 . To get the speed, multiply by 40m/s. This equals 34.64, which is approximately 35m/s.

Hope it helpss :)

3 0

1 year ago

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