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3 years ago

What is the equivalent resistance of a circuit that contains two 50.0 resistors connected in parallel with a 12.0 V battery?

Physics

2 answers:

Pachacha [2.7K]3 years ago

8 0

1/Re=1/R1 +1/R2= 2x1/R1=2/R1=2/R2=2/50=1/25
1/Re=1/25, so the answer is Re= 25

the answer is B. 25.0

sineoko [7]3 years ago

8 0

Answer:

the other person is right, the answer is 25.0

Explanation:

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Which Quantity is a vector Quantity? A] Acceleration B] Mass C] Speed D] Volume

vlabodo [156]

Well I am positive that the answer is B

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3 years ago

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Suppose a ceiling fan has a mass of 7.5 kg and is 9.7 m above the ground. What is the gravitational potential energy of the ceil

EleoNora [17]

The gravitational potential energy (G.P.E) of the ceiling fan is 712.95 Joules.

Given the following data:

Mass of ceiling fan = 7.5 kg

Height = 0.7 m

Scientific data:

Acceleration due to gravity = 9.8 $m/s^2$

To calculate the gravitational potential energy (G.P.E) of the ceiling fan:

<h3>What is gravitational potential energy?</h3>

Gravitational potential energy (G.P.E) can be defined as the energy that is possessed by an object or body due to its position (height) above planet Earth.

Mathematically, gravitational potential energy (G.P.E) is given by this formula;

$GPE = mgh$

Where:

G.P.E is the gravitational potential energy.

m is the mass of an object.

g is the acceleration due to gravity.

h is the height of an object.

Substituting the given parameters into the formula, we have;

$GPE = 7.5 \times 9.8 \times 9.7$

GPE = 712.95 Joules.

Read more on potential energy here: brainly.com/question/8664733

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2 years ago

As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.39-m-

grin007 [14]

Answer:

The strength of magnetic field is $6.21 \times 10^{-5}$ T

Explanation:

Given:

Length of rod $l = 1.39$ m

Velocity $v = 3.37$ $\frac{m}{s}$

Induced emf $= 0.291 \times 10^{-3}$ V

According to the faraday's law

Induced emf = $Blv$

We have to find strength of the magnetic field,

$B = \frac{0.291 \times 10^{-3} }{1.39 \times3.37}$

$B = 6.21 \times 10^{-5}$ T

Therefore, the strength of magnetic field is $6.21 \times 10^{-5}$ T

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3 years ago

What beat frequencies are possible with tuning forks of frequencies 255, 258, and 260 hz ?

zavuch27 [327]

Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.

The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.

A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.

Learn more about beat frequency here : brainly.com/question/14157895

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2 years ago

A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling

ivann1987 [24]

Answer:

42.99°

Explanation:

$F_h$ = Kinetic friction force

$F_{\theta}$ = Pulling force at angle $\theta$

$N_h$ = Weight of the box = 150 N

Kinetic friction force

$F_h=\muN_h$

Pulling force at angle $\theta$

$F_{\theta}=\muN_{\theta}$

N = Pulling force

According to question

$\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N$

Applying Newton's second law in the vertical direction we get

$N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}$

The angle is 42.99°

8 0

3 years ago