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worty [1.4K]
3 years ago
13

Which of the following is true for a parallel circuit?

Physics
2 answers:
siniylev [52]3 years ago
5 0
B. The voltage is the same across all resistors in the circuit.
Anarel [89]3 years ago
3 0
Your answer would be B, as in bee.
Get it? 

-Gives thumbs up- 
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Can you help me??? This is too hard.
evablogger [386]

Answer:

Vacuum. A sound vacuum was created, i believe.

4 0
3 years ago
The general formula for an acid is
Umnica [9.8K]

Answer:

To write the general formula for an acid, we fix one atom which is hydrogen because this atom is common to all the acids. General formula for acid is written by HX. where H represents Hydrogen atom.

Explanation:

4 0
2 years ago
Could someone please help me with this question?​
Elenna [48]
1 because the the mid night summer is dark
3 0
3 years ago
An object has a weight of 9 n when it is in air and 7.2 n when it is submerged into water. what is the specific gravity of the o
White raven [17]

The specific gravity of the object’s material is 5.09.

<h3>To calculate the specific gravity of the object:</h3>

Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water

Buoyant Force in water(Fb) = density of water x g x volume of the   body(Vb)

1.8 = 1000 x 9.81 x Vb

Vb = 1.8/9810 cubic meter

Now, in the air;

Weight of body = mg = 9 N

Mass of body,m = 9/9.81 Kg

So,

Density of body = m/ Vb

= 9/9.81 ÷ 1.8/9810

= 5094.44 kg per cubic meter

The specific gravity of body = density of body ÷ density of water

= 5094.44 ÷ 1000

= 5.09

Therefore, Specific gravity of body = 5.09

Learn more about Specific gravity here:

brainly.com/question/13258933

#SPJ4

6 0
1 year ago
What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 116 km/h and emitting a steady 950 Hz so
Arte-miy333 [17]

To solve this problem we will apply the concepts related to the Doppler Effect, defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be written as

f_{obs} = f(\frac{v_w}{v_w-v_s})

Here,

f_s= Frequency of the source

v_w = Speed of the sound

v_s= Speed of source

Now the velocity we have that

v_s = 116km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 32.22m/s

Then replacing our values,

f_{obs} = (950Hz) (\frac{345m/s}{345m/s-32.22m/s})

f_{obs} = 1047.86Hz

Therefore the frequency of the observer is 1047.86Hz

8 0
3 years ago
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