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meriva
2 years ago
6

A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

Physics
2 answers:
melisa1 [442]2 years ago
8 0

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m

Hence the length by which the spring got stretched will be 0.08 m.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ4

leva [86]2 years ago
3 0

8n/m this is because 200×2500 all over 100 . 0's will cancile 0's living 200×2500×1 ,25 will go into 200 8 times ×1 =8n/m

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A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re
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A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law. 

I= \frac{V}{R}

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Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
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However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}


\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059
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Get the reciprocal of both sides and divide:

R_{T} =  \frac{1}{0.0613} =16.32

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
V = IR
V = (1.8)(16.32)
V = 29.376V

The emf of the battery is 29.376V


B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
I= \frac{V}{R}
I= \frac{29.376}{65}
I= 0.45A

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
I= \frac{V}{R}
I= \frac{29.376}{25}
I= 1.18A
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
I= \frac{V}{R}
I= \frac{29.376}{170}
I= 0.17A

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.
4 0
3 years ago
On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
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Answer:

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Answer:

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v = 7.24 m /s .

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