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DerKrebs [107]
2 years ago
9

what is the result of mixing 15 garm of water 80 degree celsius with 10 gram of ice -10 degree Celsius ? give specific heat capc

tiy of ice 0.5 calorie per gram Celsius and letent heat of fusion of ice 80 calorie per gram.​
Physics
1 answer:
strojnjashka [21]2 years ago
5 0

Answer:

50

Explanation:

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10. Juan wants to see how air expands when it is heated. He is able to use any of the following supplies - a balloon, a heat lam
zmey [24]

Inflate the balloon, measure the width of the balloon, heat the balloon using the heat lamp, and then measure the width again.

4 0
3 years ago
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Electromagnets in radios can be used to A) play a CD. B) create sound. C) change the station. D) make a remote control.
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Change the station....
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3 years ago
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anice is watching her granddaughter drive a Barbie Jeep with a 6 V battery and an electric motor with 5 ohm of resistance. How m
Scrat [10]

Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

Power is given by P=\frac{V^2}{R}, here V is voltage and R is resistance

So P=\frac{V^2}{R}=\frac{6^2}{5}=7.2watt

5 0
3 years ago
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When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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