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vlabodo [156]
2 years ago
11

8.0 Kg of water is heater from 20°C to 40°C. The specific heat capacity of water is 4200 J/Kg °C. What is the energy supplied to

the water?
Physics
1 answer:
koban [17]2 years ago
6 0

Answer:

672000 J.

Explanation:

From the question,

q = cm(t₂-t₁)............................ Equation 1

Where q = heat energy supplied to the water, c = specific heat capacity of water, m = mass of water, t₂ = final temperature, t₁ = Initial temperature.

Given: m = 8.0 kg, c = 4200 J/kg°C, t₂ = 40°C, t₁ = 20°C

Substitute these values into equation 1

q = 8×4200×(40-20)

q = 8×4200×20

q = 672000 J.

Hence the energy suplied to the water is 672000 J.

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Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.
Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}

\dfrac{1}{v}=\dfrac{3}{25}

v=8.33\ cm

We need to calculate the image distance from right lens

The object distance will be

u = 15-8.33 = 6.67\ cm

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}

\dfrac{1}{v}=\dfrac{1167}{3335}

v=2.86\ cm

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0
3 years ago
What will an object weigh on the moon's surface if it weighs 190 n on earth's surface?
Greeley [361]
Turn your protracted to a 90 degree angle
3 0
3 years ago
Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.
Yuki888 [10]
Scott traveled 10 miles
5 0
2 years ago
Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius..
Troyanec [42]

Answer:

Answer:

= 338.2 m/s

Explanation:

vs= 331 + T (0.6m/s)

= 331 + 12 °C (0.6m/s)

= 331 + 7.2m/s

= 338.2 m/s

Explanation:

6 0
3 years ago
A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b
choli [55]

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =\dfrac{distance}{speed}

              time =\dfrac{545}{28}

              time =19.46 s

velocity of the car when it crosses the truck

S = ut + \dfrac{1}{2}at^2

545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

5 0
3 years ago
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