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3 years ago

Calculate the heat required, in calories, to raise

Physics

1 answer:

Goshia [24]3 years ago

5 0

Answer: $Q=4450.74\ J$

Explanation:

Given

mass of tin $m=100 gm$

Also we know that

specific heat of tin $c_{tin}=0.21 J/gm-^{\circ}C$

Melting Point of Tin $T=231.94^{\circ}C$

Heat required to raise the temperature of 100 gm tin is given by

$Q=mc\Delta T$

where $Q=heat\ added$

$c=specific\ heat\ of\ tin$

$\Delta T=change\ in\ temperature$

$Q=100\times 0.21\times (231.94-20)$

$Q=4450.74\ J$

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Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

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we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

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3 years ago

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3 years ago

Read 2 more answers

In a certain city, electricity costs $0.20 per kw·h. what is the annual cost for electricity to power a lamp-post for 8.00 hours

UNO [17]

Part a)

per day electricity power consumed when 100 W bulb is used for 8 hours

$P = 8 * 100 = 800 Wh$

for one year consumption

$E = 365 * 800 = 292 kWH$

now the cost will be given

$cost = 0.20 * 292 = $ 58.4$

now when other energy efficient light is used

$P = 8 * 25 = 200 Wh$

for one year consumption

$E = 365 * 200 = 73 kWH$

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$cost = 0.20 * 73 = $ 14.6$

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3 years ago

Which best describes the electric field created by a positive charge?

Vikentia [17]

Answer:

Which best describes the electric field created by a positive charge? ... Its rays point toward the charge.

Explanation:

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3 years ago

A satellite is put in a circular orbit about Earth with a radius equal to 35% of the radius of the Moon's orbit. What is its per

Debora [2.8K]

Answer:

0.21 lunar month

Explanation:

the radius of moon = r₁

time period of the moon = T₁ = 1 lunar month

The radius of the satellite = 0.35 r₁

Time period of satellite

The relation between time period and radius

$T\ \alpha\ \sqrt{r^3}$

now,

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{r_2^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{0.35^3r_1^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{1}=\sqrt{0.35^3}$

T₂ = 0.21 lunar month

hence, the time period of revolution of satellite is equal to 0.21 lunar month

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3 years ago