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Oksi-84 [34.3K]

3 years ago

12

Planet, , galaxy, universe. What is missing in the ordered sequence of the cosmos? A. Sun, B. Earth, C. Nebulae, D. Solar system

,

2 answers:

Artemon [7]3 years ago

8 0

D; solar system, because the planets are inside it.

Julli [10]3 years ago

3 0

Solar system because the planets are in the solar system.

:)

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The burning of 1.0 g of coal produces 35 000 joules of energy. How many kcals are produced?

sergeinik [125]

I got the 1.0g of coal producves 35,000 joules which is also 8.36 kcals

8 0

3 years ago

The law of conservation of matter states that during a chemical reaction

Sveta_85 [38]

This law states that, despite chemical reactions or physical transformations, mass is conserved — that is, it cannot be created or destroyed — within an isolated system

4 0

3 years ago

A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.

alexira [117]

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

4 0

1 year ago

Using this information...

Pepsi [2]

$19.2\:\text{m/s}$

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

$v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)$

$\:\:\:\:\:= 19.2\:\text{m/s}$

6 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago