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3 years ago

14

Is it true that at freezing point particle are vibrating so fast they break free?

2 answers:

Deffense [45]3 years ago

8 0

At freezing point, particles comes closer to each other once their potential energy decreases. Due to this, they don't get enough space to vibrate back and forth. Hence your statement is false

kondor19780726 [428]3 years ago

4 0

It is false. The effect of freezing is almost the exact opposite

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A car moves 50km in the first 30 minutes and 50km in the second 30 minutes and continues in this way which motion is this ------

egoroff_w [7]

Answer:

Uniform rectilinear movement (m. r. u.)

Explanation:

It is a continuous movement without acceleration, that is, it moves at a constant speed. The speed does not change over time, for this reason, there is no change in acceleration.

Vf = final velocity = 50/30 [km/s] = 1.67 [km/s]

Vo = initial velocity = 50/30 [km/s] = 1.67 [km/s]

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3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

Pls help i will be so happy thank you

Anvisha [2.4K]

Answer:

below given.

Explanation:

1) linear/direct

2) Indirect

3) linear/direct

4) quadratic

5) Inverse

6) linear/direct

7) Inverse / Indirect

8) Inverse / Indirect

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3 years ago

A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$

$\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}$

$\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}$

$a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}$

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

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3 years ago

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insens350 [35]

Answer:

thick wire and cold temperatures

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3 years ago

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