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3 years ago

when a cup of hot chocolate cools from 90c to 80c which of the following is happening to the molecules of the liquid

Physics

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

I think that the liquids molecules are slowing down. Hope this helps!

Explanation:

Please vote me Brainliest

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If the angular magnification of an astronomical telescope is 42 and the diameter of the objective is 72 mm, what is the minimum

andriy [413]

Answer:

Minimum diameter=1.714

Explanation:

Angular magnification=fob/fey=42

fob=40*fey

Hence minimum diameter=diameter of objective/angular magnification

minimum diameter=72/42=1.714

4 0

3 years ago

A bowling ball has a mass of 5 kg what happens to its momentum when its speed increases from 1m/s to 2m/s?

Mama L [17]

Here, Initial momentum = mu = 5*1 = 5 Kg m/s
Final momentum = mv = 5*2 = 10 Kg m/s

So, Momentum has been increased from 5 Kg m/s to 10 Kg m/s. Hence, Your Final answer is option B

Hope this helps!

8 0

3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

Ill mark as brainliest.

rewona [7]

Answer: 75V

Explanation:

Given that,

total resistance (Rtotal) = 150Ω

Current (I) = 0.5A

Change in electric potential (V) = ?

Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.

So, apply the formula V = I x Rtotal

V = 0.5A x 150Ω

V = 75V

Thus, the change in electric potential across the circuit is 75 Volts

8 0

3 years ago

Calculate the most probable speed of an ozone molecule in the stratosphere

Marysya12 [62]

Answer:

$v_{mp}=305.83 m/s$

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

$v_{mp}= \sqrt{\frac{2RT}{M} }$

plugging the values we get

$v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }$

$v_{mp}=305.83 m/s$

7 0

4 years ago