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3 years ago

7

What happens to air as it warms ,expands and become less dense

1 answer:

AleksandrR [38]3 years ago

6 0

Air expands as it warms. Therefore warm air is less dense than cool air. The warm air from the first floor apartments rises to the second floor. People on the second floor require less heating to keep their <span>apartments comfortable.
-Hope this helps </span>

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Determine the volume displaced and then calculate the density of this 54 g sample of brass.

inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density $d_{w}$ of water is : $1000kgm^{-3}=1gcm^{-3}$

Buoyant force exerted by a liquid on an object with $V_{imm}$ of it's volume immersed is :

$F_{B}=V_{imm}*d_{l}*g$

where ,

$F_{B}$ is the buoyant force
$d_{l}$ is the density of the liquid
$g$ is the acceleration due to gravity

Thus at equilibrium:

$m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}$

from these , we get the density of brass to be $1gcm^{-3}$

which is not possible

7 0

3 years ago

Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®

Furkat [3]

C. Kevlar

:) ++ * ++ !

5 0

3 years ago

Read 2 more answers

Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

Ganezh [65]

As we know that the formula of kinetic energy will be

$KE = \frac{1}{2} mv^2$

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

$KE = \frac{1}{2}(2)(1^2)$

$KE = 1 J$

7 0

3 years ago

A tiny particle with charge + 5.0 μC is initially moving at 55 m/s. It is then accelerated through a potential difference of 500

Vadim26 [7]

Answer:

ΔK.E = 2.5 × 10⁻³ J

Explanation:

Given data in the question, we have:

Charge of the particle, q = 5.0 μC = 5 × 10 ⁻⁶ C

Initial speed of the particle, v = 55 m/s

The potential difference, ΔV = 500 V

Now, the gain in kinetic energy is given as

ΔK.E = q × ΔV

on substituting the values in the above formula, we get

ΔK.E = 5 × 10 ⁻⁶ C × 500 V

or

ΔK.E = 2.5 × 10⁻³ J

8 0

4 years ago

How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of

lakkis [162]

Answer: The time required to deposit such amount of Mg is 886secs.

Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.

He also stated that;

96500C(1Farday) of electricity is required to deposit 1 mole of any metal.

For $Mg^2+$ +2e- ==>Mg

193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)

Which implies;

19300C will liberate 24g

xCwill liberate 0.110g

Where x is the amount of electricity to deposit 0.110g

x = (19300 × 0.110)/24

x = 884.6C

Recall that Q =It

Where Q is the quantity of electricity, I is the current and t is the time taken

884.6= 0.998 × t

t= 884.6/0.998

t= 886.3secs.

Therefore the time require is 886.3s.

8 0

3 years ago