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notka56 [123]
2 years ago
7

When a certain amount of heat is supplied to 1KG of insulated aluminium. The temperature of the aluminium rises by 1°C

Physics
2 answers:
laiz [17]2 years ago
7 0
Heat energy I think
Bumek [7]2 years ago
6 0

Answer:

The specific heat of aluminium is 897J/kg k . This values is almost 2.3 times of specific heat of copper .

Aluminium has higher thermal capacity than water

Explanation:

who believes me ......... more ......

Anybody don't know hey

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Define what is boyleâs law?
lawyer [7]

Answer:

Volume of given mass of gas is inversely proportional to pressure of gas

Explanation:

Boyle's law: It states that the volume of given mass of gas is inversely proportional  to the pressure of gas at constant temperature.

Mathematical representation:

Suppose, a gas of mass m

T=Constant temperature

V=Volume of gas

P=Pressure of gas

Then, V\propto\frac{1}{P}

8 0
3 years ago
Which is the only mechanism that can introduce new alleles in a gene pool? genetic drift random mating mutation
Vikentia [17]
Mutation because they change the genes
8 0
3 years ago
An object is thrown into the air at 60m/s, straight up. What is its velocity at the highest point?
Cloud [144]

Explanation:

Whenever an object is at its highest point, the velocity and acceleration of the object is zero.

6 0
3 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
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