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3 years ago

Can someone explain the equation Q = M C delta T or Q = MCΔT Thanks!

2 answers:

3 0

Q = heat energy , m=mass , c=specific heat , delta T= change in temperature

as you know heat is a form of energy which is <em>usually</em> measured in Joules according to the SI. and also we usually use kilograms for mass.

so you need to know the mass, specific heat, and change in temperature in order to find out the heat energy :)

adell [148]3 years ago

3 0

Q is the quantity of heat energy used to change the temperature of specific mass of certain matter
M is the mass of matter u wanna to change its temperature
C is the specific heat, each substance has its own "C" as specific heat is the quantity of heat to change mass of 1 kg by 1 kalvin deg.
ΔT is the the difference in temp. from and to how much u wanna to change the temp. of your matter
so if u wanna to change 1L water (for example) which = to 1kg
u wanna to increase its temp. by 2 degrees
and the "c" is ≈ 4200 then
the quantity of energy u need to do that is = MCΔT = 1*4200*2 = 8400

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What will be the final velocity of a rock if we drop it off of a bridge and it strikes the ground 2.8s later (ignoring air resis

mars1129 [50]

Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time

Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s

The acceleration of the Earth when dropping something would be 9.8 m/s

Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)

5 0

2 years ago

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&

Katen [24]

Answer:

a) T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 = $\frac{T_x}{T}$

cos 30 = $\frac{T_y}{T}$

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m $\frac{v^2}{r}$ (2)

a) we substitute in 1

T = $\frac{mg }{cos 30}$

T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

T = 2.26 N

b) from equation 2

v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

v = 1.68 m / s

8 0

2 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

3 0

3 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

steposvetlana [31]

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

6 0

3 years ago

A pear hangs in a tree at a height of 1.8 m. The pear has a mass of 0.2 kg. The pear falls out of the tree and lands on the grou

TiliK225 [7]

a) PE=mgh=0.2*9.8*1.2=2.352 J

b) KE=PE=2.352 J

c) $v=\sqrt{\frac{2KE}{m}}=4.85$ m/s

6 0

3 years ago