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3 years ago

Un vaso de agua fría está a 8 °c ¿ cuanto medirá en °f y °k? *

1 answer:

8 0

Esta calculadora de calor específica es una herramienta que determina la capacidad de calor de una muestra calentada o refrigerada. El calor específico es la cantidad de energía térmica que necesita suministrar a una muestra que pesa 1 kg para aumentar su temperatura en 1 K. Siga leyendo para aprender a aplicar la fórmula de capacidad de calor correctamente para obtener un resultado válido.

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How many joules of heat must be transferred to a 410-g aluminum pizza pan to raise its temperature from 32oC to 232oC? The speci

xxTIMURxx [149]

Answer:

recall that heat absorbed released is given by

Q = mc*(T2 - T1)

where

m = mass (in g)

c = specific heat capacity (in J/g-k)

T = temperature (in C or K)

*note: Q is (+) when heat is absorbed and (-) when heat is released.

substituting,

Q = (480)*(0.97)*(234 - 22)

Q = 98707 J = 98.7 kJ

Explanation:

3 0

3 years ago

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

Do magnets have to touch each other in order to experience a magnetic force

romanna [79]

No they do not they just need to be in each other's magnetic field

8 0

4 years ago

Read 2 more answers

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago

An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached

Kruka [31]

The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring.

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
F = kx
270 N/m x 0.38 m = 102.6 N

b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.

8 0

3 years ago