All categories

2 years ago

3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.

Physics

1 answer:

Scrat [10]2 years ago

3 0

Answer:

514 cal

Explanation:

In order to calculate the lost heat by the amount of water you first take into account the following formula:

$Q=mc(T_2-T_1)$ (1)

Q: heat lost by the amount of water = ?

m: mass of the water

c: specific heat of water = 1cal/g°C

T2: final temperature of water = 11°C

T1: initial temperature = 12°C

The amount of water is calculated by using the information about the density of water (1g/ml):

$m=\rho V=(1g/ml)(514ml)=514g$

Then, you replace the values of all parameters in the equation (1):

$Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal$

The amount of water losses a heat of 514 cal

You might be interested in

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the

Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

$m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N$

We can find the static fricton force as follow,

$F=\mu_s * N$

Where N is the normal force (mg)

$F=0.3*50*9.8\\F=147N$

Static friction force at 147N is greater than the force applied hence body does not move.

$F=\mu_k N = 0.2*50*9.8= 98N$

3 0

3 years ago

An observer standing near a window 5 m high observe that an object falling downwards is passing across the window in 0.5 s. Find

Valentin [98]

Answer:

Explanation:

An object in free fall, NOT experiencing parabolic motion, has an equation of

$h(t)=\frac{1}{2} gt^2+h_0$ which says:

The height of an object with respect to time in seconds is equal to the pull of gravity times time-squared plus the height from which it was dropped. Normally we use -9.8 for gravity but you said to use 10, so be it.

For us, h(t) is 5 because we are looking for the height of the window when the object is 5 m off the ground at .5 seconds;

g = 10 m/s/s, and

t = .5sec

$5=\frac{1}{2}(-10)(.5)^2$ +h and

5 = -5(.5)² + h and

5 = -5(.25) + h and

5 = -1.25 + h so

h = 6.25

That's how high the window is above the ground.

8 0

2 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$