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QveST [7]
3 years ago
12

Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem

perature. Th is is one of the reasons why desert regions, although very hot during the day, are bitterly cold at night. Th e heat capacity of air at room temperature and pressure is approximately 21 J K−1 mol−1. How much energy is required to raise the temperature of a room of dimensions 5.5 m × 6.5 m × 3.0 m by 10°C? If losses are neglected, how long will it take a heater rated at 1.5 kW to achieve that increase given that 1 W = 1 J s−1?
Physics
2 answers:
LenKa [72]3 years ago
6 0

Answer:

The energy is 2.201x10⁴kJ and the time is 1.467x10⁴ s

Explanation:

please, the solution is in the attached Word file

Download docx
Darina [25.2K]3 years ago
4 0

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

V= 5.5\times 6.5\times 3\ m^3

V=107.25\ m^3

V=107.25\times 1000 L

V= 107250 L

At STP number of moles given as

n=\dfrac{V}{V_{at\ S.T.P.}}

V=22.4 L at S.T.P.

n=\dfrac{107250}{22.4}\ moles

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

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A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

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2 years ago
What is the frequency of a wave if the speed is 4300 m/s and the wavelength is 32 m?
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2 years ago
2. Lenses such as those in microscopes and telescopes depend on which property of<br><br> light?
Marizza181 [45]

Answer:

The eyepiece comprises a converging lens that is a magnifying lens. The lens has a short focal length,

This lens magnifies this image.

Explanation:

In lenses such as those in microscopes and telescopes, the objective forms an image with the following features:

1. Image is real

2. Image is diminished in size

3. Also, the image formed is upside-down.

The eyepiece comprises a converging lens that is a magnifying lens. The lens has a short focal length,

This lens magnifies this image.

5 0
2 years ago
A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4
Svetllana [295]

The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

R.

b. Which direction does μ go?

Answer:

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.

7 0
3 years ago
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