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tresset_1 [31]
2 years ago
6

A wooden cube with the mass of 1kg is placed on a frictionless plane that makes an angle of 30° with the floor.

Physics
1 answer:
Anni [7]2 years ago
8 0

Answer:

I dont know

Explanation:

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Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide deta
Paladinen [302]

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

6 0
3 years ago
The resistance of a lamp that draws 0.5A of current when it is operated by a 12V battery will be Ω
lisabon 2012 [21]
According to Ohm's Law, the resistance, current and voltage are related as:

V = IR
⇒
R = V/I

V is given to be 12 Volts.
I is given to be 0.5 Ampere 

So, resistance will be:

R = 12/0.5 = 24 ohms
5 0
2 years ago
Read 2 more answers
Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
3 years ago
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

or, if taking significant digits into account,

m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
3 years ago
Using the statistical definition of entropy, what is the entropy of a system where W = 4?
Oliga [24]
     <span> S = k ln(W)
= 1.38*10^-23 J/K * ln(4)
= 1.91*10^-23 J/K</span>
so it will be c
5 0
3 years ago
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