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erastova [34]
3 years ago
10

Which is an example of a passive solar energy system?

Physics
1 answer:
Ganezh [65]3 years ago
5 0

An example is a sunspace or solar greenhouse

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Electromagnetic radiation consists of particles called:
Westkost [7]

Photons are particles of electromagnetic radiation.

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3 years ago
do you think it is ever possible for an earthquake to occur or for a volcano to erupt in a place that is unexpected ? explain
olga2289 [7]
Yes.....The continetal plates shifting would cause either disaster...
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4 years ago
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

4 0
3 years ago
A van is traveling on a road at a speed of 55 km/h. A girl sitting near the driver of the van throws a paper airplane to a boy a
aivan3 [116]
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.

The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h.  No rounding is necessary.</span>
5 0
4 years ago
Read 2 more answers
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
Fantom [35]
In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
4 years ago
Read 2 more answers
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