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lara [203]
4 years ago
15

Convert 22°c to fahrenheit

Physics
2 answers:
posledela4 years ago
7 0
The answer is:
71.6 <span>°F</span>
Nuetrik [128]4 years ago
4 0
F = 1.8C + 32

F = 1.8 (22) + 32

F = 39.6 + 32

F = 71.6°
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Ilya and Anya each can run at a speed of 8.90mph and walk at a speed of 3.80mph . They set off together on a route of length 5.0
jekas [21]

Answer:

a) t = 56.34 minutes

b) v = 5.32mph

c) t = 47.22 minutes

d) v = 6.35 mph

Explanation:

Given;

The speed of Ilya and Anya are

Run = 8.90mph

Walk = 3.80mph

distance = 5.00 miles

a) anya walk half the distance and run half the distance.

Speed = distance/time

time = distance/speed

t1 = 5×0.5/3.80 = 0.658hr

t2 = 5×0.5/8.90 = 0.281hr

t = t1 + t2 = 0.658+0.281 = 0.939hr × 60minutes/hr

t = 56.34 minutes

b) average speed = total distance/total time taken

v = 5.00miles/0.939hr = 5.32mph

c) Ilya walk half the time and run half the time.

t1 = t2 = t/2

Distance = speed × time

Substituting the values.

5 = 3.80(t/2) + 8.90(t/2)

5 = 12.70(t/2)

t = 10/12.70

t = 0.787 hr × 60 minutes/hr

t = 47.22 minutes

d) average speed = 5miles/0.787hr

v = 6.35 mph

5 0
3 years ago
A bowling ball with a mass of 9 kg is moving at a speed of 4 m/s. What is its
Ilia_Sergeevich [38]
I’m pretty sure it’s B
4 0
3 years ago
if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were dis­covered?
Anna11 [10]

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

The surprising thing was that quasars and active galaxies have a lot of mass in the center of the very small volume of the space.

Therefore, the surprising thing about quasars was that due to this mass and energy they are 100 times more luminous than Milky Way which means they have high recession velocity and a very large amount of red-shifting.

To learn more about quasars, refer: brainly.com/question/9965257

#SPJ4

8 0
1 year ago
A square loop of length 0.1 m on each side is sitting in a magnetic field that is increasing at a rate of 0.1 T/second. What can
Zolol [24]

Answer:

The induced emf can be found by Faraday’s Law.

\epsilon = -\frac{d\Phi_B}{dt}

\Phi_B = BA = B(0.1)^2 = 0.01B

The magnetic field is increasing at a rate of 0.1T/s. So,

\frac{dB}{dt} = 0.1

Finally,

\epsilon = \frac{dBA}{dt} = 0.01\frac{dB}{dt} = 0.01\times 0.1 = 0.001 V

Explanation:

Faraday’s Law states that a change in the magnetic flux induces an emf in the circuit. The magnetic flux is the multiplication of magnetic field and the area of the loop. The area of the loop is simple, and the change of magnetic field as a function of time is given in the question.

The minus sign in front of the Faraday’s Law means that the induced current always opposes the change of the magnetic flux. Since we do not know the direction of the magnetic field in this question, we cannot find the direction of the induced emf on the loop.

4 0
3 years ago
Air at 20ºC with a convection heat transfer coefficient of 20 W/m2·K blows over a pond. The surface temperature of the pond is 2
dimulka [17.4K]

Answer:

The heat flux between the surface of the pond and the surrounding air  is<em> 60 W/</em>m^{2}<em> </em>

Explanation:

Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;

q = Q/A ...............................1

since we are working with the convectional heat transfer coefficient equation 1 become;

q = h (T_{sf} -T_{sd}) ........................2

where q is the  heat flux;

Q is the heat energy that will be transferred;

h is the convectional heat coefficient = 20 W/m^{2}.K;

T_{sf} is the surface temperature = 23^{o}C 23°C + 273.15  = 296.15 K;

T_{sd} is the surrounding temperature = 20^{o}C = 20°C + 273.15  = 293.15 K;

The values are substituted into equation 2;

q = 20 W/m^{2}.K (  296.15 K - 293.15 K)

q = 20 W/m^{2}.K ( 3 K)

q =  60 W/m^{2}

Therefore the heat flux between the surface of the pond and the surrounding air  is 60 W/m^{2}

3 0
4 years ago
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