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Contact [7]
3 years ago
7

In which scenario does radiation occur?

Physics
2 answers:
Zanzabum3 years ago
4 0
Pretty sure a, because it’s a stove and it uses electricity like almost a normal appliance for the kitchen would, the rest are just a natural force without any radiation to occur in them, electricity is the same as technology and believe it or not technology like phones give off radiation, so yeah. hope this was helpful!
labwork [276]3 years ago
4 0

Answer:

b. drying towels at the beach

Explanation:

i got it correct on my test

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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr
nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

\frac{E}{t} = VI

solving for t

t = \frac{E}{VI}

t = \frac{m_w C_w \Delta T}{VI}

t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}

t = 444.125 sec

5 0
3 years ago
Read 2 more answers
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a spa
Zarrin [17]

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

4 0
2 years ago
Amy swims 500 m (0.5 Km) in 6 minutes (0.1 hour). What was her speed in Kilometers per hour? A. 8.3 Km/hr B. 5 Km/hr C. 50 Km/hr
kondor19780726 [428]
Speed = Distance ÷ Time so divide .5 km by .1h. .5 km÷.1h=5 km/h, so the answer is B. 5km/h.
8 0
2 years ago
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Which of the following is true about organizing items and information during scientific investigations?
LekaFEV [45]
There are usually multiple methods for organizing items and information in a scientific investigation.
5 0
3 years ago
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NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
2 years ago
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