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mr_godi [17]
1 year ago
5

Hi im really confused:

Physics
1 answer:
Vitek1552 [10]1 year ago
3 0

The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ

U =  c_{p} m ΔT

U = Energy

c_{p} = Specific heat

m = Mass

ΔT = Change in temperature

ρ = Density

V = Volume

ρ = 1000 g / m³ (Dry air )

c_{p} = 1 J / g K

ΔT = 5 °C

V = 4 * 4 * 3

V = 48 m³

m = ρ V

m = 1000 * 48

m = 48000 g

U = 1 * 48000 * 7

U = 336000 J

U = 336 kJ

Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ

To know more about energy required to raise temperature

brainly.com/question/12372041

#SPJ1

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3 years ago
The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in
maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

4 0
4 years ago
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3 years ago
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I think it's A.

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A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
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The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

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The time of motion of the ball is calculated as follows;

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