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Advocard
26 days ago
12

A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's

the efficiency of the fan?
Physics
1 answer:
Ostrovityanka [3.2K]26 days ago
6 0

Explanation:

Efficiency can be expressed as the ratio of useful output to the total power consumed. E = \frac{useful}{total}

The fan delivers 500W as useful output while wasting 300W. Thus, the overall power consumed equals 800W (500 + 300).

You might be interested in
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
1 month ago
If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of th
kicyunya [3294]

Answer:

Explanation:

at 23 degrees Celsius, the diameter measures 4.511 mm

GIVEN DATA:

diameter of hole  = 4.500 mm

T_1 = 23.0 degrees Celsius

T_2 = - 78.0 degrees Celsius

the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}

the diameter at 23 degrees Celsius is stated as

d = d_o (1+ \alpha \delta T)

 = 4.5 (1+2.4*10^{-5} *(23-(-78)))

   = 4.511 mm

the diameter of the rivet after temperature change is given as

d= d_o + \Delta d

 = d_o(1+ \alpha \delta T)

= 0.4500 *(1+2.4*10^{-5} *(23-(-78)))

= 0.4511 cm

8 0
25 days ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Keith_Richards [3271]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0
1 month ago
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. immediat
serg [3582]
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
4 0
1 month ago
Read 2 more answers
Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl
serg [3582]

Answer:

b) TA = TB = TC

Explanation:

  • When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
  • During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
  • The equilibrium state will be established once this equation is satisfied:

       \Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )

  • Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

       (400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC

  • At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
  • When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
  • Therefore, option b) is correct.
8 0
1 month ago
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