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Nataly_w [17]
4 years ago
15

A resistor is connected in series with a power supply of 23.90 V. The current measure is 0.60 A. What is the resistance (in Ω) o

f the resistor?
Physics
1 answer:
Yuliya22 [10]4 years ago
8 0

The resistance in this circuit is 39.8 ohms.

Explanation:

Any circuit having resistor, battery and ammeter connected in series will obey the ohm's law in basic case. So according to the Ohm's law, the current flowing in the circuit through the ammeter will be equal to the voltage shown in the voltmeter or battery and resistor is the proportionality constant. So with this law

V = IR

So, Resistance R = V/I

As the voltage is given as 23.90 V and the current is given as 0.6 A, then resistance is

R = 23.90/0.6 = 39.8 ohms.

So, the resistance in this circuit is 39.8 ohms.

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Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume
REY [17]

Explanation:

The given data is as follows.

          mass, m = 75 g

      T_{1} = 0^{o}C

      T_{2} = 27^{o}C

      Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

            q = m C \times (T_{1} - T_{2})

               = 75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C

               = 8464.5 J/mol

               = 8.46 kJ ......... (1)

Also, it is given that T_{3} = -20^{o}C = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

        q = mC \times (T_{3} - T_{1})

           = 0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K

           = -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

               8.46 kJ - 46.32 kJ

             = -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0^{o}C to ice at -20.0^{o}C is -37.86 kJ.

4 0
3 years ago
The latent heat of vaporization of H₂O at body temperature (37°C) is 2.42 x 10⁶ J/kg. To cool the body of a 60.4-kg jogger [aver
Darya [45]

Answer:

<h2>0.094 kg</h2>

Explanation:

        Latent heat of vaporization of H_{2}O at 37°C is 2.42\times10^{6}\text{ }\frac{J}{kg}.

        When the sweat on our body evaporates, it absorbs energy from our body to overcome it's Latent heat of vaporisation. Thus our body cools down when sweat evaporates.

       So, Energy absorbed by sweat to evaporate = Energy lost by body

Specific heat capacity of human body = 3500\text{ }\frac{J}{kg\text{ }C^{o}}. Jogger weights 60.4 kg. Body temperature decreases by 1.08\text{ }C^{o}

       Energy absorbed from body = mS\Delta T=3500\times60.4\times 1.08 =228312\text{ }J

       228312\text{ }J=\text{Energy absorbed by sweat}=mC=m\times2.42\times10^{6}\\m=0.094\text{ }kg

∴ 0.094 kg of sweat has evaporated from the body.

8 0
3 years ago
A steel rod is pulled in tension along its center axis (the axial direction) with a stress that is less than the yield strength.
Kamila [148]

Answer:

D. Axial stress divided by axial strain

Explanation:

 Lets take rod is pulled by force P

Stress σ = P/A

We know that

σ = ε E

E=  Lets take rod is pulled by force P

Stress σ = P/A

We know that

σ = ε E

σ=Axial stress

ε  =Axial Strain = ΔL/L

E= σ/ε

E= Axial stress/Axial Strain

So the modulus of elasticity  is the ratio of axial stress to axial strain.

σ=Axial stress

ε  =Axial Strain = ΔL/L

E= σ/ε

E= Axial stress/Axial Strain

So the modulus of elasticity  is the ratio of axial stress to axial strain.

The option D is correct.

6 0
3 years ago
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen
xxTIMURxx [149]

Answer:

A) first laser

B) 0.08m

C) 0.64m

Explanation:

To find the position of the maximum you use the following formula:

y=\frac{m\lambda D}{d}

m: order of the maximum

λ: wavelength

D: distance to the screen = 4.80m

d: distance between slits

A) for the first laser you use:

y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\

for the second laser:

y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m

hence, the first maximum of the first laser is closer to the central maximum.

B) The difference between the first maximum:

\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm

hence, the distance between the first maximum is 0.08m

C) you calculate the second maximum of laser 1:

y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m

and for the third minimum of laser 2:

y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m

Finally, you take the difference:

1.12m-0.48m=0.64m

hence, the distance is 0.64m

3 0
3 years ago
A bird flies from the ground to a height of 200 m if the bird has a mass of 1.5 kg how much gravitational potential energy does
Serjik [45]
Gravitational potential energy is equal to (mgh) so 1.5•9.8•200=2940
Or C
3 0
3 years ago
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