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Solnce55 [7]
3 years ago
14

A

Physics
1 answer:
OleMash [197]3 years ago
7 0

Answer:

R= 5.4 ohms

Explanation:

Given that

V= 9 V

Power ,P= 15 W

Lets take resistor resistance = R

We know that Power given as

P=\dfrac{V^2}{R}

V=Voltage

P=Power

R=Resistance

R=\dfrac{V^2}{P}

Now by putting the all the values in the above equation

R=\dfrac{9^2}{15}

R= 5.4 ohms

Therefore the resistance of the resister will be 5.4 ohms.

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A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s
djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, u=32 m/s

Acceleration of the driver is, a=-1.5 m/s²

Time taken to reach final velocity is, t=10.8 s.

The final velocity is given using the Newton's equations of motion as:

v=u+at, where, v is the final velocity.

Now, plug in the given values and solve for v.

v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}

Therefore, his final velocity is 15.8 m/s.

5 0
3 years ago
Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in
iVinArrow [24]

Answer:

The value is  P_2 = 40.54 \ psla

Explanation:

From the question we are told that

 The initial pressure is  P_1 = 14\  psla

  The initial temperature is  T_1 =  50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283  \  K

   The final temperature is  T_2 =  320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433  \  K

Generally the equation for adiabatic process is mathematically represented as

         PT^{\frac{\gamma}{1- \gamma} } =  Constant

=>      P_1T_1^{\frac{\gamma}{1- \gamma} } =  P_2T_2^{\frac{\gamma}{1- \gamma} }

Generally for a monoatomic gas  \gamma =  \frac{5}{3}

So

           14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }

=>       14 * 283^{-2.5} =P_2 * 433^{-2.5}

=>       P_2 = 40.54 \ psla

8 0
2 years ago
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
A motorcycle is speeding along at a velocity of120kph.Then, it changes to a velocity of 150 kph for 2 minutes.What is the motorc
Taya2010 [7]

Answer:

Explanation:

a=v-u/t

a=acceleration

v=final velocity

u=initial velocity

t=tme taken

we need to convert from kph to ms⁻¹

v= 150*1000/60*60= 41.67ms⁻¹

u= 120*1000/60*60= 33.33ms⁻¹

t= 2*60= 120s

a=41.67-33.33/120

a=8.34/120

a=0.0694ms⁻²

8 0
3 years ago
Which of the following is not an advantage that attractive people enjoy in interpersonal relationships?
Licemer1 [7]

Psychology on Egenuity Oct 5th 2018 says answer is C


5 0
3 years ago
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