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iVinArrow [24]
2 years ago
14

When measuring the potential difference across a component, you need

Physics
2 answers:
Lorico [155]2 years ago
6 0

Answer:

Measuring potential difference Potential difference is measured using a device called a . Just like ammeters, some types have a pointer on a dial, but most have a digital display. However, unlike an ammeter, you must connect the voltmeter in parallel to measure the potential difference across a component in a circuit.

NISA [10]2 years ago
4 0
To measure the potential difference across a component, a voltmeter must be placed in parallel with that component in order to measure the difference in energy from one side of the component to the other. Potential difference is also known as voltage and is measured in volts (V).
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Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes
Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, v_1=65\ km/h

Radius of the circular arc, r_1=95\ m

Car 2 has twice the speed of Car 1, v_2=130\ km/h

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

According to given condition,

\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}

\dfrac{65^2}{95}=\dfrac{130^2}{r_2}

On solving we get :

r_2=380\ m

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0
3 years ago
A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t
Ganezh [65]

Answer:

The value is \eta  =  54.4 \%

Explanation:

From the question we are told that

    The work input is  W = ( 245 + A ) \  J

     The heat delivered is Q =  (142 + B) \  J

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          \eta  =  \frac{W}{Q_t}

Here  Q_t is the total out energy produce by the heat engine and this is mathematically represented as

           Q_t= Q + W

=>         Q_t=  245 + A + 142 + B

=>         Q_t=  390 + A+B

So

               \eta  =  \frac{245 + A }{390 + A+ B}

=>          \eta  =  0.544

=>          \eta  =  0.544 *100

=>          \eta  =  54.4 \%

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