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3 years ago

How is voltmeter connected in the circuit to measure the potential difference between two points?

Physics

1 answer:

34kurt3 years ago

7 0

Voltmeter is used to find the potential difference between two points.

We always connect it in parallel to the points where we need the potential difference.

Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.

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What can be defined as the rate at which velocity changes

taurus [48]

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: $a=\frac{v_f-v_i}{\Delta t}$

where $\Delta t$ is the time that took to change from the initial velocity $v_i$ to the final velocity $v_f$

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3 years ago

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A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

Allushta [10]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$

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3 years ago

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Which description of energy changes in nuclear reactions is correct?

Kryger [21]

The answer for this question is letter "B.Fission releases energy, and its products have greater stability."

Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition.

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3 years ago

Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th

bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

the mass of planet B to be, Mₓ = M

the radius of planet A to be, R₁

the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁² - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂² [where, R₂ = 2R₁]

= GM/4R₁² -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

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2 years ago

ONE EASY QUESTION

Brrunno [24]

Answer:

its tension force which acts in a string

Explanation:

need a thanks and thats it

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3 years ago