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4 years ago

When the instantaneous voltage and current in an ac circuit are in phase, we know that?

1 answer:

7 0

We know that the centripetal acceleration is zero, since the summation of forces is the mass times the acceleration in the x- component.

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If a running system has a total change in heat of 295 joules, and it's running at a temperature of 402 kelvin, what is the entro

Liula [17]

The answer is either 75 or 55

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3 years ago

Explain why it is better to use a cable with a low resistance than to use a cable with a high

blondinia [14]

Explanation:

The primary reason for this is efficiency. In most applications, a lower resistance means less power is converted to heat and lost to the surrounding environment and more of the supplied power gets to its intended destination.

3 0

2 years ago

A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position functio

yan [13]

All you need to know is that velocity is the derivative with respect to time of position.

Therefore: ds/dt = -32t +10

To know the velocity at t=2s you just need to plug t=2 for t in the equation above.

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3 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

Name two factors that can affect the function of an enzyme

myrzilka [38]

Answer:

1. pH

2. Temperature

Hope this helps

3 0

3 years ago