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2 years ago

When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring?

Physics

1 answer:

raketka [301]2 years ago

7 0

Well first you need to know the formula for the spring force. It’s -1/2k*d

K being the constant

D being the displacement

We can actually find the force on the spring by calculating the weight of the block. It would be mass multiplied by gravity, convert the grams to kilograms first tho.

600/1000 = 0.6kg

0.6kg * 9.8 m/s^2 = 5.88N

Now plug everything into the equation, also convert the centimeters to meters:

1.2/100 = 0.012 meters

5.88 = -1/2k*(0.012)

5.88 = k*(0.006)

We could drop the negative sign because we really just want the magnitude of the spring constant.

K = 980N

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a sound wave is an example of a a. transverse wave. b. longitudinal wave. c. standing wave. d. surface wave

Leona [35]

Answer: Longitudinal wave

Explanation:

Longitudinal wave are the oscillations that are parallel to the direction of energy transfer that means the vibrations are in line with the direction where the energy is travelling.

A key feature of sound wave is that they cause sound particles to vibrate. The region where the particles are close together are called compressions and regions where particles are further apart they are called rarefactions.

The other options explanation:

-Transverse waves are where the oscillations are perpendicular to the energy of transfer.

-A standing wave is where the waves are travelling back and forth where there are some fixed points in the system whilst other vibrate with highest amplitude

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2 years ago

A train travels from over city to another its initial velocity is lower than its final velocity this is a example of

Anni [7]

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2 years ago

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

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2 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$