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3 years ago

How do you solve this ? Why is the ans C not B ?

Physics

1 answer:

melomori [17]3 years ago

4 0

Explanation:

Draw a free body diagram of the toolbox. There are two forces:

Weight force mg pulling down,

and applied force F pulling up.

Sum of forces in the y direction:

∑F = ma

F − mg = ma

45 N − 15 N = (3 kg) a

a = 10 m/s²

The answer should be B. It's possible the answer key has a mistake.

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A black hole is a ________

masya89 [10]

Answer:

Answer to the question:

Explanation:

A black hole is a finite region of space within which there is a mass concentration high and dense enough to generate a gravitational field such that no material particle, not even light, can escape it.

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3 years ago

1. The geologic time scale divides time into years and centuries, true or false

hodyreva [135]

Answer:

The answer is "False"

Explanation:

The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.

Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.

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2 years ago

Which lists three organic biological molecules?

igor_vitrenko [27]

Answer:

Explanation:

I'm learning it in science.

8 0

3 years ago

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

If the old current output was 425 kV, the ratio of current was:

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

Then, the ratio of the new output over the old output is:

$\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86$

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

7 0

2 years ago

This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s

Dennis_Churaev [7]

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

5 0

3 years ago