PHYSICS HIGH SCHOOL

What is the acceleration of a 6.4 kilogram bowling ball if a force of 12 N is applied to it?

Answers

Answer 1

Answer: F=ma
12=6.4a
1.875=a
A=1.875

Answer 2

Answer:

Answer:

1.875 m/sec2

Explanation:

Acceleration = Force/Mass

Acceleration = 12/6.4

12/6.4 = 1.875

Units = m/sec2

1875 m/sec2

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Which statement is true about a polarized object?

Answers

The statement which is true about a polarized object would be "the number of the positive and negative charges can be the same." Polarized objects became polarized and it has its center of positive charges separated from negative charges.

Answer:

For plato users: C.The number of positive and negative charges can be the same.

Why was the concern over global cooling replaced with a concern over global warming?

Answers

During 1970s, same observations were seen as what we have observed today pertaining to our climate. Journals were discussing that there would be warming because of greenhouse gases emissions. Also, it was observed between the years 1970 to 1990 that there was a steady surface temperature increase. Due to this, people are now fixated with global warming rather than on global cooling.

T or f Halogens are the basis of all fossil fuels.

Answers

The statement about “Halogens are the basis of all fossil fuels.” is false. It is the carbon that is the basis of all fossil fuels.

What happens along a stationary front?a. Cold air rises over warm air.
b. The flow of air is neither toward the warm air mass nor toward the cold air mass.
c. Air moves so rapidly upward that hurricanes form.
d. Wind stops completely.

Answers

Answer: b. The flow of air is neither toward the warm air mass nor toward the cold air mass.

A stationary front forms between two air masses. A stationary front results when the warm front or cold front air stops moving. This occurs due to the fact that warm front and cold front air masses being opposite to each other but neither of them are able to repel the other. This affects the climatic conditions of the region. The weather is often cloudy along a stationary front and also supported with fall of rain and snow especially if the air in the front is cold with low atmospheric pressure.

Therefore, along a stationary front the flow of air is neither toward the warm air mass nor toward the cold air mass.

Final answer:

A stationary front is formed when a cold air mass and a warm air mass meet, but neither is strong enough to displace the other. The air flow is generally neither towards the cold nor warm air mass, often resulting in prolonged cloudiness and precipitation

Explanation:

Along a stationary front, option b best describes what happens. Generally, stationary fronts occur when a cold air mass and a warm air mass meet, but neither is strong enough to move the other. As a result, the flow of air is typically neither toward the warm air mass nor toward the cold air mass. Instead, both air masses essentially stay where they are, often resulting in prolonged periods of cloudiness and precipitation in the area surrounding the front.

Learn more about Stationary Front here:

brainly.com/question/32838744

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All electromagnetic waves moving through a vacuum move at:

Answers

at the speed of light wich is approx 3 x 108 meters/ second

all electromagnetic waves move through a vacuum at about 3*108 meters a second hope that this helps some if not sorry

stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answers

Answer:

Approximately $5.66\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ and that air resistance is negligible.

Explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative since the stone is accelerating downward.)

The duration of the flight can be found in the following steps:

Find velocity right before landing given displacement, initial velocity, and acceleration.
Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ :

$v$ is the final velocity right before landing (needs to be found,)
$u = 15.5\; {\rm m\cdot s^(-1)}$ is the initial velocity,
$a = (-9.81)\; {\rm m\cdot s^(-2)}$ is acceleration, and
$x = (-75.0)\; {\rm m}$ is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find $v$ :

$\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \n &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \n &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}$ .

In other words, the velocity of this stone has changed from the initial value of $u = (15.5)\; {\rm m\cdot s^(-1)}$ to $v \approx (-40.05)\; {\rm m\cdot s^(-1)}$ during the flight. Divide the change in velocity by acceleration $a = (-9.81)\; {\rm m\cdot s^(-2)}$ (the rate of change in velocity) to find the duration of the flight:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \n &\approx 5.66\; {\rm s}\end{aligned}$ .

In other words, the stone would be in the air for approximately $5.66\; {\rm s}$ .

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