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3 years ago

In which parking situations should you use your parking brake?

Physics

2 answers:

vredina [299]3 years ago

6 0

Answer:

Always!!

Explanation:

I hope it help! :D

VMariaS [17]3 years ago

5 0

Answer:
-You should use your parking brake when on a hill/mountain with a huge drop because your car can’t slide off the edge.
-When there is a flood where your car is sitting because the water can’t carry your car

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What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b

kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

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2 years ago

What property or properties of a wave determine it's speed?

mash [69]

The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.

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4 years ago

If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th

alex41 [277]

Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
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V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s

6 0

3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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3 0

1 year ago

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a

emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

$\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\ I_2= 0.125 \ foot-candles$

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

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6 0

2 years ago