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2 years ago

6

Give an example of speed making a difference in the amount of kinetic energy. Tell how you know that kinetic energy amount is di

fferent in your example.
Please help this is due today

2 answers:

kykrilka [37]2 years ago

5 0

Answer:

1. A car moving 40 mph has four times as much kinetic energy as one moving 20 mph, while at 60 mph a car carries nine times as much kinetic energy as at 20 mph. Thus a modest increase in speed can cause a large increase in kinetic energy.

PLZ MARK BRAINLIEST THANK AND FIVE STAR:)

kari74 [83]2 years ago

5 0

Answer:

a car moving 40mph has four times as much kinetic energy as one moving 20 mph, while at 60 mph a car carries nine times as much kinetic energy as at 20 mph.

if you need any more info just dm me

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A proton with an initial speed of 600,000 m/s is brought to rest by an electric field. Part A Part complete Did the proton move

rodikova [14]

Answer: Part A the right sentence is: Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential

Part B. aproximatelly 98 times ΔV V

Part C. the initial kinetic energy of the proton is 1.87 10^3 eV

Explanation: Part A. The field stops the proton so the lines of electric fild must be directed in opposite direction of its movement. This means that the proton moves to a higher potential. Part B The kinetic energy of the is transformed in electric potenctial for the proton.

Part C. Energy in J divide the charge of electron gives the energy in eV.

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2 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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1 year ago

The main purpose of a service panel in a house is to A. keep the meter working correctly. B. connect the service drop to the hou

cricket20 [7]

I think the best answer would be the last option. The main purpose of a service panel in a house is to keep electrical hazards localized. Service panels are used to identify circuit loads and isolate these loads so when you want to check the equipment you can easily identify the line. Also, it would serve as the isolation point to avoid any electrical hazards.

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3 years ago

Read 2 more answers

A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

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3 years ago

Why are the parts of an atom that electrons occupy called electron clouds?

Anna [14]

Because it's literally impossible to tell exactly where something that size is
located at any particular time.

And that's NOT because it's so small that we can't see it. It's because any
material object behaves as if it's made of waves, and the smaller the object is,
the more the size of its waves get to be like the same size as the object.
When you get down to things the size of subatomic particles, it doesn't make
sense any more to try and talk about where the particle actually "is", and we only
talk about the waves that define it, and how the waves all combine to become a
cloud of <em><u>probability</u></em> of where the particle is.

I know it sounds weird. But that's the way it is. Sorry.

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3 years ago