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4 years ago

8

1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee

n them to be ___ the initial force

1 answer:

Vedmedyk [2.9K]4 years ago

7 0

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

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4 years ago

A car accelerates from 13 m/s to 25 m/s in 5.0 s. assume constant acceleration. what was its acceleration?

natima [27]

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3 years ago

Match each form of Energy with it's definition. Energy of motion Question 7 options: mechanical energy kinetic energy chemical e

Komok [63]

Answer:

a fireplace or burn gasoline in a car's engine.

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Kinetic energy

Kinetic energy is the motion of waves, electrons, atoms, molecules, substances, and objects.

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3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

4 years ago

Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.

weeeeeb [17]

Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

distance moved by the coaster, d = 225 ft = 68.58 m

final velocity of the coaster, v = 80 mi/h = 35.76 m/s

The time taken for the coaster to drop down the hill is calculated as;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\ \times \ 68.58}{9.8} } \\\\t = 3.74 \ s$

The work done by friction on the car is calculated as;

$W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76 }{3.74} \times \ 68.58\\\\W = 524,582 \ J$

Therefore, the work done by friction on the car is 524,582 J.

6 0

3 years ago