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Zolol [24]
3 years ago
7

A car travels for 10s at a steady speed of 20m/s along a straight road. The traffic lights ahead change to red, and the car slow

s down with a constant deceleration, so that it stops after an additional 8 seconds.
Use your graph to deduce how far the car has travelled during the 18 seconds described.
Physics
1 answer:
pentagon [3]3 years ago
5 0
Okay. Draw a graph. Just a square wuth 10 lines in it. And then you can do the problem on that. Its really simple. A car Travels for 10 S at a steady speed of 20 M/S along a Street Rd. The traffic lights ahead and change to read in the car shows down with a constant the deceleration, so that if it stops after in a dish and on eight seconds.

Use your graph yo deduce how far the car has traveled during the 18 seconds described.
18x8x20x10x88=
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A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m lo
frosja888 [35]

Answer:

13.78 m/s

Explanation:

Given that:

The mass of the girl & her bicycle = 42 kg

The speed at the top of the hilll= 4 m/s

The height of the hill = 14.3 m

The length of the hill (distance) = 112 m

The frictional force = 20 N

To find the speed at the bottom of the hill; we need to carry out the following processes.

The workdone by gravity = mass × acceleration due to  gravity × Δh

= 42 × 9.8 × ( 14.3 - 0 )

= 5885.88 joules

The workdone by the friction = Force × distance = - 20 × 112    (since she is riding down the hill)

The workdone by the friction = -2240 Joules

The initial Kinetic friction = 1/2 mv²

= 1/2 × 42 × 4²

= 336 Joules

The final kinetic energy = Initial Kinetic energy + total work  

The final kinetic energy = (336 + 5885.88 - 2240) Joules

The final kinetic energy = 3981.88   Joules

Using the final kinetic energy =   1/2 mv²

3981.88  = 1/2 × 42 × v²

3981.88  = 21 v²

v² = 3981.88/21

v² =189.61

v = \sqrt{189.61}

v = 13.78 m/s

Therefore, the speed at the bottom = 13.78 m/s

4 0
3 years ago
You happen to know that the coefficient of static friction between your patio table and the ground is 0.42. You decide you want
Deffense [45]

Answer:

If the force applied is larger than 185.2 N, yes.

Explanation:

In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:

F_f = \mu mg

where

\mu=0.42 is the coefficient of static friction

m=45 kg is the mass of the table

g=9.8 m/s^2 is the gravitational acceleration

Substituting,

F_f=(0.42)(45 kg)(9.8 m/s^2)=185.2 N

So, we are able to move the table if we push with a force larger than 185.2 N.

4 0
3 years ago
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Firdavs [7]

Initial velocity (u) = 2 m/s

Acceleration (a) = 10 m/s^2

Time taken (t) = 4 s

Let the final velocity be v.

By using the equation,

v = u + at, we get

or, v = 2 + 10 × 4

or, v = 2 + 40

or, v = 42

The final velocity is 42 m/s.

5 0
2 years ago
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andreev551 [17]

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3 years ago
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Electrically inert metal ball A is connected to the ground by a wire. What happens to the charge of this ball if you bring a neg
kaheart [24]

Explanation:

They will repel, meaning that they are made of an electrical conductor.

7 0
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