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liq [111]
3 years ago
5

A roller coaster starts at the top of a straight track that is inclined at 30degrees with the horizontal. This causes it to acce

lerate at a rate of 4.9m/s^2(1/2g).
a. What is the roller coaster's speed after 3 seconds?
b. How far does it travel during that time?

Physics
2 answers:
Dahasolnce [82]3 years ago
8 0
Hope this helps you.

wlad13 [49]3 years ago
6 0

Answer:

Part a)

v_f = 14.7 m/s

Part b)

d = 22.05 m

Explanation:

Part a)

Initial speed of the roller coaster

v_i = 0

acceleration of the roller coaster is given as

a = 4.9 m/s^2

now for the speed after 3 s is given as

v_f = v_i + at

v_f = 0 + 4.9 \times 3

v_f = 14.7 m/s

Part b)

Distance traveled by the roller coaster

d = (\frac{v_f + v_i}{2})t

d = (\frac{14.7 + 0}{2})(3)

d = 22.05 m

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cricket20 [7]

Answer:

a=4.44\frac{m}{s^2}

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

d=vt\\d=20\frac{m}{s}(0.5s)=10m

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}

7 0
3 years ago
Whitch two options are forms of kinetic energy?
valentina_108 [34]

Answer:the witch has nothing to do with the problem

Explanation:

7 0
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Which of the following is an effect of the electric force?
stich3 [128]

As we know that two charges exert force on each other when they are placed near to each other

The force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

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r = distance between two point charges

also we know that two similar charges always repel each other while two opposite charges always attract each other

so here correct answer would be

<em>A. A positive and negative charge attract each other.</em>

6 0
3 years ago
Read 2 more answers
Joe the house painter stands on a uniform oak board weighing 600 N and held up by vertical ropes at each end. Joe has been dieti
Mamont248 [21]

Answer

given,

weight of the oak board = 600 N

Weight of Joe = 844 N

length of board = 4 m

Joe is standing at 1 m from left side

vertical wire is supporting at the end.

Assuming the system is in equilibrium

T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

 T₁ + T₂ = 1444...........(1)

taking moment about T₂

 T₁ x 4 - 844 x 3 - 600 x 2 = 0

 T₁ x 4 = 3732

 T₁ = 933 N

from equation (1)

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3 0
3 years ago
A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)
frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}

Time taken by light = 2.8583×10⁻⁶ seconds

3 0
3 years ago
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