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wolverine [178]
3 years ago
11

Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s

Physics
1 answer:
nevsk [136]3 years ago
7 0
B, i got the same question
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An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a
zalisa [80]
The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to g=9.81 m/s^2. The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
2aS=v_f^2-v_i^2=v_f^2
from which
v_f= \sqrt{2aS}= \sqrt{2\cdot 9.81 m/s^2 \cdot 40 m}=28 m/s
5 0
3 years ago
Dans un tube en U contenant du mercure ,on verse de l'autre côté de l'acide sulfurique de densité 1,84 et de l'autre côté de l'a
Iteru [2.4K]

Explanation:

Unclear question. The clear rendering reads;

"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?

6 0
3 years ago
A person is attracted toward the centerof the earth by a 500 n gravitational force. the force with which the earth is attracted
nignag [31]
500 N  is the answer, you just tell that the moon is attracted towards the person because of the Earth's huge mass. 
4 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
Scientific notation of 86400
Mandarinka [93]
8.64×10^4

this is 86400 in scientific notation

3 0
3 years ago
Read 2 more answers
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