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never [62]
2 years ago
6

A body of mass 2kg moves round a circle of radius 5m with a constant speed of 10m/s. Calculate the force toward the centre of th

e circle​
Physics
1 answer:
antiseptic1488 [7]2 years ago
5 0

Answer:

F = 5

Explanation:

F = m x v^2/r = 2 x 10^2/5 =200/5 = 40 (N)

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Find the horizontal component and the vertical component ​
aleksandr82 [10.1K]

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

8 0
2 years ago
What step did Mendel take to be sure that his pea plants did not self-pollinate?
Ipatiy [6.2K]

Answer:

it is made up of some theory

Explanation:

it is an artificial pollination

8 0
2 years ago
How does the Earth release energy back into the atmosphere?
JulsSmile [24]

In general, the Earth releases energy back to the atmosphere through reflection, evaporation, and radiation. The Earth gets energy from the sunlight, part of which it absorbs, while part it reflects backwards, thus giving energy to the atmosphere. Also, the heating up of the Earth by the absorbed sunlight, radiates back in the lower layers of the atmosphere, again giving back energy to it. The water vapor is another way in which the Earth gives back energy tot he atmosphere as through the evaporation, the water vapor gets into the lower parts of the atmosphere and gives energy to it.

8 0
3 years ago
a roller coaster is traveling at 13m/s when it approaches a hill that is 400m long. heading down the hill, it accelerates 4.0m/s
Degger [83]
We will apply the equation:
2as = v² - u²
v = √(2as + u²)
v = √(2 x 4 x 400 + 13²)
v = 58 m/s

hope this helps
7 0
2 years ago
A 50.0 kg student climbs 5.00m up a rope at a constant speed. If the student's power output is 200.0 W, how long does it take th
hammer [34]

Answer:

The time taken by the student to climb is 12.25 seconds and work done is 2450 J.            

Explanation:

Given that,

Mass of the student, m = 50 kg

The student climbs to a height of 5 meters at a constant speed.

The student's power output is 200.0 W, P = 200 W

The power of an object is given by work done divided by time taken. So,

P=\dfrac{W}{t}

(b) W is work done,

W=mgh\\\\W=50\times 9.8\times 5\\\\W=2450\ J

(b)

t=\dfrac{W}{P}\\\\t=\dfrac{2450}{200}\\\\t=12.25\ s

So, the time taken by the student to climb is 12.25 seconds and work done is 2450 J.

3 0
3 years ago
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