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Licemer1 [7]
3 years ago
14

Có thể một lực hướng tâm tạo ra công trên một vật hay không? Hãy giải thích

Physics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

Explanation:

Công là sản phẩm của một lực và độ dời theo phương của lực đó. Vì lực hướng tâm luôn tác dụng vuông góc với chuyển động nên Không thể xảy ra công việc nào.

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A black lift of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35
horsena [70]

Answer:

Explanation:

refractive index of ember = sin of angle of incidence / sin of angle of refraction

= sin 35 / sin24

= .5735 / .4067

= 1.41

This is refractive index of ember with respect to water

refractive index of ember with respect to water

= wμe = μe / μw

μe = wμe x  μw

= 1.33 x 1.41

= 1.87

refractive index of ember with respect to air = 1.87 .

6 0
3 years ago
How fast would a 2-kilogram object need to move to have the same kinetic
siniylev [52]

Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

8 0
2 years ago
When is orange is the new black episodes come out
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3 0
3 years ago
How many species go extinct every day??
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5 0
2 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
2 years ago
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