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Radda [10]
2 years ago
13

when an object is placed near a concave mirror, at what position does it forms a magnified and erect image.​

Physics
1 answer:
alexira [117]2 years ago
4 0

Answer:

Between the principal focus and the pole of the mirror

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A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end
konstantin123 [22]

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

3 0
3 years ago
A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ​∘ ​​ with the horizontal. What is
Nastasia [14]

Answer:

Explanation:

Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.

In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.

7 0
3 years ago
What are the variables in Gay-Lussac's law? pressure and volume pressure, temperature, and volume pressure and temperature volum
Yanka [14]

Answer:

pressure and temperature (assuming volume is constant)

Explanation:

3 0
3 years ago
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4 0
3 years ago
A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12
bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction  

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair  

μ_s=F_{max}/F_{n}  

The F_n is equal to the weight multiplied by its gravity

∴F_{n}=mg  

Thus the coefficient of static friction changes as

μ_s=F_{max}/mg

μ_{s} = =165N/((25kg)\times(9.80 m/s^2 ) )

= 0.67

3 0
3 years ago
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