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3 years ago

A 2 cm tall object is placed 30cm in front of a concave mirror with focal length of 25 cm. What is the height of the image?

Physics

1 answer:

viktelen [127]3 years ago

8 0

Answer:

$- 10$ cm

Explanation:

$h_{o}$ = height of the object = 2 cm

$h_{i}$ = height of the image

$d_{o}$ = distance of the object from the mirror = 30 cm

$f$ = focal length of the mirror = 25 cm

$d_{i}$ = image distance

using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{30} + \frac{1}{d_{i}} = \frac{1}{25}$

$d_{i} = 150$ cm

using the equation

$\frac{h_{i}}{h_{o}} = \frac{- d_{i}}{d_{o}}$

$\frac{h_{i}}{2} = \frac{- 150}{30}$

$h_{i} = - 10$ cm

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A car needs to generate 75.0 hp in order to maintain a constant velocity of 27.3 m/s on a flat road. What is the magnitude of th

MaRussiya [10]

Answer:

2048.63 N

Explanation:

Force: The product of mass and the acceleration of a body. The S.I unit of Force is Newton(N).

From the question

P = F×V.................... Equation 1

Where P = power generated by the car, F = Total resistive force acting on the car, V = Velocity of the car.

make F the subject of the equation

F = P/V................ Equation 2

Given: P = 75.0 hp = (75×745.7) = 55927.5 W, V = 27.3 m/s

Substitute into equation 2

F = 55927.5/27.3

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4 years ago

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

Shtirlitz [24]

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

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3 years ago

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

aev [14]

Answer:

A) G = m³/kg.s²

B) E = kg.m²/s²

Explanation:

The given relation is:

F = Gm₁m₂/r²

where, the units of all variables are:

F = Force = kg.m/s²

m₁ = m₂ = mass = kg

r = distance = m

G = Gravitational Constant = ?

Therefore,

kg.m/s² = G(kg)(kg)/m²

(kg.m/s²)(m²/kg²) = G

G = m³/kg.s²

The given equation is:

E = mc²

where, the units of all variables are:

m = mass = kg

c = speed = m/s

E = Energy = ?

Therefore,

E = (kg)(m/s)²

E = kg.m²/s²

This is the correct answer, which is not present in any option.

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4 years ago

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a

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Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

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For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

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Sin³θ = 0.2523

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So, the separation between the two charges, d = 2 x L x Sin θ

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goldfiish [28.3K]

Answer:

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