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cupoosta [38]
3 years ago
14

an object 5 cm in size is placed at 30 cm in front of a concave mirror of focal length 45 CM at what distance from the mirror sh

ould a screen be placed in order to obtain a shop image find the magnification size and nature of the image​
Physics
1 answer:
Makovka662 [10]3 years ago
3 0

Answer: Given: h₀=5cm

p=30cm

hi

f=45cm

require: q=?

hi=?

formulas:

1/f=1/p+1/q

hi/h₀=q/p

calculations:1/q=1/f-1/p

1/q=1/45-1/30

1/q=0.022=0.033

1/q=  =    -0.011

q=1/-0.011

q=   -90.91cm

hi=p × h₀/q

hi=30ₓ5/-90.91

hi=150/-90.91

hi= -1.65cm

image is virtual , erect and diminished

please mark as brainliest plzzzzzzzzzz

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Answer:

is a reflection.

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Explanation:

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Which of the following is a true statement about work?
Over [174]

Answer:

It measure the effort it takes for an object to move

Explanation:

Work done is the force applied to move a body through a particular distance.

It can also to be said to be a measure of the effort it takes for an object to move.

The unit of work done is in Joules.

 To find work done, we use the expression below:

       Work done  = Force x distance

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3 years ago
Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0
3 years ago
An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes
Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = \frac{1}{2}( 60 cm - 10 cm )

A = \frac{1}{2} × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed V_{max} = ωA

and ω = 2π/T

so maximum speed V_{max} = \frac{2\pi }{T}A

so we substitute

so maximum speed V_{max} = \frac{2\pi }{3.3} × 0.25 m

so maximum speed V_{max} = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

5 0
3 years ago
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