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Vladimir [108]
3 years ago
9

If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced

?
A.
real, upright, and magnified
B.
virtual, inverted, and smaller
C.
virtual, upright, and magnified
D.
real, inverted, and magnified
E.
real, upright, and smaller
Physics
1 answer:
pshichka [43]3 years ago
3 0

Answer: I think its D

Explanation: Hope this was helpful...

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Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)
4 0
2 years ago
The power in an electric circuit varies inversely with the resistance. If the power is 2,200 watts when the resistance is 25 ohm
valentinak56 [21]

Answer:

The value of resistance when power is 1100 watts = R_{2} = 50 ohms

Explanation:

Power P_{1} = 2200 Watts

Resistance R_{1} = 25 ohms

Power P_{2} = 1100 Watts

Resistance R_{2} = we have to calculate

Given that the power in an electric circuit varies inversely with the resistance

⇒ P ∝ \frac{1}{R}

⇒ \frac{P_{2} }{P_{1} } = \frac{R_{1} }{R_{2} }

⇒ \frac{1100}{2200} = \frac{25}{R_{2} }

⇒ R_{2} = 50 ohms

This is the value of resistance when power is 1100 watts.

6 0
3 years ago
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

400000\ \text{N/C}

Explanation:

q_1 = Charge at 3000 m = 40 C

q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}

Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

7 0
3 years ago
Which has more gravitaatoinal potental energy: a bird on the ground or the same bird in the tree? why?
julsineya [31]
The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.

Considering also the formula for Gravitational Potential Energy GPE = mgh

For the bird on the ground, h =0,  therefore GPE = m*9.8*0 = 0

For that on the tree = mgh = m*9.8*h

Of course the one on the tree has a value greater than zero.  
7 0
3 years ago
Does anyone know how to solve this?
kompoz [17]

Answer:

110 m

Explanation:

Draw a free body diagram of the car.  The car has three forces acting on it: normal force up, weight down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substitute:

-mgμ = ma

-gμ = a

Given μ = 0.40:

a = -(9.8 m/s²) (0.40)

a = -3.92 m/s²

Given that v₀ = 30 m/s and v = 0 m/s:

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2 (-3.9s m/s²) Δx

Δx ≈ 110 m

8 0
3 years ago
Read 2 more answers
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