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Anastasy [175]
3 years ago
12

PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is

9.0 cm. What is the height of the image?
Physics
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

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How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
777dan777 [17]
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
W= qV_i - qV_f
where q is the charge of the proton, q=1 e = 1.6\cdot 10^{-19}C, with e being the elementary charge, and V_i = +125 V and V_f = -55 V are the initial and final voltage.

Substituting, we get (in electronvolts):
W=e(125 V-(-55 V))=180 eV
and in Joule:
W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J

5 0
3 years ago
A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I
Fofino [41]

Answer:

B

Explanation:

Water level remains unchanged

The first thing to realize is that the buoyancy force is the same as, or equal to the weight of the wood, this same force is also the same as or equal to the weight of the water displaced by the wood. In the two cases, the weight of the wood will be unaffected nonetheless, and thus the water level will remain the same.

Therefore, the answer is B, the water level remains unchanged.

6 0
3 years ago
Students in science class were asked to do an experiment with a can of soda. They placed an unopened can of soda on a digital sc
ElenaW [278]
Because the gas has left which once you open the soda can u can see the fizzy which is gas coming out of it
5 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
2 years ago
A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the
topjm [15]
<h2>5.3 km</h2>

Explanation:

       This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

       Let us denote each of the individual displacements by a vector. Consider the unit vectors \vec{i}\textrm{ and }\vec{j} as the unit vectors in the direction of East and North respectively.

       By simple calculations, we can derive the unit vectors \vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2} in the directions North, 45^{o} South of West and 60^{o} North of West respectively.

       So Total displacement vector = Sum of individual displacement vectors.

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       Magnitude of Displacement = |-4.25\vec{i}+3.165\vec{j}|=5.3km

∴ Total displacement = 5.3km

4 0
3 years ago
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