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Tamiku [17]
3 years ago
7

An object is placed so that the image formed is a real image of the same size as the object. What is the position of the object?

A. at the focal point B. at a distance of twice the focal length C. greater than the focal length but less than twice the focal length D. at a distance less than the focal length
Physics
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

Explanation:what’s the answer ?!

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An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

Where:

I is the current (2 A)

Δt is the time (20 s)

2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
3 years ago
A hydraulic cylinder causes the distance between points A and O to decrease at a constant rate of 3 inches per second. a) Determ
Ierofanga [76]

Answer:

a) The speed of the slider is 4.28 in/s

b) The velocity vector is 2.33 in/s

Explanation:

a) According to the diagram 1 in the attached image:

r_{C/A} =12*cos55i-12*sin55j\\r_{C/A}=6.883i-9.829j

Also:

v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j

If we comparing both sides of the expression:

-3+6.883w_{AC}=0\\w_{AC}=0.435rad/s

v_{C}=9.829*0.435=4.28in/s

b) According to the diagram 2 in the attached image:

r_{C/A}=12cos50i-12sin50j=7.713i-9.192j\\r_{B/C}=-3.856i+4.596j

v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j

Comparing both sides of the expression:

7.713w_{AC}-3=0\\w_{AC}=0.388rad/s\\v_{C}=3.575i

v_{B}=v_{C}+w_{AC}r_{B/C}\\v_{B}=3.57i+\left[\begin{array}{ccc}i&j&k\\0&0&0.388\\-3.856&4.59&0\end{array}\right]  \\v_{B}=3.57i+(0-1.78)i-(0+1.499)j\\v_{B}=1.787i-1.499j\\|v_{B}|=\sqrt{1.787^{2}+1.499^{2}  } =2.33in/s

6 0
4 years ago
I need to know everything about mars. I need to do a full 3 page essay. please help!
zheka24 [161]

Answer:

Mars was the Roman god of War along with an agricultural guardian. He is most closely related to the god Ares of Greek Mythology. In Roman mythology, he was second in importance to Jupiter, Rome's god of the Skies and Weather. Jupiter was the king of the Roman Pantheon, husband of the queen of gods Juno. He was also Mars' father. Unlike his Greek Counterpart, Ares who was most known for his hot headed temper and associated with hate and anger, Mars was part of the Romans <em>Archaic Triad</em>, sort of like the Big Three of Greek religion. The members of said Triad included Mars, Jupiter, and Quirinus, who had no Greek equivalent. Mars was most commonly depicted in posed of valor and strength, carrying swords or shields. He wore common Roman armor, including the plumed helmet. He was pictured as a strong leader of the Roman Army. The fourth planet from the Sun was given the name Mars when it was first discovered because it was red, much like the main color the Roman god was affiliated with.

This was mostly just random facts but i hope it helped some with your essay :)

4 0
3 years ago
A Decay Chain
Strike441 [17]

Answer : The energy released in first step of thorium-232 decay chain is 7.974\times 10^{-13}J

Explanation :

First we have to calculate the mass defect (\Delta m).

The balanced reaction is,

^{232}Th\rightarrow ^{228}Ra+^{4}He

Mass defect = Sum of mass of product - sum of mass of reactants

\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})

\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg

conversion used : (1amu=1.66\times 10^{-27}kg)

Now we have to calculate the energy released.

Energy=\Delta m\times (c)^2

Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2

Energy=7.974\times 10^{-13}J

The energy released is 7.974\times 10^{-13}J

Therefore, the energy released in first step of thorium-232 decay chain is 7.974\times 10^{-13}J

7 0
3 years ago
1. Is there a relationship between the volume of water displaced and the total volume of the block that has anything to do with
strojnjashka [21]

Answer:

The volume of the block is equal to the volume of water displaced by the block.

Explanation:

Volume refers to the amount of space occupied by a given object (in this case the block). When an object such as the block is immersed in water, it displaces its own volume of water. This volume of water displaced is equal to the volume of the block. Hence we can write;

Final Volume of water - Initial Volume of water= Water Displaced = Volume of the block

Recall that the density of a body is given by;

Density= mass/volume

If we obtain the volume of the block by measuring the volume of water displaced by the block, then we weigh the block using a weighing balance, we can obtain the density of the block easily from the relationship shown above.

8 0
3 years ago
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