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sukhopar [10]
2 years ago
15

Identify a form of potential energy.

Physics
1 answer:
brilliants [131]2 years ago
5 0

Answer:

Gravitational potential energy (ex. height)

Elastic potential energy (ex. stretched rubber band)

Electrical potential energy (ex. negative charges close together want to repel each other)

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Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the
Dafna11 [192]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

\qquad \sf  \dashrightarrow \:m_2s_2

it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

\qquad \sf  \dashrightarrow \:m_1 =  \dfrac{m_2s_2}{s_1}

Now, consider specific heat of composite body be s'

According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{  \cancel{m_2}(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{  (\frac{s_2 + s_1}{s_1} )}

\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

6 0
2 years ago
Read 2 more answers
What is anything that has mass and volume?
Reika [66]
Matter, substance. Material howya call it.
3 0
3 years ago
What do we mean when we say that two light rays striking a screen are in phase with each other? What do we mean when we say that
galina1969 [7]

Answer: When the electric field due to one is a maximum, the electric field due to the other is also a maximum, and this relation is maintained as time passes. They alternatively reinforce and cancel each other.

Explanation:

In a wave, the phase, is an arbitrary time reference, used to locate a given point of the wave in time, within a cycle.

Two waves can travel at the same speed, or even have the same wavelength, but this is not enough to be sure that at a given point in time, both waves will be in their maximum, as it only can be determined from the phase of the waves.

So, only when the waves reach at the same point in time at the same amplitude, we can say that they arrive in phase, in a constructive interference.

8 0
3 years ago
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
Lyrx [107]

Answer:

A) 17.7 m/s

B) 15.98 m

C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - \frac{1}{2} gt², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

7 0
3 years ago
A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i
scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original  so it has 8.8 cm with the same orientation as original  and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f  where p and q represents the distance to the mirror  for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0
3 years ago
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