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Mekhanik [1.2K]
4 years ago
14

An object is placed 12.5 cm from a lens of focal length 22.0 cm. What is the image distance?

Physics
1 answer:
Tanzania [10]4 years ago
6 0

Answer:11.5

Explanation:

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What is one joule work
natta225 [31]
It's a Newton Meter Those two are multiplied to get a joule.
8 0
3 years ago
A cosmic catastrophic event occurred that caused the tilt of the Earth's axis relative to its plane of orbit to increase from 23
Gnom [1K]

Answer: The elimination of seasonal variations

Explanation:

Since the cosmic catastrophic event which occurred led to the tilt of the Earth's axis relative to the plane of orbit to increase from 23.5° to 90°, the most obvious effect of this change would be the elimination of seasonal variations.

It should be noted that seasonal variation refers to the variation in a time series that's within a year which is repeated. The cause of seasonal variation can include rainfall, temperature, etc.

7 0
3 years ago
A 14n force is applied for 0.33 seconds, calculate the impulse
Shalnov [3]

Answer:

4.62 N-s

Explanation:

recall that the formula for impulse is given by

Impulse = Force x change in time

in our case, we are given

Force = 14 N

change in time = 0.33s

Simply substituting the above into the equation for impulse, we get

Impulse = Force x change in time

Impulse = 14 x 0.33

= 4.62 N-s

5 0
3 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
I have no clue what to do please help.
Yakvenalex [24]

Answer:

Step one : read the directions

Step two: complete the assignment

Explanation:

3 0
2 years ago
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