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IRINA_888 [86]
3 years ago
13

Please answer the questions here in the attachment I attached below. They are all multiple choice.

Physics
1 answer:
Radda [10]3 years ago
8 0

Answer:

abcbc

Explanation:

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What body parts were scientists wanting to image that prompted the development of the CT scanner
zalisa [80]

Answer:

The head

Explanation:

4 0
2 years ago
What is one common electrostatic phenomenon​
Mademuasel [1]

Answer:

There are many examples of electrostatic phenomena, from those as simple as the attraction of the plastic wrap to one's hand after it is removed from a package to the apparently spontaneous explosion of grain silos, the damage of electronic components during manufacturing, and photocopier & laser printer operation

6 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
A lamp draws a current of 0.1 A when it is connected to a 122-V source. (
Setler79 [48]
The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
4 0
3 years ago
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
Oliga [24]

Answer:

The value of change in internal l energy of the gas = 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

6 0
3 years ago
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