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borishaifa [10]
3 years ago
11

Which of the following represents an upright image? A. -do B. +m C. -m D. +do

Physics
1 answer:
Tom [10]3 years ago
5 0

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

m = \frac{y'}{y}

where we have

y' = size of the image

y = size of the object

There are two possible situations:

- When m is positive, y' has same sign as y: this means that the image image is upright

- When m is negative, y' has opposite sign to y: this means that the image is upside down

Therefore, the correct option representing an upright image is

B. +m

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Answer:

Te ayudo con una de prueba $

Explanation:

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3 years ago
the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​
34kurt

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

6 0
3 years ago
Without friction, what happens? Check ALL
Fudgin [204]
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7 0
3 years ago
An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the
Lerok [7]
Answer:

Explanation:
Kinematics equation for first Object:

but:
The initial velocity is zero

it reach the water at in instant, t1, y(t)=0:


Kinematics equation for the second Object:
The initial velocity is zero

but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s


The velocity is negative, because the object is thrown downwards
7 0
2 years ago
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha
belka [17]

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

8 0
3 years ago
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