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3 years ago

Which of the following best describes an action-reaction pair?

2 answers:

3 0

Our weight is greater on the Earth because of its stronger gravity. .... track A? (b) Is the speed gained by ball B going down the extra dip the .... The force of friction between your back foot and the floor pushes you forward. .... (a) Two force pairs act; Earth's pull on the apple (action), and the apple's pull on the ...

Liula [17]3 years ago

3 0

Answer:

The Moon Pulls on Earth, and Earth pulls back on the moon.

Explanation:

Newton's third law of motion describes action and reaction forces. It tells that an object applies an action force on another and another object also exerts a reaction force on the first object. Both the action and reaction forces are of equal magnitude but they act in opposite direction.

Option (A) is an example of action- reaction pair i.e. "The Moon Pulls on Earth, and Earth pulls back on the moon". Earth exerts a gravitational force on the Moon. The gravitational pull of the Earth on the Moon is much more stronger than the pull of Moon on the Earth.

So, the correct option is (A).

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Can someone explain please ???

Radda [10]

Answer: A

Explanation:

6 0

3 years ago

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image' $h_{i}$ ' to the height of the object ' $h_{o}$ '. It is denoted by letter 'm'.

Mathematically, it can be written as

$m= \frac{h_{i}}{h_{o}} ------------(1)$

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance $(d_{o}) = + 8 mm$ .

Image distance $(d_{i}) = + 16 mm$

Object's height $(h_{o}) = + 4 mm$

Image's height $(h_{i}) =?$

Now, apply equation (3)

$\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$

$Or, \frac{h_{i}}{4 mm} = - \frac{+16 mm}{+8 mm}$

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0

3 years ago

A hydrogen atom is in its ground state when its orbital electron:

Harlamova29_29 [7]

It is in its ground state when its orbital electron is at its lowest energy amount.

8 0

3 years ago

what happens to the current in a circuit if the resitance of the components in the circuit is increased

Anit [1.1K]

Answer:

The current decreases.

Explanation:

Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is $V = IR$ , where V is voltage, I is current and R is resistance.

Rearranging this equation, you get:

$I = \frac{V}{R}$

and

$R = \frac{V}{I}$

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation $R = \frac{V}{I}$ , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

Thus, it can be seen that the current will decrease as resistance increases and vice versa.

7 0

3 years ago

Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi

insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

s−15

return trip is then :

s+15

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0

3 years ago

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