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2 years ago

5

if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

1 answer:

Paul [167]2 years ago

6 0

The concepts of force<span>, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops </span>pushing<span>? ... F </span>net<span> =10 N </span>2<span> N. = 8 N (to the right) a = F </span>net<span> m. = 8 N. 5 kg. =1.6 m s. </span>2<span> ... </span>Two equal forces<span> act on an </span>object<span> in the directions shown. </span>If<span> these ... </span>Two<span> connected carts </span>being accelerated by a force<span> F applied by.</span>

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4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going

vaieri [72.5K]

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

$mgh = \frac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 15}$

$v=\sqrt{294}$

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

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2 years ago

When it comes to how ray lines are drawn, what makes the convex lens and concave mirror similar to each other?

Advocard [28]

Answer:

Convex lens and convex mirrors are similar that

1. They have the same image characteristics at various object positions

2. They possess a positive focal length

3. Both their ray lines converge to a particular focal point

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2 years ago

I NEED HELP ASAP WHAT IS THE DISTANCE MOVED IN 15 SECONDS

Oduvanchick [21]

Answer:

5 I think

Explanation:

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2 years ago

Read 2 more answers

Select the correct answer.

aleksklad [387]

A. Alternating current

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2 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago