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Shalnov [3]
2 years ago
10

A 90N force towards postive x-axis, A 50N force towards negative x-axis applied on 5kg object. Find:- A) the net force on the ob

ject and B) the acceleration of the object​
Physics
1 answer:
Ne4ueva [31]2 years ago
5 0

Answer:

a. Net force = 40 Newton.

b. Acceleration = 8m/s²

Explanation:

Given the following data;

Force on postive x-axis = 90N

Force on negative x-axis = 50N

Mass = 5kg

a. To find the net force;

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

The net force is equal to the difference between the two forces applied because they are acting in opposite directions.

Net force = 90 - 50

Net force = 40N

b. To find the acceleration;

Acceleration = net force/mass

Acceleration = 40/5

Acceleration = 8m/s²

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I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

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frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

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    y₂ = v₀₂ t ’+ ½ g t’²

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    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

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