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yarga [219]
3 years ago
7

I need help in physics please help me

Physics
1 answer:
artcher [175]3 years ago
3 0

The first step is to represent the vectors shown in the image in Cartesian coordinates.

For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

To write this vector in Cartesian coordinates we must find its component in x (address i) and in the y axis.

x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)

So:

C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)

So:

B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j

Finally the sum of B + C is made component by component in the following way:

F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j

Finally the magnitude of f is:

|F| = \sqrt{(-2,898)^2 + (-7.5)^2}

| F | = 8.04

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When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do
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128.9 N

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F=\frac{\Delta p}{\Delta t}

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The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

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Substituting into the original equation, we find the force exerted on the golf ball:

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7 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
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

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    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

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    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

3 0
3 years ago
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