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3 years ago

Help Please! Been stuck for some time

Physics

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other

Explanation:

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How many types of waves are there? What are the names?

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Explanation:

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3 years ago

Suppose that two sources of sound produce waves of the same exact amplitude and frequency, but are out of phase by one-half of a

tia_tia [17]

Answer:

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3 years ago

Read 2 more answers

A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .

Ray Of Light [21]

A) Net force in component form: $F=100.0i-300.0j$

B) Magnitude of the net force: 316.2, direction: $-71.6^{\circ}$

Explanation:

The three forces given in this problem are:

$F_1=300i+500j$

$F_2=-200i$

$F_3=-800 j$

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

$F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100$

- Along the y-direction:

$F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300$

So, the net force in component form is

$F=100.0i-300.0j$

The magnitude of a vector F is given by Pythagorean's theorem:

$|F|=\sqrt{F_x^2+F_y^2}$

where in this problem,

$F_x=100$ is the x-component

$F_y=-300$ is the y-component

Substituting,

$|F|=\sqrt{(100)^2+(-300)^2}=316.2$

The direction instead is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}$

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

brainly.com/question/4945130

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#LearnwithBrainly

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3 years ago

A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force

vivado [14]

Answer:

a) 57.0 kg b) 24.2 m

Explanation:

a) According Newton's second law, the applied force is equal to the product of the mass times the acceleration.

As the force is constant, the acceleration is constant too.

In this case, as we have as givens the distance and the time, and also we know that the block is starting form rest, we can get the acceleration as follows:

d = 1/2 * a * t² ⇒ a = 2d / t² ⇒ a= 2* 13.0 m / (4.5)² s² = 1.28 m/s²

Replacing in the Newton's 2nd Law equation:

F = m*a ⇒ m = F/a = 73.0 N / 1.28 m/s = 57.0 Kg

b) At t=4.5 s, applying the definition of acceleration, we can get the value of the velocity at that time, as follows:

v= a* t = 1.28 m/s * 4.5 s = 5.76 m/s

If the worker stops pushing at the end of the 4. 5 s, this means (neglecting friction) that from that time omwards, no net force acts on the block, so it continues moving at constant speed.

In order to get the distance moved in the next 4.20 sec, as it is moving at constant speed, we neeed just to apply the definition of velocity:

v= Δx / Δt ⇒ Δx = v* Δt = 5.76 m/s * 4.2 m = 24.2 m

So, the total distance traveled during all the time (9.1 s) is just the sum of the 13.0 m advanced during the time when there was a constant force applied, and the last 24.2 m at constant speed, as follows:

d = 13.0 + 24.2 = 37.2 m

3 0

3 years ago