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a 10 kg block is attached to a light chord that is wrapped around the pulley of an electric motor. at what rate is the motor doi

defon

Answer:

The rate of work is 360 W

Explanation:

Given:

Speed of block $v = 3 \frac{m}{s}$

Upward acceleration of block $a = 2 \frac{m}{s^{2} }$

Mass of block $m = 10$ kg

First Find the force act on block due to gravity

$F = M (a + g)$

Where $g = 10 \frac{m}{s^{2} }$

$F = 10 \times (10+2)$

$F = 120$ N

For finding at what rate motor doing work when it is pulling the block upward,

$P = F.v$

$P = 120 \times 3$

$P = 360$ W

Therefore, the rate of work is 360 W

6 0

4 years ago

What does waining mean?

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(of the moon) have a progressively smaller part of its visible surface illuminated, so that it appears to decrease in size.

Explanation:

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3 years ago

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A child sits two meters from the center of a merry-go-round. How far does she travel during one revolution?

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3 0

3 years ago

9. An object is launched at a velocity 40m/s in adirection making an angle of 60° upward with the horizontal.

Orlov [11]

Answer:

a) 61.224 m

b) t = 7.070 seconds

c) horizontal component = 20 m/s; vertical component = 34.641 m/s

Note: I rounded all of these values to the nearest thousandth but if you want the precise values please read the explanation below.

Explanation:

<h2><u>Horizontal and Vertical Components:</u></h2>

Let's start this problem by solving for the horizontal and vertical components of the initial velocity vector.

We can solve for these x- and y-components by using the formulas:

h. component: $v_i \times cos \theta$
v. component: $v_i \times sin \theta$

Where $v_i$ is the initial velocity (here it's given to us: 40 m/s) and $\theta$ is the angle at which the object is launched above the horizontal (it's also given to us: 60°).

Substitute these given values into the formulas to solve for the horizontal and vertical components:

h. component = $40 \times $cos(60)$
v. component = $40 \times $sin(60)$

Input these values into a calculator and you will get:

h. component = 20 m/s
v. component = 34.641 m/s

<h2><u>Time of the Object:</u></h2>

Now we want to solve for the time t of the object before finding the maximum height of the object. In other words, the max height of the object is its vertical displacement at half of the time t we're about to find.

In order to solve for t, we can use one of the constant acceleration equations we are given in Physics. This equation is:

$v_f=v_i+at$

The time t is always solved for by using the vertical (y-direction) motion of the object in projectile motion, so therefore, we are going to be using this equation in terms of the y-direction.

$(v_f)_y = (v_i)_y + a_yt$

Time is the same regardless of the x- or y- direction.

Now, we don't necessarily know the final velocity of the projectile, but we do know its final velocity in the y-direction at the very top of the trajectory, which is 0 m/s.

We can use this to our advantage and solve for only half of the time t, then multiply it by 2 at the end to get the full time that the object is in the air.

We have already solved for $(v_i)_y$ , which is the vertical component. We know that an object in projectile motion has an acceleration of -g in the y-direction, so we use -9.8 m/s² for a.

$0=(40\times $sin(60)) + (-9.8)t$

Subtract the vertical component from both sides of the equation.

$-(40 \times $sin(60))= -9.8t$

Divide both sides of the equation by -9.8 in order to solve for t.

$\text{t}=3.534797566\ \text{seconds}$

Remember that this is only half of the time that the object spends in the air. However, this is the time that it takes for the object to reach its maximum height, which we will use later. For now, let's say that the time of the object is 2t.

$\text{2t = 7.069595132 seconds}$

<h2><u>Maximum Height of the Object:</u></h2>

In order to find the maximum height of the object, let's use another kinematic equation for constant acceleration:

$$x_f=x_i+v_it+\frac{1}{2} at^2$

Since we are still dealing with the y-direction, we can change this equation to be in terms of y.

$$(x_f)_y = (x_i)_y + (v_i)_yt + \frac{1}{2} a_yt^2$

The displacement in the y-direction, or the vertical displacement, can be modeled by subtracting $(x_i)_y$ from both sides of the equation.

$$\triangle x_y = (v_i)_yt + \frac{1}{2} a_yt^2$

In order to solve for the maximum height of the object, we want to use the time t that it takes for the object to reach its highest point, which we found was ~3.53 seconds. This is true because the object essentially follows the movement of a parabola.

We know the vertical component $(v_i)_y$ , and we know the acceleration in the y-direction is -g, so let's substitute these values into the formula for vertical displacement and solve for $\triangle x_y$ .

$$\triangle x_y = (40 \times \text{sin}(60))(3.534797566) + \frac{1}{2} (-9.8)(3.534797566)^2$
$$\triangle x_y = (122.4489796) + \frac{1}{2} (-9.8)(3.534797566)^2$
$$\triangle x _y = (122.4489796) -(4.9 \times 3.534797566^2)$
$$\triangle x_y = (122.4489796) - (61.22448978)$
$$\triangle x_y = 61.2244898$

The maximum height of this object in projectile motion is 61.224 m.

(This answer exceeded the character limit if I included the "Helpful Shortcuts" section, so I included it as an attachment in case you're interested.)