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2 years ago

Suggest one reason why the bricklayer needs a higher energy diet than the computer operator

Physics

1 answer:

sashaice [31]2 years ago

7 0

Answer:

he needs more because he is doing more exercise this means he is working his body more and he needs more energy than someone he is sitting in an office. The computer operator would not need as much energy as the brick layer.

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Karen is running forward at a speed of 9 m/s. She tosses her sweaty headband backward at a speed of 20 m/s. The speed of the hea

Komok [63]

Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s

The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s

The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s

Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.

8 0

2 years ago

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Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father

-Dominant- [34]

<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>

6 0

2 years ago

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At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, its water level rises to 20.3 ml

pogonyaev

If you are asking what the volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.

3 0

2 years ago

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A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude

enot [183]

The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity $v_x=v_0 cos 30^{\circ}$ , where $v_0=35 m/s$
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ directed downwards, and with initial velocity $v_y=v_= sin 30^{\circ}$ . Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.

5 0

2 years ago