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3 years ago

Can you make the work output of a machine greater than the work input?

1 answer:

4 0

Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.

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A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago

A force of 9.00 newtons acts at an angle of 19.0° to the horizontal. What is its component in the horizontal direction?

g100num [7]

It would be helpful if you draw the figure of the problem. You will see that a right triangle would be constructed by the problem where 19.0 is the angle between the hypotenuse and the base of the triangle. It is said that the force acting is said to be 9.0 N at the said angle to the horizontal. Using trigonometric relations,

cos 19 = adjacent / hypotenuse = horizontal component / 9
horizontal component = 8.51 N

5 0

3 years ago

Read 2 more answers

What is meant by infinite slew rate

Dennis_Churaev [7]

Am infinite slew rate means that the changes in the output voltage occur immensely when the input voltage changes.
Slew rate is measurement of the response of an operational amplifier. For an ideal operational amplifier, time delay is negligible. Hence it has an infinite slew rate.
In simpler terms it means that it can provide output voltage simultaneously with the input voltage changes.

Hope this helps :)

3 0

3 years ago

A particle passes through the point P=(−3,1,0)P=(−3,1,0) at time t=3t=3, moving with constant velocity v⃗ =⟨5,3,−2⟩v→=⟨5,3,−2⟩.

eimsori [14]

Answer:

The parametric equation for the position of the particle is $(-18+5t,-8+3t,6-2t)$ .

Explanation:

Given that,

The point is

$P=(-3,1,0)$

Time t = 3

Velocity $v=(5,3,-2)$

We need to calculate the parametric equation for the position of the particle

Using parametric equation for position

$r(t)=r_{0}+v(t)t$ ....(I)

at t = 3,

$P=r(t)$

Put the value into the formula

$(-3,1,0)=r_{0}+(5,3,-2)\times3$

$(-3,1,0)=r_{0}+(15,9,-6)$

$r_{0}=(-18,-8,6)$

Put the value of r₀ in equation (I)

$r(t)=(-18,-8,6)+(5,3,-2)t$

$r(t)=(-18+5t,-8+3t,6-2t)$

Hence, The parametric equation for the position of the particle is $(-18+5t,-8+3t,6-2t)$ .

3 0

3 years ago

When an object is at rest, it might have several forces acting on it. however,?

SSSSS [86.1K]

For the object to be moving, there must be a value for the forces acting on this object in a certain direction to cause the object to move.

On the other hand, when an object is at rest, despite the possibility of having several forces acting on it, these forces balance out to a net force of zero which keeps the body still with no motion.

4 0

3 years ago