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3 years ago

What type of device forms images by changing the speed at which light

Physics

2 answers:

snow_tiger [21]3 years ago

8 0

Answer:

It's Lens

Explanation: I did the assignment

Nady [450]3 years ago

3 0

Answer:

Any Lens

Explanation:

I Hope it's right if not so Sorry :)

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I need some help with #2 and #3 please. Thank you.

Serjik [45]

Well for number 2, the difference is one tells people to get ready, and one tells people it's there already.(mainly watch is a better version of a warning.

6 0

3 years ago

an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o

gladu [14]

The output of the machine is

(output work) = (output force) x (distance)

450 N-m = (output force) x (3 meters)

Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)

= 150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.

5 0

3 years ago

A 15,000 kg rocket traveling at +230 m/s turns on its engines. Over a 6.0 s period it burns 1,000 kg of fuel. An observer on the

lesya692 [45]

Answer:

a) $v = 312.791\,\frac{m}{s}$ , b) $a = 13.333\,\frac{m}{s^{2}}$

Explanation:

The problem is asking the rocket velocity and acceleration at t = 6 s.

a) The general equation of the rocket is:

$v=v_{o} -v_{ex}\cdot \ln \frac{m}{m_{o}}$

$v = 230\,\frac{m}{s}-(1200\,\frac{m}{s} )\cdot \ln \frac{14000\,kg}{15000\,kg}$

$v = 312.791\,\frac{m}{s}$

b) The acceleration experimented by the rocket is:

$a = \frac{v_{ex}}{m_{o}}\cdot \frac{dm}{dt}$

$a = \frac{1200\,\frac{m}{s} }{15000\,kg}\cdot \frac{1000\,kg}{6\,s}$

$a = 13.333\,\frac{m}{s^{2}}$

3 0

4 years ago

Read 2 more answers

a water line starts the service with an altitude of 1200m over the sea level, what is the velocity of the water above 1050 m ove

Ilia_Sergeevich [38]

Answer:

Velocity = 94.85m/s

Explanation:

Given the following data ;

Height = 1200m

Vertical distance = 1050m

To find the time, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting into the equation, we have;

$1200 = 0(t) + \frac {1}{2}*9.8*t^{2}$

$1200 = 0 + 4.9*t^{2}$

$1200 = 4.9*t^{2}$

$t^{2} = \frac {1200}{4.9}$

$t = \sqrt{122.45}$

t = 11.07 secs

To find the velocity;

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the above equation;

$Velocity = \frac{1050}{11.07}$

Velocity = 94.85m/s

Therefore, the velocity of the water above 1050 m over the sea level is 94.85m/s.

7 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago