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arlik [135]
3 years ago
12

A magnifying glass or a hand lens is a bi-convex lens. It is convex on both sides, meaning that the glass is curved outward to

Physics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

Refracted

Explanation:

If on UsaTestPrep

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Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v
lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

<u>Vi = 94.64 m/s</u>

3 0
3 years ago
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car
Tanzania [10]

Answer:

first value+2nd +3rd

Explanation:

thug life and there

8 0
3 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
Write the equations of motions​
Sav [38]

Answer:

In physics, equations of motion are equations that describe the behavior of a physical system in terms of its motion as a function of time.[1] More specifically, the equations of motion describe the behaviour of a physical system as a set of mathematical functions in terms of dynamic variables. These variables are usually spatial coordinates and time, but may include momentum components. The most general choice are generalized coordinates which can be any convenient variables characteristic of the physical system.[2] The functions are defined in a Euclidean space in classical mechanics, but are replaced by curved spaces in relativity. If the dynamics of a system is known, the equations are the solutions for the differential equations describing the motion of the dynamics.

6 0
3 years ago
Read 2 more answers
Which device uses a rotating magnetic field to produce an electric current?
Kay [80]
<span>The correct answer is C) a motor.
In particular, we are talking about an AC motor, which produces an alternating current. In an AC motor, a coil is immersed in a rotating magnetic field. Due to the motion of the magnetic field,the angle between the direction of the field and the surface enclosed by the coil changes. As a result, the magnetic flux through the coil changes over time (the magnetic flux is given by:
</span>\phi=BAcos\theta<span>
where B is the intensity of the magnetic field, A is the area enclosed by the coil and </span>\theta<span> is the angle between the direction of B and the perpendicular to the plane of the coil). For Faraday-Newmann-Lenz law, this change in flux induces an electromotive force (emf) into the coil, according to:
</span>emf=- \frac{d \Phi}{dt}<span>
 where the numerator is the variation of magnetic flux and dt is the time interval. This emf in the coil produced an electrical current in the circuit.</span>
7 0
3 years ago
Read 2 more answers
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