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dmitriy555 [2]
3 years ago
14

How can having a mentor help individuals achieve their goals? A. Mentors can increase accountability. B. Mentors can increase mo

tivation. C. Mentors can identify areas that need additional work. D. all of the above Please select the best answer from the choices provided. A B C D
Physics
2 answers:
alexgriva [62]3 years ago
8 0

The answer is D. all of the above

lana [24]3 years ago
6 0

Your best bet is to go with D.

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The velocity of an object includes its speed and what
SCORPION-xisa [38]
I believe that the correct answer you are looking for is the distance traveled 

5 0
3 years ago
Read 2 more answers
At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values
denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2

but we know that at resonance X_L=X_C  

putting  X_L=X_C in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}  so from given expression it is clear that quality factor depends on R L and C

3 0
3 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

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
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
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