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2 years ago

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

1 answer:

6 0

I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?

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(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe

castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0

2 years ago

You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun

Rzqust [24]

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference = λ

so there is constructive interference and maximum sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

3 0

2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago

A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then proje

natka813 [3]

Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.

Explanation:

First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.

From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.

Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.

8 0

1 year ago

"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f

ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c = 25 ms = 25 × 10⁻³ s

The initial temperature $T_1$ = 273 K

The final temperature $T_2$ = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume = 0.4 mV/K × 473 K = 189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2 + (189.2 - 109.2)( 1 - $\mathbf{e^{-t/c}}$ )mV

y(t) = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 × 189.2 mV

= 170.28 mV

170.28 mV = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

170.28 mV - 109.2 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

61.08 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

0.7635 mV = ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

t = 1.44 × 25 × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = $\dfrac{189.2-170.28 }{189.2}$

The error fraction = 0.1

The error fraction = 10%

8 0

2 years ago