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Varvara68 [4.7K]
4 years ago
15

You build a 2.0-kg mound of mashed potatoes on a kitchen scale (which is a spring scale). You then use a large spoon to press on

the top of the mound at a constant force of 3.0 N. While you are doing this, the scale reading increases by 5.0%. (a) What is the acceleration of the center of mass of the potatoes? (b) Describe what happens to the shape of the mound and the scale reading as time goes by.

Physics
2 answers:
igor_vitrenko [27]4 years ago
8 0

Answer:

a) a=1m/s^2 downwards

b) The mound will be reduced in height (the acceleration is downwards) and will begin to grow sideways. The reading will remain the same until the spoon reaches the scale surface.

Explanation:

On the mound, the forces acting before the spoon are:

N1 - m*g = 0   Where N1 is equivalent to the reading of the scale.

N1 = 2*10 = 20N

After the spoon starts pressing, this value increases by 5%:

N2 = 20*105% = 21N  So, at this point the forces on the moud are:

N2 - m*g - F = m*a   Solving for a:

a = -1m/s^2   This result tells us that the center of mass of the potatoes is moving downwards. Since it cannot move because of the scale, it has no other option than to grow sideways as it is pushed.

marshall27 [118]4 years ago
5 0

Answer:

a= -1.01m/s^2

Explanation:

A) We define all the variables that are given,

In this way

Acceleration = a

m = 2.0Kg

F = 3.0N

We obtain the Normal assuming 5% of the potato reading scale, so

N = (2 + 2 * 5\%) * 9.8

N = 20.58N

We add forces to reach equilibrium, and you get

N-F-mg = ma

** The notation of the symbols is done according to the notation of the graph

So,

a = \frac {N-F-mg} {m}

a = \frac {20.58-3-2 * 9.8} {2}

a = -1.01m / s ^ 2

B) When making pressure it is understood that the shape takes the perpendicular direction of the direction of the force that is performed. If the pressure remains constant, the scale will also increase in its measurement.

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Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

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E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

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6 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

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Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

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3 years ago
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Nostrana [21]

Answer:inner core?

Explanation:

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A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
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Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

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6 0
3 years ago
A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.
UkoKoshka [18]

Answer:

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Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

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b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

6 0
4 years ago
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