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When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

Answers

Answer 1

Answer:

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

mass of the box, m = 5 kg
extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

$k = (mg)/(x) \n\nk = (5 * 9.8)/(0.05) \n\nk = 980 \ N/m$

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

$T = kx\n\nx = (T)/(k) \n\nx = (59)/(980) \n\nx = 0.06 \ m\n\nx = 60 \ mm$

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

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Answer 2

Answer:

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

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Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna

Answers

Answer:

-0.00152 V

Explanation:

Parameters given:

Diameter of the loop = 11 cm = 0.11m

Rate of change of magnetic field, dB/dt = 0.16 T/s

Radius of the loop = 0.055m

The area of the loop will be:

A = pi * r²

A = 3.142 * 0.055²

A = 0.0095 m²

The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:

EMF = - N * A * dB/dt

In this case, there's only one loop, so N = 1.

Therefore:

EMF = -1 * 0.0095 * 0.16

EMF = -0.00152 V

The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.

Answer:

The induced emf is 0.00152 V

Explanation:

Given data:

d = 11 cm = 0.11 m

$r=(d)/(2) =(0.11)/(2) =0.055m$

The area is:

$A=\pi r^(2) =\pi *(0.055^(2) )=0.0095m^(2)$

The induced emf is:

$E=-A(dB)/(dt) =-(0.0095)*(0.16)=-0.00152V$

The negative indicates the direction of E.

A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory.When it returns to its original level after encountering air resistance, its kinetic energy is __________.

A) more than 100 J.

B) Not enough information given.

C) less than 100 J.

D) 100 J.

Answers

To solve this problem we could apply the concepts given by the conservation of Energy.

During the launch given in terms of kinetic energy and reaching the maximum point of the object, the potential energy of the body is conserved. However, part of all this energy is lost due to the work done by the friction force due to friction with the air, therefore

$E_T = PE + KE -W_f$

The potential and kinetic energy are conserved and are the same PE = KE and this value is equivalent to 100J, therefore

$E_T = 100-W_f$

The kinetic energy will ultimately be less than 100J, so the correct answer is C.

Question 11 of 15When hydrogen is attached to the more highly electronegative oxygen atom in
a water molecule,
A. the electronegative atom becomes strongly positive
B. the hydrogen atom becomes partially positive
O C. the oxygen atom becomes partially negative

If answer is right WILL GIVE BRAINLIEST
D. the hydrogen atom becomes partially negative

Answers

Answer:

I'm leaning twards A

Explanation:

I think B but I’m not sure

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