Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

Answers

Answer 1

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal


Related Questions

From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:

a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down

Answers

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?

Answers

Answer:

The time taken is  [tex]\Delta t = 1.5 \ s[/tex]

Explanation:

From the question we are told that

   The mass of the ball is  [tex]m = 1.25 \ kg[/tex]

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  [tex]\theta = 1 rev = 2 \pi \ radian[/tex]

Generally the displacement for one  complete revolution is mathematically represented as

       [tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]

Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]

     [tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]

     [tex]\alpha = 0.9698 \ s[/tex]

Now the displacement for two  complete revolution is

         [tex]\theta_2 = 2 * 2\pi[/tex]

         [tex]\theta_2 = 4\pi[/tex]

Generally the displacement for two complete revolution is mathematically represented as  

     [tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]

=>   [tex]t^2 = 25.9187[/tex]

=>   [tex]t= 5.1 \ s[/tex]

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     [tex]\Delta t = t_2 - t[/tex]

substituting values

      [tex]\Delta t = 5.1 - 3.60[/tex]

     [tex]\Delta t = 1.5 \ s[/tex]

           

 

The time for the ball to complete the next revolution is 1.5 s.

The given parameters;

mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2π

The constant angular acceleration of the ball is calculated as follows;

[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]

The time to complete the next revolution is calculated as follows;

[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]

[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]

Thus, the time for the ball to complete the next revolution is 1.5 s.

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if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?

Answers

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circumstances. (Note that currents above 10 mA lead to involuntarily muscle contraction.)
(a) if she is standing on a rubber mat and offers a total resistance of 300kΩ
(b) if she is standing barefoot on wet grass and has a resistance of only 4000kΩ

Answers

Answer:

A) 0.4 mA

B) 0.03 mA

Explanation:

Given that

voltage source, V = 120 V

to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.

mathematically, Ohms Law, V = IR

V = Voltage

I = Current

R = Resistance

from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have

120 = I * 300*10^3 Ω

making I the subject of the formula,

I = 120 / 300000

I = 0.0004 A

I = 0.4 mA

Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.

B, we have Resistance, R = 4000kΩ

Substituting like in part A, we have

120 = I * 4000*10^3 Ω

I = 120 / 4000000

I = 0.00003 A

I = 0.03 mA

This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA

The current through a person will be:

a) 0.4 mA

b) 0.03 mA

Given:

Voltage, V = 120 V

Ohm's Law:

It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.

Ohms Law, V = I*R

where,

V = Voltage

I = Current

R = Resistance

a)

Given: Resistance=  300kΩ

[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]

Thus, current will be, I = 0.4 mA

b)

Given: R = 4000kΩ

[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]

Thus, current will be, I = 0.03 mA

From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.

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The two metallic strips that constitute some thermostats must differ in:_______
A. length
B. thickness
C. mass
D. rate at which they conduct heat
E. coefficient of linear expansion

Answers

Answer:

E. Coefficient of linear expansion

The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.

Answers

Answer:

T = 80√3 N ≈ 139 N

W = 160 N

Explanation:

Sum of forces on B in the x direction:

∑F = ma

80 N sin 60° − T sin 30° = 0

T = 80 N sin 60° / sin 30°

T = 80√3 N

T ≈ 139 N

Sum of forces on B in the y direction:

∑F = ma

80 N cos 60° + T cos 30° − W = 0

W = 80 N cos 60° + T cos 30°

W = 40 N + 120 N

W = 160 N

A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?

Answers

Answer:

There is no induced current on the coil.

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.

a. True
b. False

Answers

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s

Answers

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

   

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

Calculation of Wavelength

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

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Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.

Answers

Answer:

A) 1.5 V

B) 4.5 V

Explanation:

A) Batteries in parallel have the same voltage as an individual battery.

V = 1.5 V

B) Batteries in series have a voltage equal to the sum of the individual batteries.

V = 1.5 V + 1.5 V + 1.5 V

V = 4.5 V

An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)

Answers

Answer:

10.4⁰ and 169.6⁰

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ =  1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.

1. Give a specific example of a system with the energy transformation shown.
W→ΔEth

2. Give a specific example of a system with the energy transformation shown.

a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.

Answers

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?

Answers

Answer:

50 cm long

When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.

An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.

Answers

Answer:

(A). 1620 watt.

(B).0.8885.

Explanation:

So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;

Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.

(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).

(B). the rms values to find the power efficiency Pout/Pin of the inverter.

P(in) = 195 × 9.35 = 1823.3 watt.

Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.

a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?

Answers

Answer:

8.78 Amps

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A

Hence the current when the unit/heater is operating is 8.78 Amp

A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11​

Answers

Explanation:

Change in Velocity = final velocity - initial velocity

Change in velocity = 30km/h - (- 45km/h )

= 75 km/h due west

If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​

Answers

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?

Answers

Answer:

7.86×10⁻⁷ m

Explanation:

Using,

v = λf.................. Equation 1

Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.

make λ the subject of the equation

λ = v/f............... Equation 2

Note: All electromagnetic  wave have the same speed which is 3×10⁸ m/s.

Given: f = 3.818×10¹⁴ Hz

Constant: v = 3×10⁸ m/s

Substitute these values into equation 2

λ  =  3×10⁸/3.818×10¹⁴

λ  = 7.86×10⁻⁷ m

Hence the wavelength of the electromagnetic radiation is  7.86×10⁻⁷ m

The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]

Given the following data:

Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]

Scientific data:

Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]

To determine the wavelength of this electromagnetic radiation:

Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;

[tex]Wavelength = \frac{Speed }{frequency}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]

Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]

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A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?​

Answers

Is there any other type of information?

Explanation:

Using Equations of Motion :

[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

how does a system naturally change over time

Answers

Answer:

The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.

Explanation:

Answer:

Decrease in entropy

Explanation:

Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.

In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.

On the other hand, in open system both the matter and energy move into and out of the system.

Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?

Answers

Answer: Only Tech B is correct.

Explanation:

First, tech A is wrong.

The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.

Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.

A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60

Answers

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?

Answers

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

Frequency of oscillation based problem:

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]

Frequency of oscillation = 3.15 Hz

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Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.

Answers

Answer:

se the principle of induction.

place the two metallic spheres together,  now we bring the positively charged bar closer to the first sphere.

The charge that was induced in the sphere is distributed as infirm as possible,

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive

Explanation:

For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.

Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.

Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.

The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.

Please help!
Much appreciated!​

Answers

Answer:

your question answer is 22°

I’m pretty sure the answer is 22

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).

Required:
Find the work done.

Answers

Answer:

the net work is zero

Explanation:

Work is defined by the expression

        W = F. ds

Bold type indicates vectors

In this problem, the friction force does not decrease, therefore it will be zero.

Consequently for work on a closed path it is zero.

The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero

Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form _______. A. pure elements B. molecules C. metals D. the periodic table

Answers

Answer:

Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form pure elements

The answer is option A

Answer:

its molecues

Explanation:

UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons

Answers

Answer:

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

Explanation:

First we calculate the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)

E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 10.35 eV

Now, from Einstein's Photoelectric equation we know that:

Energy of Photon = Work Function + K.E of Electron

10.35 eV = 4.82 eV + K.E

K.E = 10.35 eV - 4.82 eV

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".

Kinetic energy

According to the question,

Speed of light, c = 3 × 10⁸ m/s

Wavelength, λ = 120 nm or,

                        = 1.2 × 10⁻⁷ m

Plank's Constant, h = 6.626 × 10⁻³⁴ J.s

Now,

The energy of photon will be:

→ E = [tex]\frac{hc}{\lambda}[/tex]

By substituting the values,

     = [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]

     = [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]

     = 10.35 eV

By using Einstein's Photoelectric equation,  

Energy of Photon = Work function + K.E

                   10.35 = 4.82 + K.E

                       K.E = 10.35 - 4.82

                             = 5.53 eV or,

                             = 8.85 × 10⁻¹⁹ J

Thus the response above is correct.

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A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

Answers

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q

Answers

Answer:

Upper plate Q/3

Lower plate 2Q/3

Explanation:

See attached file

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