If a car is moving with a velocity v, the distance the car takes to come to a halt is ___________ the distance that same car would take if it was starting with a velocity 2v.

Answer:

D

Explanation:

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction.

When a bus suddenly stops, a passenger tends to keep moving forward due to inertia (Newton's first law). The force causing you to fall forward in this instance is the insufficient static friction between you and the floor of the bus, which isn't enough to counteract your forward momentum.

The force acting on you that causes you to fall forward when the bus comes to an immediate stop is B) the force due to static friction between you and the floor of the bus. This is because of Newton's first law, also known as the law of inertia, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless acted upon by an external force.

While you're standing in the moving bus, both you and the bus are moving forward. When the bus suddenly stops, your body tends to keep moving forward due to inertia, causing you to fall forward. Here, the static friction between you and the floor of the bus isn't enough to counteract your momentum, leading to your forward fall.

It's important to note that other forces such as gravity and normal force are in effect too. Gravity pulls you downward, and the bus floor exerts an upward normal force to counterbalance it. These two forces are balanced and don't contribute to your forward movement. The force causing the unbalanced motion (falling forward) is the lack of sufficient friction to oppose your inertia.

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Answer:

27000 Nm

Explanation:

The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m

M = FL = 3000 * 9 = 27000 Nm

So the moment about A is 27000 Nm

Geographically throughout this area of Mexico, Central America Caribbean is located the Cocos plate. This area is scientifically known as the Central American subduction zone.

In order for a volcano to form, there is usually a clash between the technical plates that generates the elevation of the ground and the connection with ducts that release the magma from the earth. If this entire area is a subduction area, it will also be a land stress release area where volcano lines will be formed, that is, it is a convergent plate boundary area

The relationship between the epicenters of the earthquake and the position of the volcano should be explained below.

When the Mexico area should be geographically located, so the Caribbean of central America should be located on the Cocos plate. So this area should be called as the subduction zone. At the time when the volcano should be created so there is normally clash that lies between the technical plats where it generated the ground elevation and the linked with the ducts due to this, it releases the magma from the earth.

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Answer:

The calculated vectors are:

[tex]\vec{A}-\vec{B}=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=(-3,-2,-11)[/tex]

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

[tex]\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)[/tex] (1)

[tex]\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)[/tex] (2)

[tex]\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)[/tex] (3)

The operations in these cases are:

[tex]\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)[/tex]

Answer:

Part A: (3, -5, -4)

Part B: (-5, 4, 0)

Part C: (-6, 5, 3)

Part D: (-3, -2, -11)

Part E: (17, -12, -6)

Explanation:

This problem involves addition and subtraction of vectors. This can be done by adding and subtracting the respective components of each vector as the case may be.

The full descriptive solution can be found in the attachment below.

Answer:

Electric field strength=8.81×10⁶V/m

Explanation:

Given Data

voltage v= 74.0 mV

Membrane thickness d=8.40 nm

To find

Electric field strength E=?

Solution

Electric field strength =voltage/Membrane thickness

[tex]E=v/d\\E=\frac{74.0*10^{-3} }{8.40*10^{-9} }\\ E=8.81*10^{6}V/m[/tex]

Answer:

Please refer to the attachment below since we need to prove that the period of motion is 2π*(sqrt(h/g))

Explanation:

Answer:

The Proof for T=2π (sqrt(h/g) for a floating block exhibiting SHM is shown in the pictures attached below

Explanation:

Answer:

a) the mole fraction of air in the exiting stream is Xa=0.25 (25%) and for methane  Xm=0.75 (75%)

b) the volumetric flow rate is 49.33 L/s

Explanation:

Assuming ideal gas behaviour, then

for air

Pa*Va=Na*R*Ta

for methane

Pm*Vm=Nm*R*Tm

dividing both equations

(Pa/Pm)*(Va/Vm)= (Na/Nm)*(Ta/Tm)

Na/Nm = (Pa/Pm)*(Va/Vm) * (Tm/Ta) = (0.2/0.2)*(20/5)*(300/400) =  1*4*3/4 = 3

Na=3*Nm

therefore the moles of gas of the outflowing stream are (assuming that the methane does not react with the air):

Ng= Na+Nm = 4*Na

the mole fraction of A is

Xa= Na/Ng= Na/(4*Na) = 1/4 (25%)

and

Xm= 1-Xa = 3/4 (75%)

also for the exiting gas

Pg*Vg=Ng*R*Tg  = Na*R*Tg + Nm*R*Tg = Pa*Va * (Tg/Ta) + Pm*Vm * (Tg/Tm)

Vg = Va * (Pa/Pg)*(Tg/Ta) + Vm *(Pm/Pg)* (Tg/Tm)

Vg = 20 L/min * (0.2/0.1)*(370/400) + 5 L/min * (0.2/0.1)*(370/300) = 49.33 L/s

To develop this problem we will start using the concept of maximum speed for this type of systems. The maximum velocity can be described as the product between the Amplitude and the Angular velocity. At the same time, said angular velocity can be found through the relationship between linear and "angular wavenumber" velocity. The Angular wavenumber is a wave number defined as the number of radians per unit distance. Finally with the value of the angular velocity found we will proceed to find the maximum speed.

The maximum speed is given by

[tex]v_{max} = A\omega[/tex]

Here,

A = Amplitude

[tex]\omega[/tex]= Angular velocity

The angular velocity can be described as the number of radians per unit distance

[tex]\omega = vk[/tex]

[tex]\omega = v (\frac{2\pi}{\lambda})[/tex]

[tex]\omega = 112(\frac{2\pi}{12.16})[/tex]

[tex]\omega =57.8714rad/s[/tex]

Then,

[tex]v_{max} = 0.12 *57.8714[/tex]

[tex]v_{max} = 6.94m/s[/tex]

Therefore the maximum speed a point on the medium moves as this wave passes is 6.94m/s

Answer:

Explanation:

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.

Answer:

5.994 kgm/s

Explanation:

Momentum: This can be defined as the product of the mass of a body to the velocity of that body. The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv.................................... Equation 1

Where M = momentum of the baseball, m = mass of the baseball, v = velocity of the baseball.

Given: m = 155 g = (155/1000) kg = 0.155 kg. v = 87 miles per hour = 87(1600/3600) m/s = 38.67 m/s.

Substituting into equation 1

M = 0.155(38.67)

M = 5.994 kgm/s.

Thus the momentum of the baseball = 5.994 kgm/s

The magnitude of the momentum of a baseball with a mass of 155 g thrown at 87 mph is approximately 6.03 kg·m/s after converting units and applying the momentum equation.

To calculate the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour, we use the formula p = mv, where p is the momentum, m is the mass in kilograms, and v is the velocity in meters per second. First, we need to convert the mass of the baseball from grams to kilograms (155 g to 0.155 kg) and the speed from miles per hour to meters per second (87 mph to 38.88 m/s, using the conversion factor 1 mile = 1600 meters and 1 hour = 3600 seconds).

Then, we substitute the values into the momentum equation to find the momentum: p = (0.155 kg)(38.88 m/s) = 6.0264 kg·m/s. Therefore, the magnitude of the momentum of the baseball is approximately 6.03 kg·m/s when rounded to two decimal places.

Answer:

Explanation:

Given

Area of capacitor Plates [tex]A=L\times L[/tex]

distance between plates is d

capacitance C is given by

[tex]C=\frac{\epsilon A}{d}[/tex]

[tex]C=\frac{\epsilon \cdot L^2}{d}[/tex]

Provided V is Voltage

[tex]Charge(Q)=capacitance(C)\times Voltage(V)[/tex]

If L is doubled

Capacitance [tex]C'=\frac{\epsilon \cdot (2L)^2}{d}[/tex]

[tex]C'=4\times \frac{\epsilon \cdot L^2}{d}[/tex]

Electric field is given by

[tex]E=\frac{Q}{\epsilon _0A}[/tex]

[tex]E_i=\frac{Q}{\epsilon _0L^2}---1[/tex]

[tex]E_f=\frac{Q}{\epsilon _0(2L)^2}---2[/tex]

divide 1 and 2 we get

[tex]\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}[/tex]

[tex]\frac{E_f}{E_i}=\frac}{1}{4}[/tex]

Part of the problem involves not only answering the question with the correct degree of accuracy, but also making approximations to the correct answer. For example, for the first case we know that the speed is equivalent to the distance traveled in a given time. Therefore it would be defined as

]1)  [tex]V = \frac{x}{t}[/tex]

x = Displacement

t = Time

The displacement value is 100km and the time value is 5 hours. Therefore the speed value would be

[tex]V = \frac{100km}{5h} = 20km/h[/tex]

We know that the answer margin is within 5% of the value, then 5% of 20 would be 1.   That is, we have a margin of error of '1km / h' to answer the question. Any value that falls within that range can be added or subtracted from the response and the response will be valid. Values included within this value would be

[tex]V_1 = 20km/h +1km/h = 21km/h[/tex]

[tex]V_2 = 20km/h - 1km/h = 19km/h[/tex]

[tex]V_3 = 20km/h + 0.5km/h = 20.5km/h[/tex]

[tex]V_4 = 20km/h -0.5km/h = 19.5km/h[/tex]

[tex]V_5 = 20km/h +0.01km/h = 20.01km/h[/tex]

2) For the second case the margin of tolerance for the response is 5%, so if we multiply the given value we would have a response of.

[tex]x = 6.5*2 = 13[/tex]

5% of 13 is 0.65. Therefore, any value that falls within that range will be a correct answer. The value could then be

[tex]x_1 = 13+0.65 = 13.65[/tex]

[tex]x_2 = 13-0.65 =12.35[/tex]

Incorrect values will be

[tex]x_3 = 13+1 = 14[/tex]

[tex]x_4 = 13-1=12[/tex]

When you slide a coin across the floor, it slows down and eventually stops due to the force of friction. The friction converts the coin's kinetic energy into thermal energy, resulting in an increase in temperature.

When you slide a coin across the floor, it eventually slows down and stops due to the force of friction acting on it. Friction is a force that opposes the motion of objects in contact, and it causes the coin to lose kinetic energy, slowing it down. As the coin slows down, its kinetic energy is converted into thermal energy, increasing the temperature of the coin and the surface it slides on. This is why a sensitive thermometer shows an increase in temperature when the coin slides.

Answer:

Explanation:

Given

Charge on metal sphere [tex]Q=12\mu C[/tex]

no of electrons [tex]n=5.9\times 10^{13}[/tex]

Charge on each electron [tex]q=-1.6\times 10^{19}\ mu C[/tex]

Charge by Possessed by Electrons [tex]Q_2=-1.6\times 10^{19}\times 5.9\times 10^{13}[/tex]

[tex]Q_2=-9.44\mu C[/tex]

Net Charge on Sphere [tex]Q_{net}=Q+Q_2[/tex]

[tex]Q_{net}=12-9.44=2.56\mu C[/tex]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]

Part b

[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]

(a) The required average speed of the truck during collision is 11 m/s.

(b) The required time interval for the collision is 0.058 s.

(c)  The required magnitude of the average force exerted by the wall on the truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e) The required approximations are:

The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.

Given data:

The mass of the truck is, m = 2100 kg.

The speed of the truck is, v = 22 m/s.

The distance crumpled by the truck is, d = 0.62 m.

(a)

Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,

[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]

Thus, the required average speed of the truck during collision is 11 m/s.

(b)

Now, apply the first kinematic equation of motion as,

[tex]v' = v + at[/tex]

Here, a is the linear acceleration and t is the time interval for the collision.

Solving as,

[tex]0 = 22 + at\\\\ a = -22/t[/tex]

Now, apply the second kinematic equation as,

[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]

Thus, we can conclude that the required time interval for the collision is 0.058 s.

(c)

The expression for the magnitude of average force exerted by wall on truck is,

[tex]F_{av.} = \dfrac{mv}{t}[/tex]

Solving as,

[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]

Thus, the required magnitude of the average force exerted by the wall on truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d)

The ratio of force of the truck and the gravitation force on the truck is,

[tex]= \dfrac{F_{av}}{mg}[/tex]

Here, g is the gravitational acceleration.

Solving as,

[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]

Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e)

The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.

This is because:

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Answer:

The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude [tex]5.92\times10^4\ N/C[/tex] points straight down and if the mass of the droplet is [tex]2.93\times10^{-15} kg[/tex]

We need to calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2[/tex]

[tex]a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}[/tex]

[tex]a=2.688\ m/s^2[/tex]

We need to calculate the charge carried by the droplet

Using formula of electric filed

[tex]E=\dfrac{F}{q}[/tex]

[tex]q=\dfrac{ma}{E}[/tex]

Put the value into the formula

[tex]q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}[/tex]

[tex]q=1.330\times10^{-19}\ C[/tex]

Hence, The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

Answer: the maximum compression force = 59.47 kip

the minimum compression force = 43.33 kip

Explanation:

In this we required to do force balance in order to get maximum and minimum compression force in the column.

the picture below explains the steps to solve the question with diagrams to ease understanding.

Free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that can be performed by a thermodynamic system at a constant temperature and pressure. It is fulfilled that if the energy change is less than zero it will mean that the relationship will proceed towards the product, while if the relationship is greater than zero the reaction will proceed towards the reactant. Therefore the correct option is D.

The free energy change of a reaction can determine the reaction direction

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge [tex]q=5\sin4\pi t\ mC[/tex]

Voltage [tex]v=3\cos4\pi t\ V[/tex]

Time t = 0.3 sec

We need to calculate the current

Using formula of current

[tex]i(t)=\dfrac{dq}{dt}[/tex]

Put the value of charge

[tex]i(t)=\dfrac{d}{dt}(5\sin4\pi t)[/tex]

[tex]i(t)=5\times4\pi\cos4\pi t[/tex]

[tex]i(t)=20\pi\cos4\pi t[/tex]

(a).We need to calculate the power delivered to the element

Using formula of power

[tex]p(t)=v(t)\times i(t)[/tex]

Put the value into the formula

[tex]p(t)=3\cos4\pi t\times20\pi\cos4\pi t[/tex]

[tex]p(t)=60\pi\times10^{-3}\cos^2(4\pi t)[/tex]

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})[/tex]

Put the value of t

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})[/tex]

[tex]p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)[/tex]

[tex]p(t)=187.68\ mW[/tex]

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

[tex]E(t)=\int_{0}^{t}{p(t)dt}[/tex]

Put the value into the formula

[tex]E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}[/tex]

[tex]E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}[/tex]

[tex]E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}[/tex]

[tex]E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)[/tex]

[tex]E(t)=57.52\ mJ[/tex]

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

To solve the problem, the power at a certain moment (t=0.3s) is calculated by substitifying the values of voltage and charge into the power formula. The energy delivered to the element can be calculated by integrating the power function over the period from 0 to 0.6 s.

The subject of this question is related to electrical power and energy in a circuit, and it requires a knowledge of trigonometry. The instantaneous power in an electrical circuit is given by the product of charge (q) and voltage (v). Hence we can calculate the power at t = 0.3s by substituting the given values into the power formula.

Power, p = qv = (5 sin 4πt) * (3 cos 4πt) = 15 sin 4πt cos 4πt. At t = 0.3s, power p = 15 sin 4π * 0.3 cos 4π * 0.3 = 15 sin 1.2π cos 1.2π.

For part (b), the energy delivered to the element between 0 and 0.6 s can be obtained from the integral of the power function over the given time interval, which requires integration skills in calculus.

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Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 [tex]d_1 = u t +\dfrac{1}{2}at^2[/tex]

 [tex]d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2[/tex]

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

Answer:

a) 73.48 m/s

b) 313.92 m

Explanation:

Data provided in the question:

ascending velocity = - 5 m/s      [ negative sign depicts upward movement]

Time taken by bag to hit the ground, t = 8 s

a) from the Newton's equation of motion

we have  

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

where,  

u is the initial speed  

a is the acceleration = 9.81 m/s²   (since it is a case of free fall )

s is the distance

thus,

[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]

s = - 40 + 313.92

s = 273.92 m

from

v = u + at

v is the final speed

v = -5 + (9.81)(8)

or

v = 73.48 m/s

b) Distance traveled by balloon  = Speed × Time

= 5 × 8

= 40 m

Therefore,

Altitude of the balloon

= Distance traveled by bag + Distance traveled by balloon

= 273.92 m + 40 m

= 313.92 m

The speed of the bag as it hits the ground is 5 m/s. The altitude of the balloon when the bag hits the ground is 40 meters.

To determine the speed of the bag as it hits the ground, we can use the equation for velocity:

v = u + at

Since the bag is dropped with an initial velocity of 5 m/s and there is no acceleration in the vertical direction, the final velocity of the bag as it hits the ground is also 5 m/s.

To determine the altitude of the balloon when the bag hits the ground, we can calculate the distance traveled by the bag using the equation for distance:

s = ut + (1/2)at^2

Plugging in the values, we get:

s = 5(8) + (1/2)(0)(8^2) = 40 meters

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Answer:

Yes it is possible to charge balloon to several thousand of volts and the balloon will also store several Joules of energy.

Explanation:

By rubbing a balloon on one's hair or on different types of clothing, the balloon either gain or loss electron.

If the balloon gains electron it becomes negatively charged, it contains more electron and subsequently charged to several thousand volts.

Also, if the balloon losses electron, it becomes positively charged. In this case it contains more proton, which makes the balloon positively charged to several thousand volts.

However, amount of joules depends on volts produced in the balloon.

Volt = Joules/coulomb,

Joules = volts*coulomb

Because the charge of the particles (electron and proton) are small, amount of joules will always be small than volts.

So, it is possible to charge balloon to several thousand of volts and the balloon will store several Joules of energy

Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

The change in the sound intensity level in dB is -13.1 dB.

The given parameters;

The relationship between intensity of sound and distance is calculated as follows;

[tex]I = \frac{k}{r^2} \\\\I_1r_1^2 = I_2r_2^2\\\\I_2 = \frac{I_1 r_1^2 }{r_2^2} \\\\I_2 = \frac{I_1 r_1^2}{(4.49r_1)^2} \\\\I_2 = \frac{I_1r_1^2}{20.16r_1^2} \\\\I_2 = \frac{I_1}{20.16} \\\\\frac{I_2}{I_1} = \frac{1}{20.16}[/tex]

The change in sound intensity in dB is calculated as follows;

[tex]\Delta \beta = 10 \ log[\frac{I_2}{I_1} ]\\\\\Delta \beta = 10 \times log [\frac{1}{20.16} ]\\\\\Delta \beta = -13.1 \ dB[/tex]

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Answer:

(a). 12 plants

(b). 3171 $

Explanation:

(a)first convert units of 100 billion kWh/year into Watts(W)

also convert the units of 1000 MW into Watts(W)

1 billion = 10^9

1 year = 365*24 = 8760 hrs

so

100 billion kWh/year = 1[tex]\frac{100*(10^9)*(10^3)}{8760}[/tex]

                                  = [tex]1.142*10^{10}[/tex]W

1000 MW                  = [tex]1000*10^{6} = 10^{9}W[/tex]

no. of plants = [tex]\frac{1.14155*10^{10} }{10^9}[/tex] = 11.4

So 12 plants required        

(b)

savings = unit price*total units

             = [tex]0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )[/tex]

             = 3170.9 =3171 $

Answer:

a) Number of generating plants N = 11.42

That means N > 11

N = 12

b) annual savings S = $1×10^10

S = $10 billion

Explanation:

Given;

Amount of energy to be saved A=100 billion kWh/year

Capacity of each generating plant C= 1000 MW

Rate in dollars of cost of electricity R= $0.10/kWh

The number N of generating plants with capacity C that can supply the Amount A of of energy cam be given as;

N = A/C ......1

And also the Annual savings S in dollars if the rate of electricity cost R is used and amount of energy A is saved is:

S = AR .....2

But we need to derive the value of A in Watts

A = 100 billion kWh/year

There are 8760hours in a year,

A = 1×10^14 ÷ 8760 W

A = 11415525114.1W or 11415525.1141kW

C = 1000MW = 1× 10^9 W

a) Using equation 1,

N = 11415525114.1/(1×10^9)

N = 11.42

That means N > 11

N = 12

b) using equation 2

S = 1×10^11 kWh × $0.10/kWh

S = $1×10^10

S = $10 billion

Answer:

[tex]1.67\times 10^{-27}kg[/tex]

Explanation:

We are given that mass of proton

[tex]1.672623\times 10^{-27}kg[/tex]

There are seven significant figures.

We have to round off.

If we round off to three  significant figures

The  thousandth place  of given mass of proton  is less than five therefore, digits on left side of  thousandth  place remains same and digits on right side of thousandth place and thousandth  replace by zero

Therefore, the mass of proton can be written as

[tex]1.67\times 10^{-27}kg[/tex]

Hence, the mass of proton is [tex]1.67\times 10^{-27}kg[/tex]

To round the mass of a proton to seven significant figures, the correct way is to round up the last significant figure if it is 5 or greater, and if it is less than 5, simply drop the remaining digits.

To round the mass of a proton, we look at the digit right after the desired number of significant figures, which in this case is the seventh significant figure. If this digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop the remaining digits.

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Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

[tex]v = \dfrac{2\pi R}{T}[/tex]

[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]

  v = 11.69 m/s

now, Force does the ring push on her at the top

[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]

[tex] N + m g = \dfrac{mv^2}{R}[/tex]

[tex] N = \dfrac{mv^2}{R}- m g[/tex]

[tex] N = m(\dfrac{v^2}{R}- g)[/tex]

[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

It is Given that

Time rotation  T= 4.30 s

Mass m= 58 kg

Now the Velocity will be calculated as

[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]

Now by balancing the forces

[tex]N=\dfrac{mv^2}{R} -mg[/tex]

[tex]N=m(\dfrac{v^2}{R} -g)[/tex]

[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]

[tex]N=422.36 \ Newton[/tex]

Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

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To solve this problem we will apply the laws of gaus that relate the Electric Flow as the charge of the object on the permittivity constant of free space. Mathematically this is

[tex]\phi = \frac{Q}{\epsilon_0}[/tex]

Rearranging to find the charge,

[tex]Q = \phi \epsilon_0[/tex]

Here

Q = Charge

[tex]\phi =[/tex] Electric Flux

[tex]\epsilon_0 =[/tex] Permittivity of free space

The total flux would be

[tex]\phi_T = \phi_1+\phi_2+\phi_3+\phi_4+...+\phi_{\infty}[/tex]

[tex]\phi = ( 1500+2200+4600-1800-3500-5400 ) N\cdot m^2 / C[/tex]

[tex]\phi = - 2400 N\cdot m^2 / C[/tex]

Replacing we have that,

[tex]Q = (-2400 N\cdot m^2/C)( 8.85*10^{-12} C^2 / N \cdot m^2)[/tex]

[tex]Q = -21240 * 10^{-12} C[/tex]

[tex]Q = - 21.24 nC[/tex]

Therefore the charge Q inside a rectangular box is -21.24nC

Final answer:

Using Gauss's Law, the charge Q inside the box can be calculated by summing the electric flux values across all six surfaces and then multiplying by the permittivity of free space, resulting in a charge of approximately -1.24 x 10^-8 C.

Explanation:

To determine the charge Q inside the rectangular box, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed within that surface. The net electric flux (Φnet) through a closed surface is equal to the charge inside (Q) divided by the permittivity of free space (ε0). Mathematically, this is expressed as Φnet = Q/ε0.

The net flux is the algebraic sum of the fluxes through each surface, so we have:

Φnet = (electric flux 1) + (electric flux 2) + (electric flux 3) + (electric flux 4) + (electric flux 5) + (electric flux 6)

Substituting the given values, we get:

Φnet = (+1500 N·m2/C) + (+2200 N·m2/C) + (+4600 N·m2/C) + (-1800 N·m2/C) + (-3500 N·m2/C) + (-5400 N·m2/C)

Summing these yields:

Φnet = -1400 N·m2/C

Assuming ε0 is the permittivity of free space (approximately 8.854 x 10-12 C2/N·m2), we can find Q:

Q = Φnet ε0

Q = (-1400 N·m2/C)(8.854 x 10-12 C2/N·m2)

Q ≈ -1.24 x 10-8 C

Thus, the charge Q inside the box is approximately -1.24 x 10-8 C.

The speed in the second segment is 36 m/s, in the third segment is 144 m/s, and the volume flow rate through the pipe is 3.38 L/s.

A: Using the principle that the volume flow rate remains constant in an incompressible fluid, we can calculate the speed in the second segment to be 36 m/s and the speed in the third segment to be 144 m/s.

B: The volume flow rate through the pipe can be calculated by applying the equation Q = Av, where Q represents the flow rate, A is the cross-sectional area, and v is the velocity of the fluid. Given the diameter changes, the volume flow rate is 3.38 L/s.

C: The primary concept involved in this problem is the relationship between cross-sectional area and fluid velocity as the diameter changes along the pipe, affecting the flow rate.

Answer

given,

near point = 18 cm

far point = 40 cm

a) The lens should form an upright, virtual image at far point from the distant object.

therefore, f = q  = -40 cm = -0.4 m

where f is the focal length.

the required power

[tex]P =\dfrac{1}{f}[/tex]

[tex]P =\dfrac{1}{-0.40}[/tex]

       P = -2.5 D

b) If the lens is used the Person's near point

The lens should form an upright, virtual image at near point from the distant object should be q = - 18 cm = -0.18 m

    [tex]p = \dfrac{qf}{q-f}[/tex]

    [tex]p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}[/tex]

     p = 32.72 cm

The person's near point is 32.72 cm