A car traveling at 105 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. (a) What was the magnitude of the average acceleration of the driver during the collision? (b) Express the answer in terms of “g’s,” where 1.00 g = 9.80 m/s^2.

Answer:

47.86 m

Explanation:

Consider the motion along the vertical direction and assume the motion in downward direction as negative

y = vertical displacement = - 0.025 m

a = acceleration = - 9.8 m/s²

t = time taken to strike the target below the center

[tex]v_{oy}[/tex] = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

y = [tex]v_{oy}[/tex] t + (0.5) a t²

Inserting the values

- 0.025 = (0) t + (0.5) (- 9.8) t²

t = 0.0714 sec

Consider the motion along the horizontal direction

x = horizontal displacement = horizontal distance between the end of the rifle and the bull's-eye

v₀ = velocity along the horizontal direction = 670 m/s

Since there is no acceleration along the horizontal direction, we have

x = v₀ t

inserting the values

x = (670) (0.0714)

x = 47.86 m

The horizontal distance between the end of the rifle and the bull's-eye is approximately 47.9 meters.

To solve this problem, we need to determine the horizontal distance from the rifle to the bull's-eye based on the given data.

Understand the Problem: The bullet is fired horizontally with a muzzle speed of 670 m/s and hits the target 0.025 m below the center due to gravity.

Use Kinematics: Gravity affects the vertical motion, and the horizontal motion is constant because there's no acceleration horizontally in the given conditions.

Calculate the Time it Takes for the Bullet to Drop 0.025 m:

The vertical distance (y) the bullet drops can be calculated using the equation of motion under gravity:

[tex]y = \frac{1}{2} g t^2[/tex]

Where:

Rearranging for [tex]t[/tex]:

[tex]t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 0.025}{9.8}} \approx 0.0714 \text{ seconds}[/tex]

Calculate the Horizontal Distance (x):

Now that we have the time [tex]t[/tex], we can find the horizontal distance using the constant horizontal velocity.

[tex]x = v \cdot t[/tex]

Where:

[tex]x = 670 \times 0.0714 \approx 47.9 \text{ meters}[/tex]

Answer:

a) [tex]\mu_s =0.40[/tex]

b) [tex]\mu_k =0.20[/tex]

Explanation:

Given:

Mass of the cabinet, m = 51 kg

a) Applied force = 200 N

Now the force required to move the from the state of rest (F) = coefficient of static friction ([tex]\mu_s[/tex])× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 200N  = [tex]\mu_s[/tex] × 499.8 N

⇒[tex]\mu_s[/tex] = [tex]\frac{200}{499.8}=0.40[/tex]

b) Applied force = 100 N

since the cabinet is moving, thus the coefficient of kinetic([tex]\mu_k[/tex]) friction will come into action

Now the force required to move the from the state of rest (F) = coefficient of kinetic friction ([tex]\mu_k[/tex])× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 100N  = [tex]\mu_k[/tex] × 499.8 N

⇒[tex]\mu_k[/tex] = [tex]\frac{100}{499.8}=0.20[/tex]

The problem requires the understanding of simple harmonic motion in physics. The position of an oscillating mass attached to a spring can be calculated using the SHM formulas and given parameters: weight of the mass, stretch of the spring, and acceleration due to gravity. The amplitude, spring constant, angular frequency, and phase constant are figured out before plugging them into the positional formula.

This question involves the concept of simple harmonic motion (SHM) in Physics. In this scenario, a mass attached to a spring and released from a certain position will oscillate in SHM. The position of the mass at specific times (t = π/12, π/8, π/6, π/4, and 9π/32 s) can be calculated using the SHM formulas.

Since the mass is released from 6 inches (or 0.5 feet) below the equilibrium position, this value represents the amplitude (A) of the oscillation. Additionally, we know that the spring constant (k) can be found from the weight of the mass and the stretch of the spring. This gives us k = W/d = 20 lbs / 0.5 ft = 40 lbs/ft.

The angular frequency (ω) of the oscillation can be derived from ω = sqrt(k/m), where m is the mass of the object. Using the information provided (considering g = 32 ft/s2), we find m = W/g = 20 lbs / 32 ft/ s2 = 0.625 slugs. Hence, ω = sqrt(40 lbs/ft / 0.625 slugs) = 8 rad/s.

Finally, using the formula for the position in a SHM (x = A*cos(ω*t + φ), where φ is the phase constant), and knowing that the mass was initially released from rest, we can determine that φ = 0. As a result, the position at the specified times can be calculated by substituting the corresponding values into the relation.

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To find the position of the mass at different times, use the equation for simple harmonic motion (SHM) and calculate the angular frequency. Then plug in the values and use the equation x = A * cos(ωt + φ) to find the positions.

The position x of the mass at different times can be found using the equation for simple harmonic motion (SHM):

x = A * cos(ωt + φ)

where x is the position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

In this case, the amplitude A is 6 inches and the angular frequency ω can be found using the formula ω = sqrt(k/m), where k is the spring constant and m is the mass. The spring constant k can be found using the formula k = (g * m)/A, where g is the acceleration due to gravity.

By plugging in the values and calculating ω, you can then find the positions x at the given times using the equation x = A * cos(ωt + φ).

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Answer:

4L

Explanation:

mass of pendulum = M, length of pendulum, L1 = L,

acceleration due to gravity = g, Amplitude = A

Use the formula for the time period for simple pendulum

[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]     ..... (11)

Let the new length be L2 for which the time period is 2T.

[tex]2T = 2\pi \sqrt{\frac{L_{2}}{g}}[/tex]          ..... (2)

Divide equation (2) by equation (1)

[tex]\frac{2T}{T} = \sqrt{\frac{L_{2}}{L}}[/tex]

So, L2 = 4 L

Thus, the length of the pendulum is 4 L.

Answer:

Work done by firefighter = 13456.57 J

Explanation:

Here work done is equal to potential energy gained by him.

Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity value and h is the height.

Mass, m = 64.4 kg

Acceleration due to gravity, g = 9.8 m/s²

Height, h = 21.3 m

Substituting

Potential energy, PE = mgh = 64.4 x 9.81 x 21.3 = 13456.57 J

Work done by firefighter = 13456.57 J

Answer: Real image

Explanation:

converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away).

Answer:

R = 0.27 ohm

i = 11.25 A

Explanation:

Resistance of a given cylindrical rod is

[tex]R = \rho \frac{L}{A}[/tex]

here we know that

[tex]\rho = \frac{1}{3 \times 10^4} = 3.33 \times 10^{-5} ohm-m[/tex]

Now we have

[tex]R = (3.33 \times 10^{-5})\times \frac{4 \times 10^{-3}}{0.5 \times 10^{-6}}[/tex]

[tex]R = 0.27 ohm[/tex]

now we know by ohm's law

[tex]\Delta V = i R[/tex]

[tex]14 - 11 = i ( 0.27)[/tex]

[tex]i = 11.25 A[/tex]

The airplane's speed relative to the air mass is 38.51 m/s. To find the airplane's speed relative to the air mass, you can use the formula sqrt(VAx² + VAy²) by calculating the magnitude of the resultant vector of VAx and VAy.

In order to determine the airplane's speed relative to the air mass, a comprehensive analysis involves breaking down velocities into their horizontal and vertical components.

The wind velocity components are established as VWx = magnitude of wind velocity * cos(angle of wind direction) in the 'x' direction and VWy = magnitude of wind velocity * sin(angle of wind direction) in the 'y' direction.

Simultaneously, the airplane's velocity relative to the air mass components are defined as VAx = magnitude of airplane velocity * cos(angle of travel direction relative to air) in the 'x' direction and VAy = magnitude of airplane velocity * sin(angle of travel direction relative to air) in the 'y' direction.

The speed of the airplane relative to the air mass is ascertained by calculating the magnitude of the resultant vector of VAx and VAy using the formula: Speed of airplane relative to air mass = sqrt(VAx² + VAy²).

After substituting the given values in sqrt(VAx² + VAy²): the airplane's speed relative to the air mass is 38.51 m/s.

Therefore, the airplane's speed relative to the air mass is 38.51 m/s.

(a) The focal length of the lens is 6.86 cm, and the lens is converging.

(b) The height of the image is 6.375 mm, and it is inverted.

To solve this problem, we need to use the lens formula and magnification formula, which are given below:

The lens formula is:

[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]

where:

The magnification formula is:

[tex]m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}[/tex]

where:

Now we can solve the problem step-by-step.

Part (a): Finding the focal length

Given:

Object distance, [tex]d_o = 16.0[/tex] cm

Image distance, [tex]d_i = 12.0[/tex] cm (since the image is on the same side as the object, [tex]d_i[/tex] is positive)

Using the lens formula:

[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]

Substituting the given values:

[tex]\frac{1}{f} = \frac{1}{16.0} + \frac{1}{12.0}[/tex]

Calculating the right-hand side:

[tex]\frac{1}{f} = \frac{3 + 4}{48} = \frac{7}{48}[/tex]

Therefore, the focal length of the lens, [tex]f[/tex], is:

[tex]f = \frac{48}{7} = 6.86[/tex] cm

Since the focal length is positive, the lens is converging.

Part (b): Finding the height and orientation of the image

Given:

Object height, [tex]h_o = 8.50[/tex] mm

Object distance, [tex]d_o = 16.0[/tex] cm

Image distance, [tex]d_i = 12.0[/tex] cm

Using the magnification formula:

[tex]m = -\frac{d_i}{d_o}[/tex]

Substituting the given values:

[tex]m = -\frac{12.0}{16.0} = -0.75[/tex]

Using the magnification formula for height:

[tex]m = \frac{h_i}{h_o}[/tex]

Therefore:

[tex]-0.75 = \frac{h_i}{8.50}[/tex]

Giving:

[tex]h_i = -0.75 \times 8.50 = -6.375[/tex] mm

Since the height of the image, [tex]h_i[/tex], is negative, the image is inverted.

Answer:

15.6 m

Explanation:

d = diameter of the wheel = 43 cm = 0.43 m

r = radius of the wheel = (0.5) d = (0.5) (0.43) = 0.215 m

w₀ = initial angular velocity = 86 rpm = 9.01 rad/s

v₀ = initial linear speed = r w₀ = (0.215) (9.01) = 1.94 m/s

w = final angular velocity = 362 rpm = 37.91 rad/s

v = final linear speed = r w = (0.215) (37.91) = 8.15 m/s

t = time taken = 3.1 s

Using the equation

v = v₀ + at

8.15 = 1.94 + a (3.1)

a = 2 m/s²

d = distance traveled by a point at the edge of the wheel

Using the equation

d =  v₀ t + (0.5) at²

d =  (1.94) (3.1) + (0.5) (2) (3.1)²

d = 15.6 m

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

[tex]P(z)=P_{0}+\rho _wgh[/tex]

Now the gauge pressure is given by

[tex]P(z)-P_{0}=\rho _wgh[/tex]

Applying values we get

[tex]P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa[/tex]

Final answer:

To keep water from entering an open diving chamber at 60 meters depth with sea water of specific gravity 1.03, the air pressure inside the chamber must be equal to the water pressure at that depth, which is 603.84 kPa.

Explanation:

To calculate the air pressure (gage pressure) inside the diving chamber that prevents water from entering, we will use the pressure-depth relationship in fluids. The gage pressure is the pressure relative to atmospheric pressure. Given that the chamber is at a depth of 60 meters underwater and the specific gravity (SG) of the sea water is 1.03, we know that the sea water is denser than pure water (whose SG is 1). To prevent water from entering the chamber, the air pressure inside the chamber must be equal to the water pressure at that depth. The pressure due to a column of liquid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity (9.8 m/s²), and h is the height (or depth) of the liquid column.

Since the water's specific gravity is 1.03, its density (ρ) can be calculated as ρ = SG * ρwater, where ρwater is the density of pure water (1000 kg/m³). So, ρ = 1.03 * 1000 kg/m³ = 1030 kg/m³. The depth (h) is 60 meters.

Now, we can calculate the pressure:

P = ρgh = (1030 kg/m³)(9.8 m/s²)(60 m) = 603,840 Pa or N/m².

To convert this to kilopascals (kPa), we divide by 1,000:

P = 603.84 kPa

This is the gage pressure inside the chamber required to prevent water from entering.

Answer:

(a): s(t)= hi + Vo * t - g* t²/2

(b): Will take the ball to reach the ground t= 11 seconds.

Explanation:

hi= 880 ft

Vo= 96 ft/s

g= 32 ft/s²

equating to 0 the equation of s(t) and clearing t, we find the time it takes for the ball to fall to the ground.

Answer:

a) [tex]s(t) = 96t - 16t^{2} + 880[/tex]

b) It will take 11 seconds for the ball to reach the ground.

Explanation:

We have an initial height of 880 feet.

And

[tex]v(t) = 96 - 32t[/tex]

a) Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

[tex]s(t) = \int {(96 - 32t)} \, dt[/tex]

[tex]s(t) = 96t - 16t^{2} + K[/tex]

In which the constant of integration K is the initial height, so [tex]K = 880[/tex]

So

[tex]s(t) = 96t - 16t^{2} + 880[/tex]

b) How long will the ball take to reach the​ ground

This is t when [tex]s(t) = 0[/tex]

So

[tex]s(t) = -16t^{2} + 96t + 880[/tex]

This is t = -5 or t = 11.

However, t is the instant of time, so it has to be a positive value.

So it will take 11 seconds for the ball to reach the ground.

Answer:

Option (b)

Explanation:

Let the ball takes time t to reach the maximum height. The final velocity at maximum height is zero.

Use I equation of motion, we get

V = u + gt

0 = 3 - 9.8 x t

t = 0.306 s

As the air friction is negligible, so the ball takes same time to reach the hand of boy. Thus the time taken by the ball in air is

T = 2 t = 2 x 0.306 = 0.61 s

Answer:

4.9 m/s²

Explanation:

F = reading of the elevator while running = 594 N

W = weight of person and elevator together when at stop = 396 N

m = mass of person and elevator together

a = acceleration of elevator

Weight of the person is given as

W = mg

inserting the values

396 = m (9.8)

m = 40.4 kg

Force equation for the motion of the elevator is given as

F - W = ma

Inserting the values

594 - 396 = (40.4) a

a = 4.9 m/s²

The appropriate solution is "-87.6 cm/s". A complete solution is given below.

According to the question, the given values are:

Amplitude,

a = 11.3 cm

Time,

t = 0.702 sec

→ The velocity (v) will be:

= [tex]\pm w\sqrt{a^2-x^2}[/tex]

or,

= [tex]-\frac{2 \pi}{t} \sqrt{a^2-(\frac{a}{2} )^2}[/tex]

= [tex]-\frac{2 \pi}{t} \frac{\sqrt{3} }{2}a[/tex]

By substituting the given values, we get

= [tex]-\frac{2 \pi}{0.702} \frac{\sqrt{3}\times 11.3 }{2}[/tex]

=  [tex]-8.946\times 9.7861[/tex]

= [tex]-87.6 \ cm/s[/tex]

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Final answer:

To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.

In simple harmonic motion, the velocity is given by the formula v = ωA sin(ωt), where A is the amplitude and ω is the angular frequency. To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.

First, we find the time at which the particle is at the given position. Since the period is 0.702 sec, the time at half the amplitude in the positive x direction is 0.351 sec (half the period).

Next, we plug this time into the velocity formula: v = ωA sin(ωt). The angular frequency ω is given by ω = 2π / T, where T is the period. Plugging in the values, we can calculate the velocity.

Answer: Addition polymerization & Condensation polymerization

Answer:

0.567 Weber

Explanation:

side, a = 0.31 m, B = 5.9 T

Area, A = (0.31)^2 = 0.0961 m^2

Magnetic flux, ∅ = B x A

∅ = 5.9 x 0.0961 = 0.567 Weber

Answer:

The original speeds of the two cars were :

[tex]6.36\frac{m}{s},12.72\frac{m}{s}[/tex]

Explanation:

Let's start reading the question and making our equations in order to find the speeds.

The first equation is :

[tex]m_{1}=2m_{2}[/tex] (I)

The kinectic energy can be calculated using the following equation :

[tex]K=(\frac{1}{2}).m.v^{2}[/tex] (II)

Where ''K'' is the kinetic energy

Where ''m'' is the mass and where ''v'' is the speed.

By reading the exercise we find that :  

[tex]K_{1}=\frac{K_{2}}{2}[/tex] (III)

If we use (II) in (III) :

[tex](\frac{1}{2}).m_{1}.v_{1}^{2}=(\frac{1}{2}).m_{2}.v_{2}^{2}.(\frac{1}{2})[/tex]

[tex]m_{1}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex] (IV)

If we replace (I) in (IV) ⇒

[tex]2.m_{2}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex]

[tex]4.v_{1}^{2}=v_{2}^{2}[/tex]

[tex]2.v_{1}=v_{2}[/tex] (V)

'' When both cars increase their speed by [tex]9.0\frac{m}{s}[/tex], they then have the same kinetic energy ''

The last equation is :

[tex](\frac{1}{2}).m_{1}.(v_{1}+9)^{2}=(\frac{1}{2}).m_{2}.(v_{2}+9)^{2}[/tex] (VI)

If we use (I) in (VI) ⇒

[tex]2.m_{2}.(v_{1}+9)^{2}=m_{2}.(v_{2}+9)^{2}[/tex]

[tex]2.(v_{1}+9)^{2}=(v_{2}+9)^{2}[/tex]

If we use (V) in this last expression ⇒

[tex]2.(v_{1}+9)^{2}=(2.v_{1}+9)^{2}[/tex]

[tex]2.(v_{1}^{2}+18v_{1}+81)=4v_{1}^{2}+36v_{1}+81[/tex]

[tex]2v_{1}^{2}+36v_{1}+162=4v_{1}^{2}+36v_{1}+81[/tex]

[tex]2v_{1}^{2}=81[/tex]

[tex]v_{1}^{2}=40.5[/tex]

[tex]v_{1}=\sqrt{40.5}=6.36[/tex]

We find that the original speed [tex]v_{1}[/tex] is [tex]6.36\frac{m}{s}[/tex]

If we replace this value in the equation (V) ⇒

[tex]2.(6.36\frac{m}{s})=v_{2}[/tex]

[tex]v_{2}=12.72\frac{m}{s}[/tex]

Final answer:

The speed of a satellite in circular orbit is calculated using the period and the radius of the orbit. Without the average altitude of the satellite's orbit, we cannot provide a numerical speed but can describe the method to find it.

Explanation:

To calculate the speed at which a satellite travels in a circular orbit around the Earth, we can use the satellite's period, which is the time it takes to complete one orbit. The formula relates the orbital speed (v), the radius of the orbit (r), and the period (T). Since we are given the mass of Earth (ME) and the period (T), we can calculate the radius of the orbit using the equation for gravitational force, F = G*ME*m/r^2, and centripetal force, F = m*v^2/r. Once we have the radius, we can determine the speed by using the formula v = 2πr/T.

Unfortunately, we're missing the average altitude (therefore the radius) of the satellite's orbit necessary to calculate the speed. Without that information, we can only outline the method to find the speed as described above but cannot provide a numerical answer.

Final answer:

The maximum coefficient of performance (COP) of the heat pump absorbing heat from the outside air at -10°C and transferring it into a home at 30°C is 0.75.

Explanation:

The coefficient of performance (COP) of a heat pump is a measure of its efficiency and is defined as the ratio of the heat delivered to the heat absorbed. In this case, we have a heat pump that absorbs heat from the outside air at -10°C and transfers it into a home at 30°C. To determine the maximum COP of the heat pump, we can use the formula:

COP = T_h / (T_h - T_c)

where T_h is the temperature of the hot reservoir (in this case, 30°C) and T_c is the temperature of the cold reservoir (in this case, -10°C).

Plugging in the values, we get:

COP = 30 / (30 - (-10)) = 30 / 40 = 0.75

Therefore, the maximum COP of the heat pump is 0.75.

The amount of energy to reach the boiling point is [tex]50*80*4.184 J=16,736J[/tex]. To pass the boiling point, [tex]40.79*\frac{50}{18.02}kJ=113,180J[/tex] are necessary (18.02 is the molar mass of water). This means that [tex]150kJ-113.180kJ-16.736kJ=20,084J[/tex] are left. This allows the steam to heat another [tex]\frac{20,084}{50*1.99}=201.8^{\circ}C[/tex]. Therefore, it ends as steam at temperature [tex]100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C[/tex]

(a) [tex]1.50\cdot 10^{-6}J[/tex]

According to the work-energy theorem, the work done by the electric force is equal to the kinetic energy gained by the particle, so we can write:

[tex]W=K_f - K_i[/tex]

where

W is the work done by the electric force

[tex]K_f[/tex] is the final kinetic energy of the particle

[tex]K_i[/tex] is the initial kinetic energy of the particle

Since the particle starts from rest, [tex]K_i = 0[/tex]. Moreover,

[tex]K_f = 1.50\cdot 10^{-6}J[/tex]

Therefore, the work done by the electric force is

[tex]W=K_f = 1.50\cdot 10^{-6}J[/tex]

(b) -333.3 V

According to the law of conservation of energy, the gain in kinetic energy of the particle must correspond to a loss in electric potential energy, so we can write:

[tex]\Delta K = -\Delta U[/tex]

Where

[tex]\Delta K = 1.50\cdot 10^{-6} J[/tex] is the gain in kinetic energy

[tex]\Delta U[/tex] is the loss in electric potential energy

So we have

[tex]\Delta U = - \Delta K = -1.50\cdot 10^{-6}J[/tex]

The loss in electric potential energy can be rewritten as

[tex]\Delta U = q \Delta V[/tex]

where

[tex]q=+4.50 nC = 4.5\cdot 10^{-9} C[/tex] is the charge of the particle

[tex]\Delta V[/tex] is the change in electric potential over the distance the charge has moved

Solving for [tex]\Delta V[/tex],

[tex]\Delta V= \frac{\Delta U}{q}=\frac{-1.50\cdot 10^{-6}}{4.5\cdot 10^{-9}}=-333.3 V[/tex]

(c) 4166 V/m

The magnitude of the electric field is given by

[tex]E=\frac{|\Delta V|}{d}[/tex]

where

[tex]|\Delta V|[/tex] is the magnitude of the change in electric potential

d is the distance through which the charge has moved

Since we have

[tex]|\Delta V|=333.3 V\\d = 8.00 cm = 0.08 m[/tex]

The magnitude of the electric field is

[tex]E=\frac{333.3}{0.08}=4166 V/m[/tex]

(d) [tex]-1.50\cdot 10^{-6}J[/tex]

The change in electric potential energy of the charge  has already been calculated in part (b), and it is

[tex]\Delta U = -1.50\cdot 10^{-6}J[/tex]

Answer:

The height of the building is 180 meters.

Explanation:

It is given that,

Time taken for a ball to strike the ground, t = 6 sec

It is released from the rest. We need to find the height of the building from which it is released. It can be calculated using the equation as follows :

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

Here, a = g and u = 0

So, [tex]s=0+\dfrac{1}{2}gt^2[/tex]

[tex]s=\dfrac{1}{2}\times 10\ m/s^2\times (6\ s)^2[/tex]

s = 180 meters

So, the height of the building is 180 meters. Hence, this is the required solution.

Answer:

Energy density = 0.0831 J/m³

Explanation:

Energy density of a capacitor is given by the expression

        [tex]u=\frac{1}{2}\epsilon E^2[/tex]

We have electric field ,

              [tex]E=\frac{V}{d}=\frac{480}{3.5\times 10^{-3}}=1.37\times 10^5V/m[/tex]

Here there is no dielectric

So energy density,

         [tex]u=\frac{1}{2}\epsilon_0 E^2=\frac{1}{2}\times 8.85\times 10^{-12} \times (1.37\times 10^5)^2=8.31\times 10^{-2}=0.0831 J/m^3[/tex]

Energy density = 0.0831 J/m³  

Explanation:

For this question the magnetic force provides the force required for the circular motion

Equating

Magnetic force = centripetal force

Bqvsinx = mv^2/r

as magnetic field is perpendicular x=90

Bqvsin90 = mv^2/r

Bq = mv/r

B= mv/rq

Then replace r= 0.6 , v= 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity

Answer:

Tangential speed = 157 m/s

Explanation:

Tangential speed = Angular speed x Radius

Angular speed = 7.85 x 10⁴ rad/s

Radius = 2 mm = 2 x 10⁻³ m

Tangential speed = Angular speed x Radius = 7.85 x 10⁴ x 2 x 10⁻³

Tangential speed = 15.7 x 10 = 157 m/s

Tangential speed = 157 m/s

The tangential speed of a point on the outer edge of the disk attached to a high-speed drill, given an angular velocity of 7.85 x 10^4 rad/s and a radius of 2.00 mm, is 157 m/s.

To determine the tangential speed of a point on the outer edge of the disk which is attached to a high-speed drill, we apply the formula that relates tangential speed (v) to the angular velocity ($$omega$) and the radius (r) of the disk: v = r  $$omega$. Given the angular velocity of the disk is 7.85  $$x$ 10^4 rad/s and the radius is 2.00 mm (which we need to convert to meters for standard units), the calculation is straightforward.

v = r  $$omega$
v = 0.002 m  $$x$ 7.85  $$x$ 10^4 rad/s
v = 157 m/s.

Thus, the tangential speed of a point on the outer edge of this disk is 157 m/s.

Answer:

it is concept that considered all matter in universe homogeneous and it is acceptable on large scale.

Explanation:

The cosmological principle is the concept that, when recognized on a large basis, the visual allocation of matter and energy is homogeneous and isotropic, since the forces are expected to act homogeneously across the universe and should therefore not produce any observed errors in the wide-scale. This eliminates any uncertainty about irregularities in the characteristics of matter and even includes the development of this matter since its creation during the Big Bang.

Final answer:

The cosmological principle states that the universe at any given time is the same everywhere on a large scale. This principle is crucial for understanding and studying the universe, as it allows us to make assumptions about its properties based on what we observe locally. It is a fundamental concept in cosmology.

Explanation:

The cosmological principle is the assumption that, on a large scale, the universe at any given time is the same everywhere, meaning it is isotropic and homogeneous. This principle is the starting assumption for nearly all theories that describe the entire universe. It implies that the universe is about the same everywhere, apart from changes with time, and that the part we can see around us is representative of the whole.

Answer:

a)4.36*10^-12W/m^2

b)L=6.39dB

c)ξ=5.73×10⁻¹¹m

Explanation:

Take speed of sound in air as 344m/s and density of air 1.2kg/m3 .

a)

[tex]I=\frac{P^2}{2pv}[/tex]

where

P=pressure

p=density of air

v=velocity

[tex]I=\frac{6*10^-5^2}{2*1.2*344} = 4.36*10^-12[/tex] W/m^2

b)

Sound intensity level in dB is defined as:  

L = 10∙log₁₀(I/I₀)  

with  

I₀ = 1.0×10⁻¹² W/m²  

Hence;  

L = 10∙log₁₀( 4.36x10^-12 W/m² / 1.0×10⁻¹² W/m²) = 6.39dB

c)

Displacement is given by :  

ξ = p/(Z∙ω) = p/(Z∙2∙π∙f)  

where  

Z = 416.9 N∙s/m³ = 416.9 Pa∙s/m

f frequency and ω angular frequency of the sound wave.  

So the amplitude of this sound wave is:  

ξ = 6×10⁻⁵Pa / (416.9 Pa∙s/m ∙ 2∙π∙ 400s⁻¹) = 5.73×10⁻¹¹m

To calculate the intensity and sound intensity level of a sound wave with a frequency of 400 Hz and a pressure amplitude of 6.0 * 10^(-10) Pa, use the given formulas. The intensity is 1.5 * 10^(-16) W/m², the sound intensity level is -60 dB, and the displacement amplitude is 5.3 * 10^(-9) m.

To calculate the intensity and sound intensity level of a sound wave with a frequency of 400 Hz and a pressure amplitude of 6.0 * 10^(-10) Pa, we can use the formulas:

(a) Intensity (I) = (Pressure Amplitude)^2 / (2 * Z)

(b) Sound Intensity Level (SIL) = 10 * log10(I / Io)

(c) Displacement Amplitude = Pressure Amplitude / (2 * π * f * v)

Using the given values, we can calculate:

(a) Intensity = (6.0 * 10^(-10))^2 / (2 * (1.21 * 10^(-3))) = 1.5 * 10^(-16) W/m²

(b) Sound Intensity Level = 10 * log10((1.5 * 10^(-16)) / (10^(-12))) = -60 dB

(c) Displacement Amplitude = (6.0 * 10^(-10)) / (2 * π * 400 * 343) = 5.3 * 10^(-9) m

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Answer:

option (b)

Explanation:

Mass of moon = m

Mass of earth = 81.25 x mass of moon = 81.25 m

Radius of moon = r

radius of earth = 3.668 x radius of moon  = 3.668 r

Frequency on earth = 2 Hz

Let the frequency on moon is f.

The formula for the frequency is given by

[tex]f = \frac{1}{2\pi }\times \sqrt{\frac{g}{l}}[/tex]

The value of acceleration due to gravity on earth is g.

ge = G Me / Re^2

ge = G x 81.25 m / (3.668 r)^2

ge = 6.039 x G m / r^2 = 6.039 x gm

ge / gm = 6.039

Now use the formula for frequency

[tex]\frac{fe}{fm} = \sqrt{\frac{ge}{gm}}[/tex]

[tex]\frac{2}{fm} = \sqrt{\frac{6.039 gm}{gm}}[/tex]

[tex]\frac{2}{fm} = 2.46[/tex]

fm = 0.814 Hz

Answer:0.4061 L

Explanation:

let initial volume of gas be [tex]V_1[/tex]=1 L

[tex]\gamma[/tex] =1.3

initial temprature[tex]\left ( T_1\right )[/tex]=273k

Now if gas is compressed adiabatically to half of its initial volume then

[tex]V_2[/tex]=0.5L

Using adiabatic relation

[tex]TV^{\gamma -1}[/tex]=constant

[tex]T_1V_1^{\gamma -1}[/tex]=[tex]T_2V_2^{\gamma -1}[/tex]

[tex]273\times \left ( \frac{1}{0.5}\right )^\left ( {\gamma -1}\right )[/tex]=[tex]T_2[/tex]

[tex]T_2[/tex]=273[tex]\times \left ( 2\right )^\left ( 0.3\right )[/tex]

[tex]T_2[/tex]=336.1k

Now the is cooled at constant pressure i.e.

[tex]\frac{V_2}{T_2}[/tex]=[tex]\frac{V_3}{T_3}[/tex]

[tex]\frac{0.5}{336.1}[/tex]=[tex]\frac{V_3}{273}[/tex]

[tex]\frac{0.5}{336.1}[/tex]=[tex]\frac{V_3}{273}[/tex]

[tex]V_3[/tex]=0.406 L

The final volume of the gas after it is cooled to 273 K at constant pressure, following an adiabatic compression to half its initial volume, will remain at 0.5 L.

The question involves an adiabatic compression of a gas followed by cooling at constant pressure. To find the final volume after the gas is cooled to 273 K at constant pressure, we can use the ideal gas law before and after the compression, considering that for adiabatic processes, PVγ is constant (where P is pressure, V is volume, and γ is the adiabatic index). After the adiabatic compression, we have a new volume of 0.5 L and a new pressure. We then cool the gas at this new pressure back to the original temperature of 273 K. By the ideal gas law (PV = nRT), at constant temperature and number of moles (n), the final volume (Vf) will be equal to the initial volume (Vi) because the pressure has remained constant after the compression step.

Therefore, even after cooling, the final volume will remain at 0.5 L because pressure does not change during the cooling process at constant pressure.