Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. A certain person can see distant objects well, but his near point is 50.0 cm from his eyes instead of the usual 25.0 cm . Suppose that this person needs ordinary glasses What focal length lenses are needed to correct his vision ?What is their power in diopters?

Answer:

Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]

Explanation:

Tortoise travels an additional distance 580 m .

Since there is no acceleration, the distance traveled by any object is d = v t , where v is the total distance and t is the total time taken.

Speed of hare = [tex]v_{h}[/tex] = 9 m/s

Speed of tortoise = [tex]v_{t}[/tex] =5 m/s

Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]

Distance traveled by tortoise = d +580 m = 5 [tex]t_{t}[/tex]

Final answer:

The distance that the hare travels from the starting line after time t is represented by the expression 9t meters.

Explanation:

To write an expression that represents the hare's distance from the starting line in meters as a function of time, we use the formula for distance traveled at constant speed d = vt, where d is distance, v is speed, and t is time. Since the hare starts running at 9 meters per second, the distance the hare travels from the starting line after t seconds can be represented by the expression 9t meters. Note that we do not include the 580-meter head start given to the tortoise in the hare's expression since we are only considering the hare's distance from the starting line and not its position relative to the tortoise.

The object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s with a constant acceleration of 5 m/s^2.

To find the distance the object will move before it achieves a velocity of 6.4 m/s, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and x is the distance. Rearranging the equation, we get x = (v^2 - v0^2) / (2a). Plugging in the values, we have x = (6.4^2 - 4.3^2) / (2 * 5) = 10.12 m. Therefore, the object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s.

Final answer:

To find how far an object with constant acceleration a = 5 [tex]m/s^2[/tex] and initial velocity of 4.3 m/s moves before reaching a velocity of 6.4 m/s, use the kinematic equation [tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0). After calculation, the object will move 5.0 m before reaching 6.4 m/s.

Explanation:

We need to calculate how far the object moves before it achieves a velocity of v = 6.4 m/s given constant acceleration a = 5 [tex]m/s^2[/tex], initial velocity v0 = 4.3 m/s, and starting position x0 = 2.7 m. We can use the kinematic equation which relates final velocity, initial velocity, acceleration, and displacement:

[tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0)

Rearranging for displacement (x - x0):

(x - x0) = ([tex]v^2[/tex] - [tex]v0^2[/tex]) / (2a)

Substituting the given values:

(x - 2.7 m) = [tex](6.4 m/s)^2[/tex] - [tex](4.3 m/s)^2[/tex] / [tex](2 * 5 m/s^2)[/tex]

(x - 2.7 m) = (40.96 [tex]m^2/s^2[/tex] - 18.49 [tex]m^2/s^2[/tex]) / (10 [tex]m/s^2[/tex])

(x - 2.7 m) = 22.47 [tex]m^2/s^2[/tex] / (10 [tex]m/s^2[/tex])

(x - 2.7 m) = 2.247 m

x = 2.7 m + 2.247 m

x = 4.947 m

To the nearest 0.1 m, the object will move 5.0 m before it achieves a velocity of 6.4 m/s.

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature[tex](T_i)=300 K[/tex]

[tex]P_i=10 atm[/tex]

[tex]P_f=2 atm[/tex]

Work done in iso-thermal process[tex]=P_iV_iln\frac{P_i}{P_f}[/tex]

[tex]P_i[/tex]=initial pressure

[tex]P_f[/tex]=Final Pressure

[tex]W=10\times 2.463\times ln\frac{10}{2}=39.64 J [/tex]

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is[tex]=P\Delta V[/tex]

[tex]V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L[/tex]

[tex]V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L[/tex]

[tex]\Delta W=1\times (12.315-2.463)=9.852 J[/tex]

[tex]\Delta q=\Delta W=9.852 J[/tex]

[tex]\Delta U=0[/tex]

Answer:

The distance in -x axis is 39.77 m

Explanation:

Given that,

Distance in +x axis = 24 m

Time = 3.5 sec

Total time = 9.3 sec

Average velocity = 0

We need to calculate the velocity in +x axis

[tex]v=\dfrac{d}{t}[/tex]

Put the value into the formula

[tex]v=\dfrac{24}{3.5}[/tex]

We need to calculate the velocity in -x axis

[tex]v=\dfrac{x}{9.3-3.5}[/tex]

We need to calculate the distance

Using formula of average velocity

[tex]\dfrac{\dfrac{24}{3.5}-\dfrac{x}{9.3-3.5}}{2}=0[/tex]

[tex]x=\dfrac{24}{3.5}\times5.8[/tex]

[tex]x=39.77\ m[/tex]

Hence, The distance in -x axis is 39.77 m.

Answer:

i)0.1213m

ii)0.2426 m

Explanation:

The formula to apply here are;

For a closed conduit at one end the fundamental frequency produced by a pipe of length L is ;

f=v/4L where v is speed of sound and L is length of conduit

707=343/4L

707*4L=343

2828L=343

L=343/2828

L=0.1213m

For an open conduit, the fundamental frequency produced by pipe of length L is given by;

f=v/2L where v is speed of sound and L is length of conduit

707=343/2L

707*2L=343

1414L=343

L=343/1414

L=0.2426 m

Answer:

The position-time graph and the velocity-time graph has been shown in the figure attached.

Explanation:

Given:

For t = 0 s to t = 20 s,

The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a red line.

For t = 20 s to t = 30 s,

The car moves with a constant acceleration whose position-time and the velocity time graph has been shown with a green line.

For t = 30 s to t = 50 s,

The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a blue line.

For t = 50 s to t = 60 s,

The car moves with a constant deceleration whose position-time and the velocity time graph has been shown with a black line.

For a constant velocity, the velocity-time graph of the particle is a straight line parallel to a time axis and the position-time graph is a straight line inclined at some positive angle with the time axis.

For a constant acceleration, the velocity-time graph is a straight line incline at a positive angle with the time axis and the position-time graph is a parabolic curved line having upward concavity.  

For a constant deceleration, the velocity-time graph is a straight line incline at a negative angle with the time axis and the position-time graph is a parabolic curved line having downward concavity.

Answer:

The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0

On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.

Answer:

The speed of the object in the last instant prior to hitting the ground is -24.1 m/s

Explanation:

The equation for the position and velocity of the object will be:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

We know that at its maximum height, the velocity of the object is 0. We can obtain the time it takes the object to reach the maximum height and with that time we can calculate the maximum height:

v = v0 + g · t

0 = 18.5 m/s - 9.8 m/s² · t

-18.5 m/s / -9.8 m/s² = t

t = 1.89 s

Now,let´s find the max-height:

y = y0 + v0 · t + 1/2 · g · t²

y = 12.2 m + 18.5 m/s · 1.89 s + 1/2 ·(-9.8 m/s²) · (1.89 s)²

y = 29.7 m

Now, let´s see how much it takes the object to hit the ground:

In that instant, y = 0.

y = y0 + v0 · t + 1/2 · g · t²

0 = 29.7 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²   (notice that v0 = 0 because the object starts from its maximum height, where v = 0)

-29.7 m = -4.9 m/s² · t²

t² = -29.7 m / -4.9 m/s²

t = 2.46 s

Now, we can calculate the speed at t= 2.46 s, the instant prior to hitting the ground.

v = v0 + g · t

v = g · t

v = -9.8 m/s² · 2.46 s

v = -24.1 m/s

Answer:

In order to avoid collision, the ferry must stop at a distance of 12.5 m from the dock.

Solution:

The initial velocity of the taxi, [tex]v_{o} = 5 m/s[/tex]

The minimum value of acceleration , [tex]a_{min} = - 1 m/s^{2}[/tex]

The maximum value of acceleration , [tex]a_{max} = 1 m/s^{2}[/tex]

Now,

When the deceleration starts the ferry slows down and at minimum deceleration of [tex]- 1 m/s^{2}[/tex], the ferry stops.

Thus, inthis case, the final velocity, v' is 0.

Now, to calculate the distance covered, 'd' in decelerated motion is given by the third eqn of motion:

[tex]v'^{2} = v_{o}^{2} + 2ad[/tex]

[tex]0^{2} = 5^{2} + 2\times (- 1)d[/tex]

[tex]d = 12.5 m[/tex]

Explanation:

(a) We need to convert following angles in radians. The conversion formula from degrees to radian is given by :

[tex]Radians=(\dfrac{\pi}{180})\times degrees[/tex]

1. If angle is 16.8°

[tex]Radians=(\dfrac{\pi}{180})\times 16.8[/tex]

16.8 degrees = 0.29 radians

2. If angle is 53.7°

[tex]Radians=(\dfrac{\pi}{180})\times 53.7[/tex]

53.7 degrees = 0.937 radians

3. If angle is 94.2°

[tex]Radians=(\dfrac{\pi}{180})\times 94.2[/tex]

53.7 degrees = 1.64 radians

(b) We need to convert following angles to degrees. The conversion formula from radian degrees to is given by :

[tex]Degrees=(\dfrac{180}{\pi})\times radians[/tex]

1. If angle is 0.258 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 0.258[/tex]

0.258 radian = 14.78 degrees

2.  If angle is 1.01 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 1.01[/tex]

0.258 radian = 57.86 degrees

3.  If angle is 7.51 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 7.51[/tex]

0.258 radian = 430.29 degrees

Hence, this is the required solution.

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

[tex]v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s[/tex]

[tex]v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s[/tex]

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

[tex]D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads[/tex]

Now, we just use the formula:

[tex]w_f^2-w_o^2=2\alpha*x[/tex]

[tex]\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2[/tex]

Explanation:

The given data is as follows.

   Initial velocity, u = [tex]3.3 rev/s \times 2 \pi rad/rev[/tex]

                           = 20.724 rad/sec

   Final velocity, v = [tex]6.4 rev/s \times 2 \times pi rad/rev[/tex]    

                           = 40.192 rad/sec

Now, we will calculate the value of angular rotation (d) as follows.

          d = No. of revolutions × [tex]2 \pi rad/rev[/tex]

         d = [tex](\frac{250 m}{2 \pi r}) \times 2 \pi[/tex]

            = [tex]\frac{250}{0.32}[/tex]

            = 781.25 rad

Also, we know that,

                [tex]v^{2} = u^{2} + 2ad[/tex]

or,              a = [tex]\frac{(v^{2} - u^{2})}{2d}[/tex]

                      = [tex]\frac{(40.192)^{2} - (20.724)^{2})}{2 \times 781.25}[/tex]

                      = [tex]\frac{1615.39 - 429.48}{1562.5}[/tex]

                     = 0.7589 [tex]rad/s^{2}[/tex]

Thus, we can conclude that the tire's angular acceleration is 0.7589 [tex]rad/s^{2}[/tex].

Answer:

The magnitude of the current circulating through the coil is 1.68 mA.

Explanation:

Given that,

Number of turns = 100

Length = 5 cm

Resistance = 12.0 Ω

Rate of magnetic field = 0.081 T/s

We need to calculate the magnitude of the current circulating through the coil

Using formula of current

[tex]I=\dfrac{e}{R}[/tex]

[tex]I=\dfrac{-NA\dfrac{dB}{dt}}{R}[/tex]

Where, A = area

N = number of turns

R = resistance

Put the value into the formula

[tex]I=\dfrac{100\times(0.05)^2\times0.081}{12.0}[/tex]

[tex]I=0.0016875\ A[/tex]

[tex]I=1.68\times10^{-3}\ A[/tex]

[tex]I=1.68\ mA[/tex]

Hence, The magnitude of the current circulating through the coil is 1.68 mA.

The question involves Faraday's law, which relates changing magnetic fields to induced currents. Given the rate of change of the magnetic field and the resistance of the coil, the magnitude of the induced current is calculated to be 1.67 mA.

This question involves the concept of electromagnetic induction in physics. According to Faraday's law, a changing magnetic field will induce an electromotive force (EMF) in a circuit. In this case, we're given that the magnetic field is decreasing at a rate of 0.081 T/s. This rate of change can be used to calculate the induced EMF, which is equal to the rate of change of the magnetic flux.

The magnetic flux (Φ) through the coil is given by Φ = NBA, where N is the number of turns in the coil, B is the magnetic field, and A is the area of the coil. So, the rate of change of magnetic flux (dΦ/dt) is N*d(BA)/dt. Because the area of the coil is constant, this simplifies to NAdB/dt. A is 0.0025 m^2, N is 100, and dB/dt is -0.081T/s, so dΦ/dt is -0.02 Wb/s.

Faraday's law tells us that the induced EMF (ε) is equal to -dΦ/dt, so ε has magnitude 0.02 V. If we define positive current as current that would generate a magnetic field opposing the decrease in the existing field (Lenz's Law), the induced EMF 'pushes' positive current in this direction. The current (I) can then be calculated using Ohm's Law (V = IR), giving I = ε/R = 0.02V / 12 Ω = 0.00167 A, or 1.67 mA.

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Answer:

a) The ball clears the crossbar by 10.6 m

b) The ball approaches the crossbar while falling

Explanation:

The position of the ball is described by the vector position r (see attached figure):

r = (x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle  

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s²)

The vector r is composed by rx and ry (see figure):

r = (rx ; ry)

a) Let´s find the time at which the ball flies a distance of  36.0 m. If at that time the vertical component of the vector r, ry, is equal or greater than 3.05 m, then the ball will clear the crossbar.

rx = x0 + v0 t cos α = 36.0 m

Since the origin of the system of reference is located where the kicker is, x0 = 0.

36.0 m = v0 t cos  α

36.0 m /(v0 cos  α) = t

36.0 m / (22.8 m/s * cos 51.0°) = t

t = 2.51 s

Now let´s calculate the height of the ball at that time:

ry = y0 + v0 t sin α + 1/2 g t²

Since the kicker is on the ground, y0 = 0

ry = 22.8 m/s * 2.51 s * sin 51.0° - 1/2 * 9.8 m/s² * (2.51 s)² = 13.6 m

Since the crossbar is 3.05 m high, the ball clears it by (13.6 m - 3.05 m) 10.6 m

b) Please see the figure to figure this out ;)

If the ball approaches the crossbar while still rising, the vertical component (vy) of the velocity vector will be positive. In change, if the ball approaches the crossbar while falling the vertical component of the velocity will be negative. See the figure.

The velocity vector is given by this equation:

v = (vx ; vy)

v = ( v0 cos α ; v0 sin α + g t)

Let´s see the vertical component at time t = 2.51

vy = v0 sin α + g t

vy = 22.8 m/s * sin 51.0° - 9.8 m/s² * 2.51 s

vy = -6.88 m/s

Then, the ball approaches the crossbar while falling.

Answer:

u = 12962.11 m/s

Explanation:

Given that,

The radius of mercury, [tex]r=2440\ km=2440\times 10^3\ m[/tex]

Mass of Mercury, [tex]M=3\times 10^{24}\ kg[/tex]

Final speed of the object, v = 2000 m/s

Let u is its initial speed when the object is far from  Mercury. It can be calculated by applying the conservation of energy as :

Initial kinetic energy + gravitational potential energy = final kinetic energy

[tex]\dfrac{1}{2}mu^2+(-\dfrac{GmM}{r})=\dfrac{1}{2}mv^2[/tex]

[tex]\dfrac{1}{2}u^2+(-\dfrac{GM}{r})=\dfrac{1}{2}v^2[/tex]

[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+\dfrac{GM}{r}[/tex]

[tex]u^2=2\times (\dfrac{1}{2}v^2+\dfrac{GM}{r})[/tex]

[tex]u^2=2\times (\dfrac{1}{2}(2000)^2+\dfrac{6.67\times 10^{-11}\times 3\times 10^{24}}{2440\times 10^3})[/tex]

u = 12962.11 m/s

So, the initial speed of the object is 12962.11 kg. Hence, this is the required solution.

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

Answer:

Equivalent capacitance, [tex]C'=3.9\ \mu F[/tex]

Explanation:

Capacitance, [tex]C_1=9\ \mu F[/tex]

Capacitance, [tex]C_2=13\ \mu F[/tex]

Capacitance, [tex]C_3=16\ \mu F[/tex]

Let C' is the equivalent capacitance of the combination of capacitors. It is given by :

[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}[/tex]

[tex]\dfrac{1}{C'}=\dfrac{1}{9}+\dfrac{1}{13}+\dfrac{1}{16}[/tex]

[tex]C'=3.99\ \mu F[/tex]

or

[tex]C'=3.9\ \mu F[/tex]

So, their equivalent capacitance is [tex]3.9\ \mu F[/tex]. Hence, this is the required solution.

Final answer:

The equivalent capacitance of a 9.0 μF, 13 μF, and a 16 μF capacitors connected in series is approximately 4.0 μF when expressed with two significant figures.

Explanation:

The question concerns the calculation of the equivalent capacitance when multiple capacitors are connected in series. To find the equivalent capacitance of the 9.0 μF, 13 μF, and 16 μF capacitors connected in series, you use the formula for capacitors in series:

1/Ceq = 1/C1 + 1/C2 + 1/C3

Substituting the given values:

1/Ceq = 1/9.0 μF + 1/13 μF + 1/16 μF

Calculating the reciprocals and summing them gives:
1/Ceq = 0.111 + 0.077 + 0.063

1/Ceq = 0.251

Now, the equivalent series capacitance, Ceq, can be calculated by taking the reciprocal of the result:
Ceq = 1 / 0.251

Ceq is approximately 3.98 μF.

To express this answer using two significant figures, the equivalent capacitance is 4.0 μF.

Answer:

(a). The wavelength of photon is 914 A.

(b). The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.

Explanation:

Given that,

Ionization energy = 13.6 eV

(a). We need to calculate the wavelength

Using formula of wavelength

[tex]E=\dfrac{hc}{\lambda}[/tex]

[tex]\lambda=\dfrac{hc}{E}[/tex]

Where, h = Planck constant

c = speed of light

E = energy

Put the value into the formula

[tex]\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{13.6\times1.6\times10^{-19}}[/tex]

[tex]\lambda=9.14\times10^{-8}\ m[/tex]

[tex]\lambda=914\ \AA[/tex]

The wavelength of photon is 914 A.

(b). We need to calculate the temperature of the black body whose spectrum peaks at wavelength

Using Wien's displacement law

[tex]\lambda_{max} T=2.898\times10^{-3}[/tex]

[tex]T=\dfrac{2.898\times10^{-3}}{\lambda}[/tex]

Put the value of wavelength

[tex]T=\dfrac{2.898\times10^{-3}}{914\times10^{-10}}[/tex]

[tex]T=31706.78\ K[/tex]

The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.

Hence, This is the required solution.

Answer:

It take the car to catch up with the truck 111.91 s.

Explanation:

To solve this problem we have to use the formula for uniformly accelerated motion (for the car) and the formula for uniform rectilinear movement (for the truck).

We apply the corresponding formula for each vehicle, so we will have two equations. As the question is how much time, time is the unknown variable that we will call t from now on.

Equation for the car is:

[tex]x_{c}=\frac{1}{2}*a*t^{2}[/tex]

Equation for the truck is

[tex]x_{t} =v*t[/tex]

We know that t will be the same for the two vehicle on the instant the car catch up the truck.

On the time t the distance x traveled for both cars are the same, so we can equate the two formulas ans isolate t.

[tex]v*t=\frac{1}{2} *a*t^{2} \\t=(2*v)/a\\t=111,91s[/tex]

Note: all unit of measurement must be the same, for speed, we need to convert 36,3mph to m/s.

36,3mph=162.27m/s we use 162.27m/s in the formulas.

It will take the car approximately 11.17 seconds to catch up with the truck that passes it traveling at a constant speed when the car starts from rest and accelerates at 2.9 m/s².

The question asks how long it takes for a car that starts from rest with a constant acceleration to catch up with a truck traveling at a constant speed. The car accelerates from rest at 2.9 m/s², while the truck travels at a constant speed of 36.3 mph, which is approximately 16.2 m/s (1 mph equals approximately 0.44704 m/s).

To solve this problem, we need to consider that the car and truck will have traveled the same distance when the car catches up to the truck. The equations for distance for the car (with acceleration) and the truck (traveling at constant speed) are:

We set the distances equal to each other:

½ * 2.9 m/s² * time² = 16.2 m/s * time

This gives us a quadratic equation:

½ * 2.9 * time² - 16.2 * time = 0

Factoring out the common term 'time', we get:

time(½ * 2.9 * time - 16.2) = 0

Ignoring the solution where time equals zero (since we want to know how long it takes after they have started moving), we get:

½ * 2.9 * time = 16.2

time = (16.2 / (0.5 * 2.9)) seconds

Solving this, time ≈ 11.17 seconds

Therefore, it will take the car approximately 11.17 seconds to catch up with the truck.

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

[tex]t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days[/tex]

Now the concentration of the pollutant after 14.872 days is calculated as

[tex]N_{t}=N_{0}e^{-kt}[/tex]

where

[tex]N_{o}[/tex] is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

[tex]N_{t}=3\times e^{-0.3\times 14.872}[/tex]

[tex]N_{t}=0.03463mg/L[/tex]

Answer:

water leak is 650 gallons

time required to full drain is 80 hrs

Explanation:

given data

volume V = 3200 gallon

rate = V(t) = 80 - t

to find out

how much water leak between 10 and 20 hour and  drain complete

solution

we know here rate is 80 - t

so here rate will be

[tex]\frac{dV(t)}{dt}[/tex] = 80 - t

and for time 10 and 20 hour

take integrate between 10 and 20

so water leak = [tex]\int\limits^ {20}_ {10} {(80-t)} \, dt[/tex]   .....................1

water leak = ( 80t - [tex]\frac{t^{2} }{2} )^{20}_{10}[/tex]

water leak = 650

so water leak is 650 gallons

and

we know here for full tank drain condition

water leak full = 80 t - [tex]\frac{t^{2} }{2} [/tex]

3200  = 80 t - [tex]\frac{t^{2} }{2} [/tex]

6400 = t² - 160 t

t = 80

so time required to full drain is 80 hrs

Answer:

18.1 MN/C

Explanation:

The gravitational force of the plastic sphere is in equilibrium with the electric force.

Mass of the plastic sphere = m = 14.2 g = 0.0142 kg

Force of gravity = F = mg = (0.0142)(9.81) = 0.139 N

This force is balanced by the electric force due to the charge 7.7 nC

Charge = q = 7.7 x 10⁻⁹ C

Electric field = E = F / q = (0.139) /(7.7 x 10⁻⁹) = 18.1 MN/C

Final answer:

The magnitude of displacement is always less than or equal to the distance traveled. Displacement is the straight line distance between the starting and ending points, while distance is the total path length traveled.  So the correct option is C

Explanation:

The magnitude of displacement is always less than or equal to the distance traveled. This is because displacement is the straight-line measurement from the initial point to the final point, regardless of the path taken, and it's a vector quantity with both magnitude and direction. On the other hand, distance is a scalar quantity that represents the total path length traveled without regard to direction. Therefore, if a path is straight and in one direction, the distance and the magnitude of displacement are equal. However, if the path involves changes in direction, the distance will be greater than the magnitude of displacement.

The correct answer to the question 'Which of the following is always true about the magnitude of a displacement?' is C. It is less than or equal to the distance traveled.

Answer:

a) Re= 574.17

b) r= 8.71 cm

Explanation:

In order to solve this problem, we will need to use the formula for the Reynolds number:

Re=ρ*v*d/μ

All of the required data is already given in the problem, but before we use the above-mentioned formula, we need to convert the data to SI units, as follows:

a) Now we proceed to calculate Reynolds number:

Re=1.06 *10³ kg/m³ * 0.0650 m/s * 0.025 m / (3.00 * 10⁻³ Pa · s)

Re=574.17

Re<2,300 ; thus the flow is laminar.

b) To answer this question we use the same equation, and give the Reynolds number a value of 4,000 in order to find out d₂:

4,000= 1.06 *10³ kg/m³ * 0.0650 m/s * d₂ / (3.00 * 10⁻³ Pa · s)

We solve for d₂:

d₂=0.174 m

Thus the radius of the capillary that would cause the flow to become turbulent is 0.174 m / 2= 0.0871 m or 8.71 cm, given that neither the speed nor the viscosity change.

However, in your question you wrote that the artery ends in a capillary so that the radius reduces its value. But the lower the radius, the lower the Reynolds number. And as such, it would not be possible for the flow to turn from laminar to turbulent, if the other factors (such as speed, or density) do not change.

Answer:

Explanation:

Initial velocity u = 36 km/h = 10 m /s

v = 0 , accn = a , Time taken to stop = t , distance covered to stop = s

v² = u² - 2as

u² = 2as

a = u² / 2s

= 10 x 10 / 2 x 17

= 2.94 ms⁻²

Force applied = mass x acceleration

= 15000 / 9.8 x 2.94

= 4500 N

b )

v = u + at

0 = 10 - 2.94 t

t = 10 / 2.94

= 3.4 s

c )

from the relation

u² = 2as

it is clear that stopping distance is proportional to u², if acceleration a is constant .

If initial speed u is doubled , stopping distance d will become 4 times or 17 x 4 = 68 m .

d )

u = at

if a is constant time taken to stop will be proportional to initial velocity.

If initial velocity is doubled , time too will be doubled. Or time will become

3.4 x 2 = 6.8 s .

Answer:

[tex]s=\pm 1[/tex]

Explanation:

The dot product of two vectors [tex]\vec{a}=a_1\vec{x}+b_1\vec{y}[/tex] and [tex]\vec{b}=a_2\vec{x}+b_2\vec{y}[/tex] is given by

[tex]\vec{a}\cdot\vec{b}=a_1\cdot a_2+b_1\cdot b_2[/tex]

The dot product of two orthogonal vector is always zero thus if [tex]\vec{a}=\vec{x}+s\vec{y}[/tex] and [tex]\vec{b}=\vec{x}-s\vec{y}[/tex], their dot product would be

[tex]\vec{a}\cdot\vec{b}=1\times1+s\times-(s)=1-s^2=0[/tex]

[tex]\implies 1-s^2=0[/tex]

[tex]\implies s^2=1[/tex]

[tex]\implies s=\pm 1[/tex]

Answer:

Explanation:

Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds

s = 1/2at²

= .5 x a x 100

=50a

Velocity after 10 s

u = at = 10a

Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.

s = ut + 1/2 gt²

50a = -10ax20 + .5 x 10x400

250a = 2000

a = 8 m s⁻² .

Velocity after 10 s

= at = 80 m/s

further height reached with this speed under free fall

h = v² / 2g

= 80 x 80 / 2 x 10

= 320 m

height achieved under acceleration

= 50a

= 50 x 8 = 400m

Total height

= 320 + 400= 720 m

velocity after falling from 720 m

v² = 2gh = 2 x10 x 720

v = 120 m/s

Answer:

a) The initial acceleration is 7.84 m/s²

b) The impact speed is 117.6 m/s

c) The height is 705.6 m

Explanation:

a) The speed from A to B is:

v = u + at

Where

u = initial speed = 0

t = 10 s

Replacing:

v = 10t (eq. 1)

The vertical distance between A to B is:

[tex]h=\frac{1}{2} at^{2} +ut=\frac{1}{2} a*(10)^{2}+0=50a[/tex] (eq. 2)

From B to C, the time it take is equal to 20 s, then:

[tex]h=vt+\frac{1}{2} at^{2}[/tex]

Replacing eq. 1 and 2:

[tex]-50a=(10a*20)-\frac{1}{2} *g*20^{2} \\-250a=-\frac{1}{2} *9.8*20^{2} \\a=7.84m/s^{2}[/tex]

b) The impact speed is equal:

[tex]v_{i} ^{2} =v^{2} +2gs[/tex]

Where

s = h = -50a

[tex]v^{2} _{i} =(10a)^{2} +2*(-9.8)*(-50a)\\v=\sqrt{100a^{2}+980a } \\v=\sqrt{(100*7.84^{2})+(980*7.84) } =117.6m/s[/tex]

c) The height is:

[tex]v_{i} ^{2} =v^{2} +2gs\\0=(10a)^{2} -2gs\\(10a)^{2} =2gs\\s=\frac{(10a)^{2} }{2g} \\s=\frac{(10*7.84)^{2} }{2*9.8} =313.6m[/tex]

htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by [tex]\frac{1}{2} mv^2[/tex]

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = [tex]\frac{1}{2} (4m)(v^2) = 2mv^2[/tex]

B) KE =[tex]\frac{1}{2} (3m)(2v)^2 = 6mv^2[/tex]

C) KE = [tex]\frac{1}{2} (2m)(3v)^2 = 9mv^2[/tex]

D) KE = [tex]\frac{1}{2} (3)(4v)^2 = 8mv^2[/tex]

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

Answer:

Electric field at radius r inside the solid sphere is

[tex]E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:

Given

The radius of the solid sphere is [tex]r_{i}=0.105 \ m\\[/tex]

The charge on the solid sphere is [tex]q_{1}=+4.4\ \mu C[/tex]

The inner radius of the shell is [tex]r_{3}=0.15\ m[/tex]

The outer radius of the shell is [tex]r_{3}=0.242\ m[/tex]

The total charge on the shell is [tex]q_{2}=-1.1\ \mu C[/tex]

PART(A)

The magnitude of electric field at radius r where [tex] r<r_{1}[/tex] \\The volumetric charge density of the solid sphere will be

[tex]\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\\[/tex]

The charge enclosed by the radius r inside the solid sphere is

rho=[tex]q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}\\[/tex]

According to gauss law

[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]

PART(B)

The electric field at radius r where [tex]r_{2}<r<r_{3}[/tex]

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

[tex]q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C\\[/tex]

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

[tex]q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C\\[/tex]

The charge enclosed by the radius r where [tex]r_{1}<r<r_{2}[/tex]

[tex]q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0\\[/tex]

According to gauss law

[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C\\[/tex]

(a) Inside the aluminum sphere, the electric field is 0 N/C due to electrostatic equilibrium.

(b) Inside the copper shell, for any radius between r2 and r3, the electric field is given by (3.95 x 10⁴) / r² N/C.

To solve this problem, we need to understand the behavior of the electric field in different regions around charged spherical conductors.

(a) Electric Field Inside the Aluminum Sphere

Inside a conductor in electrostatic equilibrium, the electric field is zero. Since the aluminum sphere is a conductor, the electric field inside it (for any radius r < r1) is zero.

Result: The magnitude of the electric field inside the aluminum sphere is 0 N/C.

(b) Electric Field Inside the Copper Shell

For the region inside the copper shell but outside the aluminum sphere (r2 < r < r3), we can apply Gauss's Law. According to Gauss's Law, the electric field at a distance r from the center due to a symmetric charge distribution can be calculated as follows:

Define the Gaussian surface: Here, it will be a spherical surface of radius r where r2 < r < r3.

Calculate the enclosed charge: The total charge enclosed within the Gaussian surface is only the charge of the aluminum sphere, q1 = +4.4 μC, because the copper shell's net charge is enclosed separately.

Apply Gauss's Law: Gauss's law states that the electric flux through a surface is proportional to the enclosed charge:

Equation: E * 4πr² = q1/ε₀

Solve for E:

E = q1 / (4πε₀ r²)

Substitute the values: q1 = 4.4 x 10⁻⁶ C, ε₀ = 8.854 x 10⁻¹² C²/(N·m²), and r is the distance within the copper shell.

[tex]E = \frac{4.4 \times 10^{-6} \, \text{C}}{4\pi \cdot (8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \cdot r^2}[/tex]

Simplifying gives:

E = (4.4 x 10⁻⁶) / (1.112 x 10⁻¹⁰ * r²)

E = (3.95 x 10⁴) / r² N/C

Hence, the magnitude of the electric field inside the copper shell, at any point where r2 < r < r3, is (3.95 x 10⁴) / r²  N/C.

Answer:

ΔH = 33.17m

Explanation:

By knowing the amount of time it takes the rocket to travel the horizontal 27m, we will be able to calculate the height when x=27m. So:

[tex]X = V_{o}*cos(60)*t[/tex]   where [tex]X=27m; V_{o}=75m/s[/tex]

Solving for t:

t=0.72s

Now we calculate the height of the rocket:

[tex]Y_{f}=V_{o}*sin(60)*t-\frac{g*t^{2}}{2} = 44.17m[/tex]

If the wall was 11m-high, the difference is:

ΔH = 33.17m

Using the equation of projectile motion, the rocket cleared the top of the wall by 38.305m.

The equation of projectile motion is given by:

y= xtanθ  +(1/2)gx²/(u²cos²θ )

Given information:

Initial velocity, u=75m/s

The angle of projection, θ = 60°

The horizontal distance, x=27m

Vertical distance will be calculated as:

y= xtanθ  +(1/2)gx²/(u²cos²θ )

y=27 tan 60° + 0.5×9.8×27²/(75²cos²60°)

y=49.305m

So, the rocket cleared the top of the wall by

49.305-11 =38.305m

Therefore, using the equation of projectile motion the rocket cleared the top of the wall by 38.305m.

To know more about projectile motion, click here:

https://brainly.com/question/12860905

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Answer:

pirate walk 259 feet

displacement vector = 111 i - 234 j

direction is along south of east at angle 64.62° ,  anticlockwise

total travel 345 feet

Explanation:

given data

walk east = 111 feet

walk south = 234 feet

to find out

How far did the pirate walk and displacement vector and distance and direction relative to east

solution

we consider here distance AB is 111 feet and than he turn right i.e south distance BC is 234 feet so

so angle BAC will be

tan θ = [tex]\frac{234}{111}[/tex]

θ = 64.62

and AC distance will be

AC = [tex]\sqrt{234^{2} + 111^{2}}[/tex]

AC = 259 feet

so pirate walk 259 feet

and

displacement vector is express as

displacement vector = AC ( cosθ i + sinθ j )

displacement vector = 259 ( cos64.62 i + sin64.62 j )

displacement vector = 111 i - 234 j

and

so direction is along south of east at angle 64.62° ,  anticlockwise