What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

[tex]a = \frac{v^2}{r}[/tex]

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

[tex]a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}[/tex]

Hence the centripetal acceleration of the race car is 10.53m/s²

Answer:

The minimum current in the rod that can cause it to float 2.96 A.

Explanation:

Given;

length of the copper rod, L = 1.43 m

mass of the rod, m = 236 g = 0.236 kg

magnetic field, B = 0.546 T

The magnetic of the rod is given by;

Fm = BIL

The downward force on the rod is given by;

F = mg

For the rod to float, the difference in the two force will be zero;

BIL - mg = 0

BIL = mg

I = mg / BL

I = (0.236 x 9.8) / (0.546 x 1.43)

I = 2.96 A

Therefore, the minimum current in the rod that can cause it to float 2.96 A.

Answer:

Explanation:

We are expected to solve for the velocity with no slip condition

we know that the expression that relate coefficient of friction and velocity is given as

μs = v^2/rg

Given

coefficient of friction μs = 0.3

radius r= 0.15

assume g=9.81m/s^2

substituting into the expression we have

0.3= v^2/0.15*9.81

v^2=0.3*0.15*9.81

v^2=0.44145

v=√0.44145

v=0.66

therefore the velocity is 0.66m/s

Answer:

578.66 N

Explanation:

The first step is to calculate the mass

mgh= 3200J

3200/9.8×5.53

3200/54.194

m = 59.047 kg

Therefore the weight can be calculated as follows

Weight = m × g

= 59.047 × 9.8

= 578.66 N

Answer:

12.5 m

Explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m

Answer:

12.898A

Explanation:

The formula for calculating the magnetic field in the loop bus expressed as;

B = Iμ0/2r

Given

Diameter d = 0.60cm

Radius r = d/2 = 0.30cm

r = 0.0030m

Permittivity of free space μ0 = 4π×10^-7

Magnetic field strength B = 2.7×10^-3T

Substitute into the formula and get I

2.7×10^-3 = 4π×10^-7I/2(0.0030)

2.7×10^-3 = 4π×10^-7I/0.0060

Cross multiply

2.7×10^-3 × 6.0×10^-3 = 4π×10^-7I

16.2×10^-6 = 4π×10^-7I

I = 16.2×10^-6/4π×10^-7

I = 16.2×10^-6/4(3.14)×10^-7

I = 16.2×10^-6/12.56×10^-7

I = 1.2898×10^{-6+7}

I = 1.2898×10

I = 12.898A

Hence the current in the loop is 12.898A

Answer:

5.6449

9 mpa

Explanation:

we are to determine mass of steam that has entered and also the pressure of steam.

After solving

Mass of steam = m2 - m1

= 15.925-10.2901

= 5.6449kg

Then the enthalpy of steam was calculated to be 3109.26

Using steam table, tl = 400⁰c

Hl = 3109.26

Supply line pressure = 9mpa

Please refer to attachment for all calculations

Analysing the Question:

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

Finding the missing force:

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

Answer:

Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.

Explanation:

To find the distance at which the first package will land we need to calculate the time:

[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]

Where:

Y(f) is the final position = 0

Y(0) is the initial position = 160 m

V(0y) is initial speed in "y" direction = 0

g is the gravity = 9.81 m/s²

t is the time=?                                          

[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]

[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]

Now we can find the distance of the first package:

[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]

Then, after 2 seconds the distance traveled by plane is (from the initial position):

[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]

Now, the distance of the second package is:

[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]

The distance between the packages is:

[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]

Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.

I hope it helps you!

Answer:

250000 [m²]

Explanation:

A unit analysis has to be performed, where the unit of length is the meter. And the unit for the area is the square meter (m²)

L = 500 [m]

Therefore if we want to convert this length to meters we must square the length.

A = 500² = 500*500 = 250000 [m²]

Answer:

it's A trust me ok

Explanation:

i can't explain

Answer:

x₁ = 0.1878 m

Explanation:

For this exercise we will use conservation of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point with fully compressed spring

         Em_f = K_e + U

         Em_f = ½ K x² + m g x

energy is conserved

         Em₀ = Em_f

        m g h = ½ K x² + m g x

       ½ K x² + mg (x- h) = 0

let's substitute

       ½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0

        3.65 x² + 0.294 (x- 0.25) = 0

        x² + 0.080548 (x- 0.25) = 0

        x² - 0.020137 + 0.080548 x = 0

        x² + 0.080548 x - 0.020137 = 0

let's solve the quadratic equation

      x = [0.080548 ±√ (0.080548² + 4   0.020137)] / 2

      x = [0.080548 ± 0.29502] / 2

      x₁ = 0.1878 m

      x₂ = -0.1072 m

These are the compression and extension displacement of the spring

Answer:

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Explanation:

From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:

[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)

[tex][l] = B\cdot [t][/tex] (Eq. 2)

Now we finally clear each constant:

[tex]A = \frac{[l]}{[l]^{3}}[/tex]

[tex]A = \frac{1}{[l]^{2}}[/tex]

[tex]B = \frac{[l]}{[t]}[/tex]

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Answer:

Thank you ☺️

Answer:

The time is [tex]t_t =   3.7583 \ s [/tex]

Explanation:

From the question we are told that

   The angle is  [tex]\theta =  30^o[/tex]

    The horizontal  distance is  [tex]d =  120 \ ft[/tex]

Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as

      [tex]v =  u  +  at[/tex]

here a =  -g  the negative sign is because the direction of motion is against gravity

So

       [tex]v =  v_i   +  at[/tex]

Here[tex] v_i [/tex] is the vertical  component of the initial  velocity of the ball which is  mathematically

represented as

        [tex]v_i = usin(\theta )[/tex]

So      

=>     [tex]0  =  usin(\theta )  -9.8t[/tex]

Generally the total time taken to travel the 120 ft  is mathematically represented as

        [tex]t_t = \frac{120}{v_h}[/tex]

Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as

     [tex]v_h  =  u cos(\theta )[/tex]

So

       [tex]t_t = \frac{120}{ u cos(\theta )}[/tex]

Generally the time taken to reach the maximum height is  

      [tex]t = \frac{t_t}{2}[/tex]

=>    [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]

=>    [tex]t = \frac{60}{ u cos(\theta )} [/tex]

So

      [tex]0  =  usin(\theta )  -9.8*   [\frac{60}{ u cos(\theta )}][/tex]

        [tex]  usin(\theta )   = 9.8*   [\frac{60}{ u cos(\theta )}][/tex]

       [tex]  usin(\theta ) *  u cos(\theta)  =60*  9.8   [/tex]

        [tex]  u^2 sin(\theta ) cos(\theta)  =60*  9.8   [/tex]

       [tex]  u^2 sin(30 ) cos(30)  =60*  9.8   [/tex]

        [tex]  u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2}  =588.6  [/tex]

         [tex]  u^2 *\sqrt{3}  =2354.4 [/tex]

          [tex]  u^2 = 1359.31 [/tex]

        [tex]  u = 36.87 \ ft/s  [/tex]

Substituting this value into the equation for total time

     [tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]

     [tex]t_t =   3.7583 \ s [/tex]

Answer:

noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).

Answer:

True.

Explanation:

        [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)

        [tex]\Delta x} = v_{avg}* {\Delta t}[/tex]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

[tex] E_{m} = E_{k} + E_{p} [/tex]

[tex] E_{m} = \frac{1}{2}v^{2} + gh [/tex]

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

[tex] E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg [/tex]

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

[tex] P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW [/tex]

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

Answer:

700 N

Explanation:

The gravitational potential energy of the system decreases by 28,000 joules

The diver jumps from a height of 40 meters

Therefore the weight in Newton can be calculated as follows

= 28,000/40

= 700 newtons

Answer:

Neurons

Explanation:

We humans have a nervous system that coordinates our behavior and transmits signals between different parts of our body.

Now, this nervous system contains a lot of nerve cells which we call Neurons. These Neurons have a cell like body and their job is to transmit signals from one part of our body to another.

Thus, the cell is called Neurons.

Answer:

Explanation:

Answer:

The value is [tex]n=  9.375 *10^{15} \  electrons [/tex]

Explanation:

From the question we are told that

  The average current is  [tex]I  =  25.0 \mu A = 25.0 *10^{-6} \  A[/tex]

Generally the quantity of charge (electron )  is mathematically represented as

       [tex]Q =  ne[/tex]

Here e is the charge on a single electron with value [tex]e = 1.60  *10^{-19} \  C[/tex]

   Generally current is mathematically represented as

     [tex]I  = \frac{Q}{t}[/tex]

=>   [tex]I  = \frac{ne}{t}[/tex]

Here t is time which is given as 1 minutes =  60  seconds

  and  n is the number of electrons

So

      [tex]25.0 *10^{-6}  = \frac{ n* 1.60  *10^{-19}}{60}[/tex]

=>    [tex] 60  * 25.0 *10^{-6} =  n* 1.60  *10^{-19}  [/tex]

=>    [tex]n=  \frac{60  * 25.0 *10^{-6} }{ 1.60  *10^{-19} }[/tex]

=>    [tex]n=  9.375 *10^{15} \  electrons [/tex]

The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons

We are given;

Average Current; I = 25 μA = 25 × 10⁻⁶ A

Formula for the current is;

I = Q/t = ne/t

where;

n is number of electrons

e is electron charge = 1.6 * 10⁻¹⁹ C

t is time = 1 minute = 60 seconds

Thus making n the subject gives;

n = It/e

n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)

n = 9.375 × 10¹⁵ electrons

Read more about number of electrons at; https://brainly.com/question/11406294

Answer:

10) The distance between the archer and the tree is 50.074 meters.

11) The speed of the banana when it hits the water is approximately 13.554 meters per second.

Explanation:

10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:

[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)

Where:

[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.

[tex]x[/tex] - Final position of the arrow, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:

[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]

[tex]x = 50.074\,m[/tex]

The distance between the archer and the tree is 50.074 meters.

11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:

[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)

Where:

[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.

[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.

Each component of the speed are obtained by using these kinematic equations:

[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)

[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)

Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.

If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:

[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]

[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]

[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]

[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]

And the speed of the banana right before hitting the water is:

[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v \approx 13.554\,\frac{m}{s}[/tex]

The speed of the banana when it hits the water is approximately 13.554 meters per second.

Answer:

Answer is C or B

Explanation:

Answer:

a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].

b) The reading after 20 minutes is approximately 70.770 ºF.

Explanation:

a) Newton's law of cooling is represented by the following ordinary differential equation:

[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)

Where:

[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.

[tex]\tau[/tex] - Time constant, measured in minutes.

[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.

[tex]M[/tex] - Medium temperature, measured in Fahrenheit.

Now we proceed to solve the differential equation:

[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]

[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]

[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]

[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]

[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)

Where:

[tex]t[/tex] -Time, measured in minutes.

[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.

b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])

[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]

[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]

[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]

[tex]\tau = 5.461\,min[/tex]

The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:

[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]

[tex]T(20) \approx 70.770\,^{\circ}F[/tex]

The reading after 20 minutes is approximately 70.770 ºF.