A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?

Answer:

whats the formula

Explanation:

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]

[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]

[tex]\alpha=\cos^{-1}(0.3846)[/tex]

[tex]\alpha=67.38^{\circ}[/tex]

We need to calculate the angle β

Using cosine law

[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]

[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]

[tex]\beta=\cos^{-1}(0.923)[/tex]

[tex]\beta=22.63^{\circ}[/tex]

We need to calculate the force on 130 cm side

Using formula of force

[tex]F_{130}=ILB\sin\theta[/tex]

[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]

[tex]F_{130}=0[/tex]

We need to calculate the force on 120 cm side

Using formula of force

[tex]F_{120}=ILB\sin\beta[/tex]

[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]

[tex]F_{120}=0.1385\ N[/tex]

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

[tex]F_{50}=ILB\sin\alpha[/tex]

[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]

[tex]F_{50}=0.1385\ N[/tex]

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.

b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.

c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.

Given the following data:

Let us assume the current is flowing in a counterclockwise direction in the right-angle triangle.

First of all, we would determine the angles by using cosine rule:

[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]

[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]

a. To the determine the magnitude of the magnetic force on the 130 cm side:

Mathematically, the force acting on a current in a magnetic field is given by the formula:

[tex]F = BILsin\theta[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]

b. To the determine the magnitude of the magnetic force on the 120 cm side:

[tex]F_{120}=BILsin\beta[/tex]

[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]

c. To the determine the magnitude of the magnetic force on the 50 cm side:

[tex]F_{50}=BILsin\alpha[/tex]

[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]

Read more: https://brainly.com/question/13754413

Answer:

The time taken is  [tex]t = 2.739 *10^{8} \ s[/tex]

Explanation:

From the question we are told that

    The speed of the spacecraft is [tex]v = 0.99c[/tex]

    where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]

    =>   [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]

    The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]

      [tex]t = 2.739 *10^{8} \ s[/tex]

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

Where:

[tex]m[/tex] - Mass of the crate, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]

[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]

The final height of the crate is:

[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]

[tex]y_{2} = 0.8\,m[/tex]

If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:

[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\mu_{k} \approx 0.548[/tex]

Then, the magnitude of the friction force is:

[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:

[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]f = 55.851\,N[/tex]

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]

Now, the final speed is cleared:

[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]

[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]

Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:

[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Answer:

The distance is  [tex]d = 0.00029065 \ m[/tex]

Explanation:

From the question we are told that

    The  first wavelength is  [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]

     The  second wavelength is  [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]

     The  difference in the  fringe pattern is  n =  1.0  

Generally the equation defining the effect of the movement of  the mirror M 2 in a Michelson interferometer is mathematically represented as

          [tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]

Here d is the mirror M 2  must be moved

substituting values

         [tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]

        [tex]d = 0.00029065 \ m[/tex]

Answer:

0.1 kg or 100 g

Explanation:

The length of the wire = 0.5 m

the field magnitude = 0.98 T

the current through the wire = 2.0 A

magnetic force due to a wire carrying current is

F = [tex]IlB[/tex]

where

F is the force

[tex]I[/tex] is the current = 2 A

[tex]l[/tex] is the length of the wire

B is the magnetic field strength

Substituting, we have

F = 2 x 0.5 x 0.98 = 0.98 N

This force balances the weight of the mass

weight = mg

where m is the mass of the wire

g is acceleration due to gravity = 9.81 m/s^2

therefore, weight = m x 9.81 = 9.81m

equating this weight with the force, we have

0.98 = 9.81m

m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g

Answer:

100 g

Explanation:

Answer:

tyygggghgtyhyrdfgyyyhjjillbxsrfvgygvnjj

Explanation:

cffczhxucuxoyitxohvojcdivbjv ohxc

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

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

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

Answer:

Hey there!

The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body

Let me know if this helps :)

Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

Answer:

1) f = 214 Hz , 2)  answer is c , 3) f = 428 Hz , 4)   f₂ = 428 Hz ,   f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

                L = λ / 2

               λ  = 2L

               λ  = 2 0.8

               λ  = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

                v =λ  f

                f = v / λ  

                f = 343 / 1.6

                f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

           λ' = 2 L ’

          as L ’<L₀

          λ' <λ₀

          f = v / λ'

          f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

          λ = 2 0.4 = 0.8 m

          f = 343 / 0.8

          f = 428 Hz

4) the different harmonics are described by the expression

         λ = 2L / n           n = 1, 2, 3

         λ₂ = L

         f₂ = 343 / 0.8

         f₂ = 428 Hz

         λ₃ = 2 0.8 / 3

         λ₃ = 0.533 m

         f₃ = 343 / 0.533

         f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

       the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

        L = λ / 4

        L = 1.6 / 4

        L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

           λ = 4L / n         n = 1, 3, 5 (2n + 1)

therefore only odd values ​​can produce standing waves

           λ₃ = 4L / 3

           λ₃ = 4 0.4 / 3

           λ₃ = 0.533

           f₃ = 343 / 0.533

           f₃ = 643 Hz

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

Learn more:

https://brainly.com/question/14573987?referrer=searchResults

Answer:

n= speed of light in vacuum/ speed of light in the other medium.

Explanation:

If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;

n= speed of light in vacuum/ speed of light in the other medium.

Hence;

n= c/v

Where;

n= refractive index of the material

c= speed of light in vacuum

v = speed of light in another medium.

Note that the refractive index is the amount by which a transparent medium decreases the speed of light.

Answer:

I think the answer is A...I'm not sure

Explanation:

A=(6,4)

B=(-2,-1)

A-B=(6-(-2)),(4-(-1))

=(6+2),(4+1)

=(8,5)

Answer:

[tex]6-(-2)=[/tex]

[tex]6+2[/tex]

[tex]=8[/tex]

[tex]4-\left(-1\right)[/tex]

[tex]=4+1[/tex]

[tex]=5[/tex]

[tex](8,5)[/tex]

[tex]\textbf{OAmalOHopeO}[/tex]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

Answer:

0.273m/s

Explanation:

first find out the meaning of 0.90×10−4m3/s

literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s

1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?

cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²

so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²

they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s

Answer:

D. evacuated tube collector due to reduced heat loss

Explanation:

Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.

During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.

Answer:

b. The current in the loop always flows in a counterclockwise direction.

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.

Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

Learn more about the magnetic field here:

https://brainly.com/question/14848188

Answer:

The intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

[tex]I(dB) = 10Log(\frac{I}{I_o} )[/tex]

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

[tex]120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2[/tex]

The intensity of sound of a whisper

[tex]20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2[/tex]

Thus, the intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......