9) Two mirrors are at right angles to one another. A light ray is incident on the first at an angle of 30° with respect to the normal to the surface. What is the angle of reflection from the second surface?

Answer:

The distance is 1.4 km.

Explanation:

Given that,

Initial velocity u= 1.7 m/s

Accelerate a= 3.0 m/s²

Time = 30 s

We need to calculate the distance

Using equation of motion

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

Where, u = initial velocity

a = acceleration

t = time

Put the value in the equation

[tex]s=1.7\times30+\dfrac{1}{2}\times3.0\times(30)^2[/tex]

[tex]s=1401\ m[/tex]

[tex]s=1.4\ km[/tex]

Hence, The distance is 1.4 km.

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of [tex]Cl_2[/tex] = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of [tex]Cl_2[/tex] = 71 g/mole

Molar mass of [tex]PCl_5[/tex] = 208.24 g/mole

First we have to calculate the moles of [tex]P[/tex] and [tex]Cl_2[/tex].

[tex]\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles[/tex]

[tex]\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2P+5Cl_2\rightarrow 2PCl_5[/tex]

From the balanced reaction we conclude that

As, 5 moles of [tex]Cl_2[/tex] react with 2 moles of [tex]P[/tex]

So, 0.352 moles of [tex]Cl_2[/tex] react with [tex]\frac{2}{5}\times 0.352=0.1408[/tex] moles of [tex]P[/tex]

That means, in the given balanced reaction, [tex]Cl_2[/tex] is a limiting reagent and it limits the formation of products and [tex]P[/tex] is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of [tex]PCl_5[/tex].

As, 5 moles of [tex]Cl_2[/tex] react with 2 moles of [tex]PCl_5[/tex]

So, 0.352 moles of [tex]Cl_2[/tex] react with [tex]\frac{2}{5}\times 0.352=0.1408[/tex] moles of [tex]PCl_5[/tex]

Now we have to calculate the mass of [tex]PCl_5[/tex].

[tex]\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5[/tex]

[tex]\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g[/tex]

Now we have to calculate the mass of product produced (actual yield).

[tex]\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100[/tex]

[tex]70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100[/tex]

[tex]\text{Actual yield of }PCl_5=20.67g[/tex]

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Try this solution, note, the resistance of each resistor is marked with green colour, the items (a) and (b) - with red colour.

By using Ohm's law,

a.) Current = 19.17 A

b.) Current = 110.6 A

We are given different voltages and currents for the two resistors.

First resistor:

Voltage [tex]V_{1}[/tex] = 3.88 v

Current [tex]I_{1}[/tex] = 43.6 A

By using Ohm's law, we can calculate [tex]R_{1}[/tex]

V = IR

Make R the subject of formula

R = V/I = 3.88/43.6

R = 0.08899 Ohms

Second resistor

Voltage [tex]V_{2}[/tex] = 4.48 v

Current [tex]I_{2}[/tex] = 14.45 A

By using Ohm's law, we can calculate [tex]R_{2}[/tex]

V = IR

Make R the subject of formula

R = V/I = 4.48/14.45

R = 0.3100 Ohms

a.) Whenever resistors are connected in series, the same current will pass through them

The total resistance in the resistors in series will be achieved by using the below formula

R = [tex]R_{1}[/tex] + [tex]R_{2}[/tex]

R = 0.08899 + 0.3100

R = 0.399 Ohms

By using Ohm's law:

V = IR

Make I the subject of formula

I = V/R = 7.65/0.399

I = 19.17 A

b.) Whenever resistors are arranged in parallel, the same voltage will be supplied to them.

The resultant resistor can be calculated by using the below formula

1/R = 1/[tex]R_{1}[/tex] + 1/[tex]R_{2}[/tex]

1/R = 1/0.08899 + 1/0.3100

1/R = 11.237 + 3.2258

R = 1/14.463

R = 0.069 Ohms

By using Ohm's law,

V = IR

Make I the subject of formula

I = V/R = 7.65/0.069

I = 110.6 A

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Answer:

c.) 25 N

Explanation:

 We find the volume of the brick, knowing that the volume of a cube is given by the formula:

[tex]l=0,1 metros \\V=l^{3}\\V=(0,1\: \: m)^{3}=0,001\: \: m^{3}[/tex]

being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

[tex]m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg[/tex]

We find the weight by multiplying the mass of the object with the gravity constant.

[tex]W=m*g=(2.5 \:Kg)*(9,81 \:m/s^{2} )=24,5 N\approx25\: N[/tex]

Final answer:

The weight of the brick is 24.5 N.

Explanation:

To find the weight of the brick, we can use the formula weight = density × volume × gravitational acceleration.

First, we need to calculate the volume of the cube. The volume of a cube is given by the formula V = side³, where side is the length of one side of the cube.

Given that the side of the cube is 10 cm, the volume of the cube is V = 10 cm × 10 cm × 10 cm = 1000 cm³.

Next, we convert the volume from cm³ to m³ by dividing by 100^3: V = 1000 cm³ ÷ (100 cm/m)³ = 0.001 m³.

Now, we can calculate the weight of the brick: weight = 2500 kg/m³ × 0.001 m³ × 9.8 m/s² = 24.5 N.

Therefore, the weight of the brick is 24.5 N, which is not one of the given options.

Answer:

250 A

Explanation:

B = 5 x 10^-5 T, r = 1 m

Let current be i.

the magnetic field due to a straight current carrying conductor is given by

B = μ0 / 4π x 2i / r

5 x 10^-5 = 10^-7 x 2 x i / 1

i = 250 A

Answer:

The sled slides d=0.155 meters before rest.

Explanation:

m= 60 kg

V= 2 m/s

μ= 0.3

g= 9.8 m/s²

W= m * g

W= 588 N

Fr= μ* W

Fr= 176.4 N

∑F = m * a

a= (W+Fr)/m

a= 12.74m/s²

t= V/a

t= 0.156 s

d= V*t - a*t²/2

d= 0.155 m

The electric force on the electron is opposite in direction to the electric field E. E points in the -y direction, so the electric force will point in the +y direction. The magnitude of the electric force is given by:

F = Eq

F = electric force, E = electric field strength, q = electron charge

We need to set up a magnetic field such that the magnetic force on the electron balances out the electric force. Since the electric force points in the +y direction, we need the magnetic force to point in the -y direction. Using the  reversed right hand rule, the magnetic field must point in the -z direction for this to happen. Since the direction is perpendicular to the +x direction of the electron's velocity, the magnetic force is given by:

F = qvB

F = magnetic force, q = charge, v = velocity, B = magnetic field strength

The electric force must equal the magnetic force.

Eq = qvB

Do some algebra to isolate B:

E = vB

B = E/v

Let's solve for the electron's velocity. Its kinetic energy is given by:

KE = 0.5mv²

KE = kinetic energy, m = mass, v = velocity

Given values:

KE = 2.9keV = 4.6×10⁻¹⁶J

m = 9.1×10⁻³¹kg

Plug in and solve for v:

4.6×10⁻¹⁶ = 0.5(9.1×10⁻³¹)v²

v = 3.2×10⁷m/s

B = E/v

Given values:

E = 7500V/m

v = 3.2×10⁷m/s

Plug in and solve for B:

B = 7500/3.2×10⁷

B = 0.00023T

B = 0.23mT

To minimize the required magnitude of the magnetic field and keep the electron moving along the x-axis, the direction of the magnetic field should be chosen to cancel out the force due to the electric field. The magnitude of the magnetic field needed is 2.59 mT.

To keep the electron moving along the x-axis and minimize the required magnitude of the magnetic field, the force on the electron due to the magnetic field should cancel out the force on the electron due to the electric field.

Answer:

12 cm and 0.4

Explanation:

f = - 20 cm, u = - 30 cm

Let v be the position of image and m be the magnification.

Use lens equation

1 / f = 1 / v - 1 / u

- 1 / 20 = 1 / v + 1 / 30

1 / v = - 5 / 60

v = - 12 cm

m = v / u = - 12 / (-30) = 0.4

Explanation:

It is given that,

Momentum of the photon, [tex]p=5.55\times 10^{-27}\ kg-m/s[/tex]

(a) We need to find the wavelength of this photon. It can be calculated using the concept of De-broglie wavelength.

[tex]\lambda=\dfrac{h}{p}[/tex]

h is the Planck's constant

[tex]\lambda=\dfrac{6.67\times 10^{-34}\ Js}{5.55\times 10^{-27}\ kg-m/s}[/tex]

[tex]\lambda=1.2\times 10^{-7}\ m[/tex]

or

[tex]\lambda=120\ nm[/tex]

(b) The wavelength lies in the group of ultraviolet rays. The wavelength of UV rays lies in between 400 nm to 10 nm.

Answer:

SI unit of k (spring constant) = N/m

Explanation:

We have expression for force in a spring extended by x m given by

         F = kx

Where k is the spring constant value.

Taking units on both sides

         Unit of F = Unit of k x Unit of x

          N = Unit of k x m

          Unit of k = N/m

SI unit of k (spring constant) = N/m

Answer:

N/m

Explanation:

The unit of k (spring constant) in SI system is N/m.

SI unit of k (spring constant) = N/m

Final answer:

The magnitude of the average induced electric field in the coil, encircling a solenoid with given dimensions and a decreasing current of 79.0 A/s, is approximately [tex]\(1.80 \times 10^{-4}\)[/tex]volts per meter.

Explanation:

The induced electric field in the coil is determined by Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In this scenario, the solenoid serves as the primary coil, and the circular coil is the secondary coil.

The magnetic flux through a single turn of the circular coil is given by[tex]\(\Phi = B \cdot A\)[/tex], where is the magnetic field strength and  is the area of the coil. The magnetic field inside a solenoid is given by[tex]\(B = \mu_0 \cdot N_s \cdot I\)[/tex].

Substituting these expressions into the formula for magnetic flux, we get [tex]\(\Phi = \mu_0 \cdot N_s \cdot I \cdot A\).[/tex] The induced emf  is then given by Faraday's law: [tex]\(\mathcal{E} = -\frac{d\Phi}{dt}\)[/tex]. Taking the derivative with respect to time and using the given values, we find[tex]\(\mathcal{E} = -\mu_0 \cdot N_s \cdot A \cdot \frac{dI}{dt}\).[/tex]

Finally, the induced electric field in the coil is given by [tex]\(E = \frac{\mathcal{E}}{A}\)[/tex]. Substituting the values into this formula provides the magnitude of the average induced electric field in the coil, approximately[tex]\(1.80 \times 10^{-4}\)[/tex]volts per meter. This calculation yields a quantitative measure of the induced electric field strength in response to the changing current in the solenoid.

Answer:

The increases depth of the water is 0.021 m.

Explanation:

Given that,

One side of square = 5.8 m

Depth = 4.2 m

Temperature = 24°C

Coefficient of volume expansion for water[tex]\beta = 210\times10^{-6}\ /^{\circ}C[/tex]

We need to calculate the volume of the pool

[tex]V= 5.8\times5.8\times4.2[/tex]

[tex]V=141.288\ m^3[/tex]

Using formula of coefficient of volume expansion

[tex]\Delta V=\Beta V_{0}\times\Delta T[/tex]

If the temperature changes by 24 degrees C during the afternoon,

[tex]5.8\times5.8\times h=210\times10^{-6}\times141.288\times24[/tex]

[tex]h=\dfrac{210\times10^{-6}\times141.288\times24}{5.8\times5.8}[/tex]

[tex]h=0.021\ m[/tex]

Hence. The increases depth of the water is 0.021 m.

To find the angle, in radians, that the ladder makes with the ground, we can use trigonometry. The length of the ladder can be found using the Pythagorean theorem, and the angle can be calculated as the arcsine of the ratio of the height of the building to the length of the ladder. The angle is approximately 0.6435 radians.

To find the angle, in radians, that the ladder makes with the ground, we can use trigonometry. The ladder, the ground, and the building form a right triangle. The ratio of the opposite side (the height of the building) to the hypotenuse (the length of the ladder) is equal to the sine of the angle. Using this information, we can calculate the angle in radians.

First, we need to find the length of the ladder using the Pythagorean theorem: l^2 = 20^2 + 15^2 = 625. Taking the square root of both sides, we find that the length of the ladder is 25 feet.

The sine of the angle can be calculated as the ratio of the height of the building to the length of the ladder: sin(angle) = 15/25 = 0.6. Taking the arcsine (inverse sine) of 0.6, we find that the angle in radians is 0.6435 (rounded to two decimal places).

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The question asks for the electric field strength at the center of a charged ring, which involves using physics concepts related to electric fields and charge distributions. An accurate calculation would require application of formulas for fields due to point charges distributed in a ring, or thorough integration techniques taught at high school level physics.

The subject of the question is Physics, and it pertains to the concept of electric fields created by charged objects. We are considering two identical rings, each with a positive charge of +10.0 nC, facing each other with a separation of 23.0 cm. What is required is to calculate the electric field strength at the center of the left ring. By the symmetry of the setup and since the charges are identical and positive, the electric field at the center of the left ring due only to it would be zero because there is no charge displacement leaving the right ring's field to consider. The electric field strength at a point due to a charged ring on the ring's axis can be calculated using the formula for electric fields due to point charges, since a charged ring can be thought of as a distribution of point charges.

However, without precise formulae for the field due to a ring or detailed integration methods, we cannot calculate the precise value for this field. A good physics course at the high school level will offer the tools necessary to derive such formulae and solve this problem accurately.

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         [tex]\omega =\frac{72}{0.445}=161.8rad/s[/tex]

Frequency

         [tex]f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min[/tex]

Revolutions per minute = 2.33

b) Centripetal acceleration

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

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   [tex]a=\frac{72^2}{0.445}=11649.44m/s^2[/tex]

Centripetal acceleration = 11649.44m/s²

To find the revolutions per minute of the tires, calculate the circumference of the tire and divide the jet's speed in meters per minute by this circumference. For the centripetal acceleration, first find the tire's angular velocity in radians per second and plug it, along with the tire's radius, into the centripetal acceleration formula.

For part (a), to calculate how many revolutions per minute (rev/min) the tires are rotating, you must determine the circumference of the tire first. The circumference (C) is given by [tex]C = \pi d[/tex], where d is the diameter. Given d = 0.89 m, the circumference is [tex]C = \pi (0.89 m)[/tex]. This value represents the distance the tire covers in one revolution.

To find out how many rev/min the tires make, we need to consider the speed of the jet which is 72 m/s. Since there are 60 seconds in a minute, the distance covered in one minute is [tex]72 m/s * 60 s/min.[/tex] Dividing this distance by the tire's circumference gives us the number of revolutions per minute:

[tex]rev/min = \(\frac{(72 m/s) \* (60 s/min)}{\\(pi)(0.89 m)}\)[/tex]

After calculating, you get the tires' rotation rate in rev/min.

For part (b), to calculate the centripetal acceleration (ac) at the edge of the tire, use the formula [tex]ac = \(r\omega^2\),[/tex]where r is the radius of the tire, and [tex]\(\omega\)[/tex] is the angular velocity in radians per second. The angular velocity can be found by converting the rev/min to revolutions per second (rev/s), and then multiplying by [tex]2\pi[/tex] to convert to radians per second.

Lastly, replacing [tex]\(\omega\)[/tex] and r in the centripetal acceleration formula will give you the edge's centripetal acceleration in m/s2.