In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the distance to a star is 3.6 light-years, what is this distance in meters? (There are 365.25 days in one year.)

Answer:

It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.

Explanation:

Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit ([tex]\tau_0=RC[/tex]) which in our case is;

[tex]\tau_0=1.3\,k\Omega\,*\,10.0 \mu F=13\,\,10^{-3}\,s[/tex] (13 milliseconds)

Recall as well the expression for the charge in the capacitor (from it initial value [tex]Q_0[/tex], as it discharges via a resistor R:

[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }[/tex]

So knowing that the capacitor started with a charge of 30 [tex]\mu C[/tex] and reduced after a time "t" to 30 [tex]\mu C[/tex] , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:

[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }\\15 \mu C=30 \mu C\,\,e^{-\frac{t}{13\,ms} }\\e^{-\frac{t}{13\,ms} }=\frac{1}{2} \\-\frac{t}{13\,ms}=ln(\frac{1}{2})\\t=-13\,ms\,*\,ln(\frac{1}{2})\\t=9.109\,\, ms[/tex]

In seconds this is approximately 0.009 seconds

Final answer:

The question involves calculating the time it takes for the charge on a capacitor to reduce to a certain value in an RC circuit, using the natural logarithm and the RC time constant formula. This is a Physics problem relevant to High School level.

Explanation:

The subject matter of the student's question is Physics, specifically relating to the discharge of a capacitor through a resistor and the calculation of time constants in RC circuits. The grade level would be High School, as it involves concepts typically taught in a high school physics course.

How to Calculate the Discharge Time

To calculate the time it takes for a capacitor to reduce its charge to a certain value, we use the formula V = V0 * e^(-t/RC), where V0 is the initial voltage, V is the final voltage, R is the resistance, C is the capacitance, t is the time, and e is the base of the natural logarithm. However, since we're dealing with the capacitor's charge rather than voltage, we can adapt the formula to Q = Q0 * e^(-t/RC). By substituting the given values and solving for t, we can find the time it takes for the capacitor's charge to decrease from 30.0 μC to 15.0 μC.

To solve for t, we would take the natural logarithm of both sides after dividing by Q0, resulting in t = -RC * ln(Q/Q0). Inserting the given values (R = 1.30 kΩ and C = 10.0 μF), and calculating the natural logarithm of (Q/Q0 = 15.0 μC / 30.0 μC), we find the time required for the charge to drop to half its initial value.

Answer:

p = 4/3   f

Explanation:

The constructor station describes the position of the image object in relation to the focal length

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distances to the object and the image respectively

 the objent distance  is

         1 / p = 1 / f - 1 / q

they indicate that the image is located at q = 4f this distance is positive because the image is after the lens

        1 / p = 1 / f - 1 / 4f

        1 / p = 3 / 4f

         p = 4f / 3

the object is positive because it is to the left of the lens, according to the optical sign convention

The sign convention is that the distance to the objects is positive if it is on the left side of the lens and the image is positive if it is on the right side of the lens.

Answer:

Forces observed by both is same as both

The frames are non-acceleration

Explanation:

Zweistein  =  Einstein

[tex](F_B + E_B)_Z[/tex] = [tex](F_B + E_B)_E[/tex]

                    as [tex](F_B + E_B)_E[/tex] tend to zero (0)

[tex](E_B)_Z[/tex] < [tex](F_B)_Z[/tex] as some quantity is subtracted from the force.

According to Einstein's Theory of Relativity, the laws of physics are the same for all observers, regardless of their motion. Applying this to the calculation of electric field strength around a sphere, both Einstein and Zweistein, assuming they are not accelerating, should compute the same value of the electric field.

If we consider the principles of physics consistently at play regardless of the observer, both Einstein and Zweistein must compute the same total force on the sphere. This is following Einstein's postulate stating the laws of physics remain the same in all inertial frames. Therefore, the electric field computed at the sphere by both Einstein and Zweistein should also be the same.

The principle of relativity implies that the laws of physics—including those of electric field strength—are the same for all observers, regardless of their speed or direction as long as they are not accelerating. Einstein's Theory of Relativity extends this to include accelerated motion and gravity.

This principle, combined with the equivalence of inertial and gravitational mass, is enough to show that the force of gravity perceived by an observer in an accelerating spacecraft must be the same as if the observer were stationary on the Earth. So, if Einsteins's Theory of Relativity applies to electric fields in the same way it does for gravitational fields, then the electric field strength calculated by Einstein and Zweistein should be the same.

To clarify, in the context of this question, the 'electric field' refers to the region around a charged particle or object where another charge placed within it would experience a force either of attraction or repulsion.

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Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

[tex]\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s[/tex]

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

[tex]\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s[/tex]

So the flow speed of each of the narrower pipe is:

[tex]v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}[/tex]

[tex]v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s[/tex]

Answer: 140cm

Explanation:

Radius of curvature r = 2f

Where f is the focal point

f = 70cm, therefore

r =2 x 70 = 140cm

Answer:

The radius of curvature is 9.35 cm.

Explanation:

Given:

u = -5 cm

v = -70 cm

The radius of curvature of the mirror can be obtained from the expression:

[tex]\frac{1}{v} +\frac{1}{u} =\frac{2}{R} \\-\frac{1}{v} -\frac{1}{u} =\frac{2}{R} \\-\frac{1}{70} -\frac{1}{5} =\frac{2}{R} \\-0.014-0.2=\frac{2}{R} \\R=-9.35cm[/tex]

Answer:

c.[tex]a_m[/tex]

Explanation:

We are given that

Acceleration due to gravity on the moon=[tex]a_m[/tex]

Acceleration due to gravity on the earth=[tex]a_e[/tex]

[tex]g_m=\frac{1}{6}g_e[/tex]

Net force due to am on an object on moon=[tex]F_{net}=ma_m[/tex]

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=[tex]F=ma_e[/tex]

[tex]F=F_{net}[/tex](given)

[tex]ma_m=ma_e[/tex]

[tex]a_e=a_m[/tex]

Hence, option c is true.

Answer:

The intensity is  [tex]I = 0.0003053 \mu W/cm^2[/tex]

Explanation:

 From the question we are told

     The  frequency of the electromagnetic wave is  [tex]f = 86.0 Hz[/tex]

     The peak value of the electric field is  [tex]E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m[/tex]

Generally  the intensity of this wave is mathematically represented as

                [tex]I = c * \frac{1}{2} * \epsilon_o E^2_o[/tex]

Where c is the speed of light with value  [tex]c = 3 *10^8 m/s[/tex]

           [tex]\epsilon_o[/tex] is the permittivity of free space with value  [tex]\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2[/tex]

Substituting values into equation for intensity

               [tex]I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}[/tex]

                 [tex]I = 3.053 *10^{-6} W/m^2[/tex]

Converting to [tex]cm^2[/tex] we divide by 10,000

                [tex]I = \frac{3.053 *10^{-6}}{10000} W/cm^2[/tex]

                [tex]= 3.053 *10^{-10} W/cm^2[/tex]

                [tex]= 0.0003053 *10^{-6} W/cm^2[/tex]

                [tex]I = 0.0003053 \mu W/cm^2[/tex]

Answer: The approximate magnetic field is 1.38 × 10^-3T

Explanation: Please see the attachment below

The approximate magnetic field at a point within the solenoid is 4.4 × 10^-4 T.

To find the magnetic field strength at a point within a solenoid, we can use the equation B = µ0nI, where B is the magnetic field strength, µ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid carries a current of 10.0 A and has 110.0 turns per meter of its length, we can substitute these values into the equation to calculate the magnetic field strength:

B = (4π × 10-7 T·m/A)(110.0 turns/m)(10.0 A) = 4.4 × 10-4 T

Therefore, the approximate magnetic field at a point within the solenoid is 4.4 × 10-4 T.

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Answer:

12,608 kPa

Explanation:

First we need to convert the density to standard units, that is kg/m^3

0.85 g/cm^3 = 0.85 × [tex]\frac{100^{3} }{1000}[/tex] = 0.85 × 1000 = 850 kg/m^3

Pressure of oil = pressure at surface + rho × g × h

= 101,000 + (850×9.81×1500)

= 12,608 kPa

The units in the part 'rho.g.h' will cancel out against each other and you will be left with the unit Pascals - which is the unit for pressure.

Hope that answers the question, have a great day!

Answer:

-6sin(2t)

Explanation:

fayemioluwatomisin is correct until the end.

x(0)=0

[tex]x(t)=C_1cos(2t)+C_2sin(2t)\\0=C_1cos(2(0))+C_2sin(2(0))\\C_1=0[/tex]

x'(0)=-12

[tex]x'(t)=2C_2cos(2t)\\-12=2C_2cos(2(0))\\-12=2C_2\\C_2=-6[/tex]

[tex]x(t)=-6sin(2t)[/tex]

The mass-spring system's equation of motion can be determined using Hooke's Law to find the spring constant, and applying the conditions of simple harmonic motion (SHM).

The question relates to the physical concept of a spring-mass system and its simple harmonic motion (SHM). The force exerted on a spring and the resulting stretch can determine the spring constant (k), using Hooke's Law F = kx, where F is the force applied and x is the displacement from the equilibrium position. In this case, we're given that a force of 880 newtons stretches the spring 4 meters, which allows us to calculate the spring constant k = 880 N / 4 m = 220 N/m.

Now, considering that the mass (m) attached to the spring is 55 kilograms and it's released with an upward velocity (v) of 12 m/s from the equilibrium position, the initial conditions for the SHM are established. The equation of motion for a spring-mass system in SHM is x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant. The velocity is the first derivative of the displacement function, v(t) = -Aω sin(ωt + φ), and the acceleration is the second derivative of the function.

The maximum velocity and acceleration occur when the displacement is zero and maximum respectively. To fully determine the equation of motion for this particular situation, additional information about the phase constant or initial displacement may be needed, unless it is assumed that the spring is released from the equilibrium point with an upward velocity, in which case the initial displacement would be zero, and the phase constant φ could be determined.

Answer:

The expression for resistance is    [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

Explanation:

Generally flow of charge at that point is mathematically given as

              [tex]J = \frac{I}{2 \pi r L}[/tex]

Where L is length of the cylinder as given the question

The potential difference that is between the cylinders is  

                 [tex]\delta V = -E dr[/tex]

 Where is the radius

Where E is the electric field that would be experienced at that point which is mathematically represented as

               [tex]E = \rho J[/tex]

Where is the [tex]\rho[/tex] is the resistivity as given the question

considering the formula for potential difference we have

                [tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]

To get V we integrate both sides

           [tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]

                [tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

According to Ohm law

           [tex]V= IR[/tex]

Now making R the subject we have

          [tex]R = \frac{V}{I}[/tex]

               Substituting for V

         [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

The resistance of a coaxial cylindrical configuration with radii R1 and R2, length L, and resistivity rho, is given by the formula R = (rho / (2 * pi * ln(R2/R1))).

The question asks for an expression for the resistance of a coaxial cylindrical configuration, where the space between two metal cylinders of length L and radii R1 and R2 is filled with a material of resistivity rho. To find this, we utilize the formula for the resistance R of a material, which is R = rho * (L/A), where A is the cross-sectional area. However, since the current flows radially through the material between the cylinders, we need to consider the formula for resistance in terms of the radii and length of the cylinders. The resistance can be expressed as R = (rho * L) / (2 * pi * L * ln(R2/R1)), simplifying to R = (rho / (2 * pi * ln(R2/R1))).

Final answer:

The width of the slit is 1.0 μm.

Explanation:

When light passes through a single slit, it undergoes diffraction, which causes interference patterns. The width of the central peak in the diffraction pattern is related to the width of the slit and the wavelength of the light. In this case, the width of the central peak is given as 5.0 mm and the wavelength is given as 600 nm.

Using the formula for the width of the central peak, we can solve for the width of the slit:

Width of slit = (wavelength * distance to screen) / (number of the peak * distance to the peak)

Substituting the given values into the formula, we find that the width of the slit is 1.0 μm.

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.

Frequency is the number of complete cycles (oscillations) of a wave that occur in one second. It is measured in Hertz (Hz), where 1 Hz means one cycle per second. X-ray radiation has a higher frequency than radio waves.

On the electromagnetic spectrum, X-rays have much higher frequencies compared to radio waves. X-rays have frequencies in the range of [tex]10^16[/tex] to [tex]10^19[/tex] Hz, while radio waves typically have frequencies in the range of [tex]10^3[/tex] to [tex]10^9[/tex] Hz. Therefore, X-ray radiation has significantly higher frequencies than radio waves.

Wavelength and amplitude are other properties of waves, but they are not always larger for X-ray radiation than for a radio wave. The wavelength of X-rays is much shorter than that of radio waves, and the amplitude can vary depending on the specific wave source and conditions.

Hence, In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.

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Answer:

0.358 kg

Explanation:

From simple harmonic motion,

T = 2π√(m/k)................ Equation 1

Where T = period of the spring, k = spring constant of the spring, m = mass suspended, π = pie

make m the subject of the equation

m = kT²/4π².................. Equation 2

Given: k = 15 N/m, T = 0.97 s, π = 3.14

Substitute into equation 2

m = 15(0.97²)/(4×3.14²)

m = 14.1135/39.4384

m = 0.358 kg.

Hence mass suspended = 0.358 kg

0.356kg

The period, T, of a mass on a spring of spring constant, k, in simple harmonic motion can be calculated as follows;

T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]         --------------------(i)

Where;

m = mass

From the question;

T = 0.97s

k = 15N/m

Taking π = 3.142 and substituting the values of T and k into equation (i) as follows;

0.97 = 2 x 3.142 x [tex]\sqrt{\frac{m}{15} }[/tex]

0.97 = 6.284 x [tex]\sqrt{\frac{m}{15} }[/tex]

[tex]\frac{0.97}{6.284}[/tex] = [tex]\sqrt{\frac{m}{15} }[/tex]

0.154 = [tex]\sqrt{\frac{m}{15} }[/tex]

Square both sides

0.154² = [tex]\frac{m}{15}[/tex]

0.0237 = [tex]\frac{m}{15}[/tex]

m = 0.356

Therefore, the mass that could be suspended on the spring to have an oscillation period of 0.97s when in SHM is 0.356kg

Answer:

Mass of the sphere is 19.2 kg

Explanation:

We have given diameter of the sphere d = 0.74 m

So radius r = 0.37 m

Initial angular velocity [tex]\omega _i=0rad/sec[/tex]

Time t = 14 sec

Angular displacement [tex]\Theta =160revolution=160\times 2\pi =1004.8rad[/tex]

From second equation of motion

[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]

So [tex]1004.8=0\times t+\frac{1}{2}\times \alpha \times 14^2[/tex]

[tex]\alpha =10.25rad/sec^2[/tex]

Torque is given [tex]\tau =10.8Nm[/tex]

Torque is equal to [tex]\tau =I\alpha[/tex], here I is moment of inertia and [tex]\alpha[/tex] angular acceleration

So [tex]10.8=10.25\times I[/tex]

[tex]I=1.053kgm^2[/tex]

Moment of inertia of sphere is equal to [tex]I=\frac{2}{5}Mr^2[/tex]

So [tex]1.053=\frac{2}{5}\times M\times 0.37^2[/tex]

M = 19.23 kg

So mass of the sphere is 19.23 kg

Answer:146.26 kN

Explanation:

Given

Outside diameter of tube [tex]d_o=54\ mm[/tex]

thickness of tube [tex]t=5\ mm[/tex]

therefore inner diameter [tex]d_i=54-2\times 5[/tex]

[tex]d_i=44\ mm[/tex]

Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]

[tex]A=\frac{\pi }{4}\times 980[/tex]

[tex]A=245\pi mm^2[/tex]

Stress developed must be less than the limited value

thus [tex]\frac{P}{A}\leq \sigma [/tex]

[tex]P\leq \sigma A[/tex]

Maximum value of [tex]P=\sigma \times A[/tex]

[tex]P=190\times 245\times \pi [/tex]

[tex]P=146.26\ kN[/tex]

Final answer:

To calculate the maximum load a stainless steel tube can support, one must first find the cross-sectional area based on its outer diameter and wall thickness. Then, using the limit of the normal stress and the area, the maximum axial load can be computed.

Explanation:

To determine the maximum load P that a stainless steel tube can support when it is used as a compression member, we need to use the formula for axial stress in a cylindrical member, which is σ = P/A, where σ is the normal stress and A is the cross-sectional area. Given that the normal stress must not exceed 190 MPa, we need to first calculate the cross-sectional area. For a tube with an outside diameter (d) of 54 mm and a wall thickness (t) of 5 mm, the internal diameter (di) would be d - 2t = 54 mm - (2 x 5 mm) = 44 mm. The cross-sectional area A can be calculated using the formula for the area of a hollow circle, A = π/4 * (d2 - di2).

A = π/4 * ((55)^2 - (44)^2)

A = 855.298 * 10^-6 m2

Once we have the area, we can calculate the maximum load by rearranging the stress formula to P = σ * A. By plugging in the values for σ and A, we can find the value of P that the member can safely support. Remember to convert diameters to meters when calculating the area in square meters for consistency with the stress units of Pascals (Pa).

P = 190 * 855.298 Pa

P = 1.62 MPa

Answer:

Velocity between points A and B will be 0.2344 m/s

Complete Question:

A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.

Answer:

a) [tex]v_{A} = 10 m/s[/tex]

b)

[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]

Explanation:

a) Velocity of point A

The velocity of point A is a combination of the translational and the rotational velocity of the racket.

[tex]v_{A} = v_{G} + v_{r}[/tex]

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = wr[/tex]

r = 25 cm = 0.25 m

w = 20 rad/s

[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]

[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]

b) Velocity of point B

At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]

[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]

In the y - direction, r = 30/2 = 15 cm = 0.15 m

r = 0.15 m

[tex]w_{x} = 1 rad/s[/tex]

[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]

In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]

[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]

[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s

Answer:

Michelson

Explanation:

The question is not clear enough. So i have attached a copy of the correct question.

Answer:

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

The magnitude of the magnetic field B is calculated using Faraday's Law and is equal to V/-c. For a square loop being pulled out from the field, the rate of change of area is proportional to the speed of withdrawal and the length of the side still in the field. Hence, c = Lv.

Part A: The magnitude of the magnetic field B can be calculated using Faraday's Law of electromagnetic induction, which states that the induced voltage in a circuit is equal to the negative rate of change of magnetic flux. Mathematically, it is expressed as V = -dB/dt, where V is the induced emf, and dB/dt is the rate of change of magnetic field. Because the area is decreasing at a constant rate, the magnetic field B is V/-c.

Part B: The rate of change of the area, denoted as c = -da/dt, for a square loop of side length L being pulled out of the magnetic field with constant speed v, could be calculated using the formula c = Lv. This is because as the square loop is pulled out with a constant velocity, the rate at which its area decreases is proportional to its speed and the length of the side that is still in the field.

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Answer:

AM radio, FM radio, microwaves, sodium light.

Explanation:

Electromagnetic radiation are waves from electromagnetic field which spread through the space or any other material medium and carries radiating energy. Examples includes X rays, radio waves, Gamma rays, etc. Exposure to high level of electromagnetic radiation could be harmful to human body, on the other hand, science as not been able to prove that exposing humans to low level electromagnetic radiation is harmful to our health.

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

[tex]F=\frac {m(v-u)}{t}[/tex]

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

[tex]F=\frac {0.3(2--5)}{0.0015}=1400N[/tex]

The average force exerted on the ball by the floor is  1400 N.

Force is the product of mass and acceleration. The S.I unit of force is Newton (N)

To calculate the average magnitude of the force exerted on the ball by the floor, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the average force exerted on the ball by the floor is  1400 N.

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Answer:

3.53 N/C

Explanation:

Electric field = F / q where F is the force in N and q is charge on the electron

F = mass of an electron × a ( acceleration in m/s)

using equation of motion to solve for the acceleration

s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0

2s / t² = a

F = me × ( 2s / t²)

E electric field = me × ( 2s / t²)  / q = me × 2s / ( t² × q)

me, mass of an electron = 9.11 × 10⁻³¹ kg

E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C

Answer: The magnitude of the electric field is 3.53 N/C

Explanation: Please see the attachments below

Answer:

a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.

b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

Explanation:

To focus an infinitely distant object, the lens should be placed at its focal length from the film or CCD. For an object at 2f, the lens should also be at a distance of 2f to focus the image correctly.

The question pertains to the principles of optics, specifically focusing using converging lenses in cameras.

(a) To focus an object that is infinitely far away, the lens should be placed at its focal length f from the film or CCD. This is because light rays from a distant object enter the lens parallel to its principal axis and converge at the focal point on the opposite side. Therefore, for an object at infinity, the image distance (di) approaches the lens's focal length (f).

(b) To focus an object at a distance of 2f in front of the lens, the lens should also be 2f from the film. This is derived from the lens formula (1/f = 1/do + 1/di) where do is the object distance and di is the image distance. For an object at 2f, solving the equation gives di = 2f.

Answer:

Answer

Explanation:

The followings are the reasons for the reduction in  the death toll from Tornadoes occurence in the 2000s compared to 1950s;

(a) Improvement in technology and rapid natural occurence prediction. Through the use of advanced radar systems the early detection of natural occurence e.g tornadoes, providing  warning signal to the residents and tracking their possible pathways is made possible.

Fast spread of information through television channels, Radio, Social media etc., help in fast spreading of information, thereby sending early signals to people thus making them prepare for its occurence.

Improvement of protective measures, provision of housing infrastructure for the displaced. With increase in technology, the protective measures are also improving.

Quick Government and medical interventions for casualties.  Conducting Tornado drills help people to practice to take cover in a specified location during tornadoes.

TV meteorologist are able to inform viewers of the approaching tornadoes and hurricanes  which can be tracked from the time of its origin i.e., from a depression to a storm.  Advanced radar data help in delineating the possible pathways of the hurricane.

Tornadoes on the other hand form in a thunderstorm or during a frontal activity. In thunderstorms there are vortices formed which are pre-cursors of tornadoes.. A vortice when touches the ground forms a tornado. Once a tornado is on the ground it becomes easy for radars to measure windpeeds. Hence a TV meteorologist can report the intensity of the Tornado.

Answer:

Spectrophotometry is the quantitative measurement of light spectra reflection and transmission properties of materials as function of the wavelength. Photometry measures the total brightness as seen by the human eye

Explanation:

Please give me brainliest

Answer:2.7m/s^2

Explanation:

mass=2.3kg

Force=6.2Newton

Acceleration=force ➗ mass

Acceleration=6.2 ➗ 2.3

Acceleration=2.7m/s^2

The resulting acceleration of the object is  2.70 m/s^2.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Given that:

Mass of the object: m = 2.3 kg.

Force applied on it: F = 6.2 Newtons.

Now from Newton's 2nd law of motion: it can be stated that:

force = mass × acceleration

acceleration = force/mass

= 6.2 Newton/2.3 kg

= 2.70 m/s^2.

Hence,  the acceleration of the object is 2.70 m/s^2.

Learn more about acceleration here:

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Answer:

B)

Explanation:

Hi there,

Notice in the prompt it is saying given mass, which means quantity of gas is constant. This eliminates option C, as this is Avogadro's Law (the more gas you have, the more space it takes up). It also states pressure remains constant, so it eliminates D, Boyle's Law (pressure is inversely related with volume), and A, The Combined Gas Law (combines Charles, Boyle's, and Gay-Lussac's Laws).

Charles' Law is telling us that as a gas is heated, the volume increases. This makes sense, as when gases are heated the average kinetic energy (energy of motion and movement) of the system is higher because of heat increase. When the system's kinetic energy is higher, it will have the tendency to occupy more space to balance out the increase in energy. Think of people who become more jittery in an already crowded space, they will push outward to accompany more space to balance.

So, answer is B.

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Answer:

Explanation:

To detrmine the time interval at which the balls are in contact.

Given information

The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.

Expression for the effective spring constant ball is shown below.

K = |F|/|x|

Here,

k is a spring constant

F is the force exerted on the ball

x is dispalcement due force

substitute 15.9 kN for F and 0.130 mm in above equation

The spring constant is 122 x 10⁶ N/m

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c).  The value of the refractive index of the medium is a measure of its "optical density":  Light spreads at maximum speed in a vacuum but slower in others  transparent media; therefore in all of them n> 1. Examples of typical values ​​of  are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which  the refracted ray disappears and all the light is reflected. As in the maximum value of  angle of refraction, from which everything is reflected, is βmax = 90º, we can  know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

 βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media  different, the incident beam is divided into three: the most intense penetrates the second  half forming the refracted ray, another is reflected on the surface and the third is  breaks down into numerous weak beams emerging from the point of incidence in  all directions, forming a set of stray light beams.

The minimum value of the index of refraction is 1, which occurs in a vacuum where the speed of light is at its maximum and not slowed by any material.

The minimum value that the index of refraction can have is for a vacuum, where light travels at its fastest and the speed of light is not slowed down by any medium. According to the definition of the index of refraction, n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material. Since the speed of light in a vacuum is the highest possible speed for light and cannot be exceeded, the index of refraction n has a minimum value of 1 in a vacuum. That is because when we plug in the values into the equation, with both c and v being equal (since v is the speed of light in a vacuum), the ratio n = c/v equals 1. Therefore, no material can have an index of refraction less than 1.