A 92.0kg baseball player slides into second base. The coefficient of friction between the player and the ground is 0.61. If the player comes to rest after 1.2s, what was his initial speed?

Explanation:

Charge on proton, q₁ = e

Charge on alpha particles, q₂ = 2e

The magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

Here, [tex]\theta=90=sin(90) = 1[/tex]

For proton, [tex]F_p=ev_pB[/tex]..........(1)

For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)

Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,

[tex]ev_pB=2ev_aB[/tex]

[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]

So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.

If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.

The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.

The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.

This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.

Therefore, the proton is moving twice as fast as the alpha particle.

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Answer:

[tex]t =4.605 *10^{-3}[/tex]

Explanation:

given data:

wavelength of emission =[tex]0.8 \mu m[/tex]

output power = 100 mW

We can deduce value of obsorption coefficient from the graph obsorption coefficient vs wavelength

for wavelength [tex]0.8 \mu m[/tex] the obsorption coefficient value is 10^{3}

intensity can be expressed as a function of thickness as following:

[tex]I(t) = I_{O} *e^{-\lambda *t}[/tex]

putting all value to get thickness

[tex]1*10^{-3} =100*10^{-3}e^{-10^{3}*t}[/tex]

[tex]0.01 = e^{10^{3}t}[/tex]

[tex]t =4.605 *10^{-3}[/tex]

Answer:

Power, P = 722.96 watts

Explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire, [tex]\rho=1.3\times 10^{-6}\ \Omega-m[/tex]

We need to find the power drawn by this heater. Power is given by :

[tex]P=\dfrac{V^2}{R}[/tex]

And, [tex]R=\rho\dfrac{l}{A}[/tex]

[tex]P=\dfrac{V^2\times A}{\rho\times l}[/tex]

[tex]P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}[/tex]

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.    

Answer:

[tex]\Delta U = 0.2072 J[/tex]

Explanation:

Potential difference between two points in constant electric field is given by the formula

[tex]\Delta V = E.\Delta x[/tex]

here we know that

[tex]E = 370 N/C[/tex]

also we know that

[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]

now we have

[tex]\Delta V = 370 (0.2) = 74 V[/tex]

now change in potential energy is given as

[tex]\Delta U = Q\Delta V[/tex]

[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]

[tex]\Delta U = 0.2072 J[/tex]

Explanation:

Force between two point changes, F₁ = 1 N

Distance between them, r₁ = 2 cm = 0.02 m

We know that the electrostatic force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

i.e.

[tex]F\propto \dfrac{1}{r^2}[/tex]

i.e

[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2[/tex]

Let F₂ is the force when the distance between the charges is 8 cm, r₂ = 0.08 m

[tex]F_2=\dfrac{F_1\times r_1^2}{r_2^2}[/tex]

[tex]F_2=\dfrac{1\ N\times (0.02\ m)^2}{(0.08\ m)^2}[/tex]

F₂ = 0.0625 N

So, the distance between the sphere is 8 N, the new force is equal to 0.0625 N. Hence, this is the required solution.

Answer:

Speed, v = 6.91 m/s

Explanation:

Given that,

Spring constant, k = 594 N/m

It is attached to a table and is compressed down by 0.196 m, x = 0.196 m

We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]

[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]

v = 6.91 m/s

So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.

Answer:

364.4 J

Explanation:

I = Moment of inertia of the forearm = 0.550 kgm²

v = linear velocity of the ball relative to elbow joint = 17.1 m/s

r = distance from the joint = 0.470 m

w = angular velocity

Using the equation

v = r w

17.1 = (0.470) w

w = 36.4 rad/s

Rotational kinetic energy of the forearm is given as

RKE = (0.5) I w²

RKE = (0.5) (0.550) (36.4)²

RKE = 364.4 J

Answer:

363.96 J

Explanation:

v = 17.1 m/s

r = 0.47 m

I = 0.55 kgm^2

Let ω be the angular velocity

ω = v / r = 17.1 / 0.47 = 36.38 rad/s

The kinetic energy of rotation is

K = 1/2 I ω^2 = 0.5 x 0.55 x 36.38 x 36.38 = 363.96 J

Answer:

23

Explanation:

First, we need to convert the hose diameter from inches to meters.

0.75 in × (2.54 cm / in) × (1 m / 100 cm) = 0.0191 m

Calculate the flow rate given the velocity and hose diameter:

Q = vA

Q = v (¼ π d²)

Q = (0.30 m/s) (¼ π (0.0191 m)²)

Q = 8.55×10⁻⁵ m³/s

Find the volume of the pool:

V = π r² h

V = π (1.5 m)² (1.0 m)

V = 7.07 m³

Find the time:

t = V / Q

t = (7.07 m³) / (8.55×10⁻⁵ m³/s)

t = 82700 s

t = 23 hr

Final answer:

To find out how much water should be added to move the bright spot by 5.0 cm, apply Snell's Law to find the angle of refraction and use trigonometry to determine the change in horizontal distance of the spot due to the change in water depth.

Explanation:

The student is asking about the behavior of light as it passes through different media, specifically how the depth of the water in an aquarium would need to change to move the bright spot created by a laser. The scenario of shining a laser at an angle into water involves Snell's Law, which describes how light bends at the interface between two media with different indices of refraction. To calculate the necessary increase in water height to move the bright spot by 5.0 cm, we need to apply the principles of refraction and trigonometry.

First, we determine the initial angle of refraction using Snell's Law:

nairsin(θair) = nwatersin(θwater)1.00sin(45°) = 1.33sin(θwater)

Upon finding the angle of refraction θwater, we can use trigonometric functions to calculate the initial horizontal distance (Xinitial) from the point of incidence to the bright spot. To find the new distance (Xfinal) after adding water, we repeat the process with the new depth. The difference in the horizontal distances gives the amount the bright spot moved, 5.0 cm. Adjusting the depth to match this movement provides the answer to the student's question.

Answer:

The normal force between the motorcycle and track at the top of the loop = 2343.75 N

Explanation:

At the top of the loop the track is directed tangential to the motion, the normal to the tangent is along radius of circle. that is here we need to find centripetal force.

We have expression for centripetal force

                  [tex]F=\frac{mv^2}{r}[/tex]

Here mass, m = 175 kg

        Velocity, v = 7.5 m/s

        Diameter, d = 8.4 m

        Radius, r = 0.5d = 0.5 x 8.4 = 4.2 m

Substituting

        [tex]F=\frac{mv^2}{r}=\frac{175\times 7.5^2}{4.2}=2343.75N[/tex]

The normal force between the motorcycle and track at the top of the loop = 2343.75 N

Answer:

4 m/s

Explanation:

M = mass of the hammer = 8.91 kg

m = mass of the metal piece = 0.411 kg

h = height gained by the metal piece = 3.88 m

Potential energy gained by the metal piece is given as

PE = mgh

PE = (0.411) (9.8) (3.88)

PE = 15.6 J

KE = Kinetic energy of the hammer

Given that :

Potential energy of metal piece = (0.219) Kinetic energy of the hammer

PE = (0.219) KE

15.6 = (0.219) KE

KE = 71.2 J

v = speed of hammer

Kinetic energy of hammer is given as

KE = (0.5) M v²

71.2 = (0.5) (8.91) v²

v = 4 m/s

Answer:

(a) 85.64°F

(b) 78.28°C

(c)  - 268.8°C  

Explanation:

(a) 29.8°C into °F

To convert °C into °F, we use the formula given below

C / 100 = (F - 32) / 180

C / 5 = (F - 32) / 9

9 x 29.8 / 5 = F - 32

F = 85.64°F

(b) 172.9°F into °C

C / 5 = (F - 32) / 9

C = (172.9 - 32) x 5 / 9

C = 78.28°C

(c) 4.2 K into °C

C = K - 273

C = 4.2 - 273 = - 268.8°C

Answer : The time taken for the concentration will be, 7.98 seconds

Explanation :

First order reaction : A reaction is said to be of first order if the rate is depend on the concentration of the reactants, that means the rate depends linearly on one reactant concentration.

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A]_o}{[A]}[/tex]

where,

k = rate constant  = [tex]0.150s^{-1}[/tex]

t = time taken for the process  = ?

[tex][A]_o[/tex] = initial concentration = 0.860 M

[tex][A][/tex] = concentration after time 't' = 0.260 M

Now put all the given values in above equation, we get:

[tex]0.150s^{-1}=\frac{2.303}{t}\log\frac{0.860}{0.260}[/tex]

[tex]t=7.98s[/tex]

Therefore, the time taken for the concentration will be, 7.98 seconds

To determine the time for the concentration of substance A to decrease in a first-order reaction, the formula t = (1/k) * ln([A]0/[A]t) is used with the known rate constant and initial and final concentrations.

The question asks how long it would take for the concentration of substance A to decrease from 0.860 M to 0.260 M given a first-order reaction with a rate constant of 0.150 s−¹ at 400 °C.

For a first-order reaction, the time (t) it takes for the concentration of a reactant to change can be found using the formula:

t = (1/k) * ln([A]0/[A]t)

Where:

In this case, we can substitute the given values into the equation:

t = (1/0.150 s−¹) * ln(0.860 M / 0.260 M) = (1/0.150 s−¹) * ln(3.3077)

The time can be calculated by completing the computation for ln(3.3077) and dividing by the rate constant.

Answer:

current = 1.51 A

Explanation:

Initially the capacitor without any dielectric is connected across AC source

so the capacitive reactance of that capacitor is given as

[tex]x_c = \frac{1}{\omega c}[/tex]

now we have

[tex]i = \frac{V_{rms}}{x_c}[/tex]

here we know that

[tex]i = 0.29 A[/tex]

now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor

so here net capacitance is given as

[tex]c_{eq} = 4.2c + c = 5.2c[/tex]

now the equivalent capacitive reactance is given as

[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]

[tex]x_c' = \frac{x_c}{5.2}[/tex]

so here we have new current in that circuit is given as

[tex]i' = \frac{V_{rms}}{x_c'}[/tex]

[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]

[tex]i' = 1.51 A[/tex]

Answer:

Heat generated by the current = 1547.89 J

Explanation:

We have equation for heat energy H = mCΔT

Mass of copper = 0.221 kg

Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°

ΔT = 38 - 20 = 18°C

Substituting in H = mCΔT

           H = 0.221 x 389.112 x 18 = 1547.89 J

Heat generated by the current = 1547.89 J

The heat generated by the electric current that heated a 221 g copper wire from 20.0 °C to 38.0 °C is calculated to be 1.542 kJ, using the specific heat capacity of copper and the formula Q = mcΔT.

An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. To calculate the heat generated by the current, we use the formula for heat energy, Q = mcΔT, where m is the mass in kg, c is the specific heat capacity in kcal/kg°C, and ΔT is the change in temperature in °C.

Given:
m = 0.221 kg
c = 0.093 kcal/kg°C
ΔT = (38.0 - 20.0) °C = 18.0 °C

Substituting the values into the formula:
Q = 0.221 kg * 0.093 kcal/kg°C * 18.0 °C = 0.368658 kcal

To convert kcal to joules (since 1 kcal = 4.184 kJ),
Q = 0.368658 kcal * 4.184 kJ/kcal = 1.542 kJ

Therefore, the heat generated by the electric current is 1.542 kJ.

Answer:

39.08 m

Explanation:

θ = 31.6 degree

u = 29.3 m/s

Let it hits at the maximum height h. It takes half of the time of flight to cover the distance.

T = 2 u Sinθ / g

t = T / 2 = u Sinθ / g = (29.3 x Sin 31.6) / 9.8 = 1.566 s

The horizontal distance covered in this time is

d = u Cosθ x t = 29.3 x Cos 31.6 x 1.566 = 39.08 m

Thus, the fire hose be located at a distance of 39.08 m from the building.

Answer:

(kg⋅m/s)

Explanation:

The unit of momentum is the product of the units of mass and velocity. In SI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s)

This question is dealing with fundamental SI units. Thus, let's list the seven basic SI Units upon which other units are expressed;

Mass - kilogram (kg)

Length - meter (m)

Time - second (s)

Amount of substance - mole (mol)

Electric current - ampere (A)

Thermodynamic temperature - kelvin (K)

Luminous intensity - candela (cd)

Fundamental SI unit of momentum is Kg.m/s

Now, we want to write the SI Unit of momentum.

From the question, we are told that Physicists often measure the momentum of subatomic particles using the formula;

MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg.m²/s²

Now, we know that unit of speed is in m/s.

Thus, in units, momentum = MeV/c = (kg.m²/s²)/(m/s)

Simplifying this gives; Kg.m/s

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Answer:

Acceleration of the ship, [tex]a=2.14\times 10^{-7}\ m/s^2[/tex]

Explanation:

It is given that,

Mass of both ships, [tex]m=39000\ metric\ tons=39\times 10^6\ kg[/tex]

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

[tex]F=G\dfrac{m^2}{d^2}[/tex]

[tex]F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}[/tex]

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{8.38\ N}{39\times 10^6\ kg}[/tex]

[tex]a=2.14\times 10^{-7}\ m/s^2[/tex]

So, the acceleration of either ship due to the gravitational attraction of the other is [tex]2.14\times 10^{-7}\ m/s^2[/tex]. Hence, this is the required solution.

The gravitational attraction between the two ships can be determined by using Newton's law of gravitation. Afterwards, the acceleration of one ship due to this force is found by applying Newton's second law of motion.

In this scenario, we're dealing with the concept of gravitational attraction between two massive bodies. We can find the gravitational force between the ships using Newton's law of gravitation:

F = G * (m1 * m2) / r²,

where F is the gravitational force, G is the gravitational constant (6.674 x 10^-11 N(m²/kg²)), m1 and m2 are the two masses, and r is the distance between them.

So, let's plug in the values. Each ship mass (m1 = m2) is 39,000 metric tons, equivalent to 3.9 x 10^7 kg. The distance between them (r) is 110 m, which we need to square. Calculating for F, we get an incredibly small value, which evidences the weak nature of gravitational force.

To find the acceleration of either ship (let's say m1), due to the gravitational force of the other, we will apply Newton's second law (F = m*a). Here, F is the gravitational force we found, m is the mass of the ship, and a is the acceleration we're looking for:

a = F / m,

Plugging the values found into this equation will result in the desired acceleration in m/s².

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Answer:

Charge, [tex]Q=1.56\times 10^{-8}\ C[/tex]

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

[tex]E=\dfrac{kQ}{r^2}[/tex]

Q is the charge on an object

[tex]Q=\dfrac{Er^2}{k}[/tex]

[tex]Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}[/tex]

[tex]Q=1.56\times 10^{-8}\ C[/tex]

So, the charge on the object is [tex]1.56\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

The object's charge q is calculated to be approximately 1.57 × 10⁻⁵ C.

To find the object's charge q, we use the formula for the electric field due to a point charge:

E = kQ/r²

Here, E is the electric field, k is the Coulomb's constant (8.99 × 10⁹ N·m²/C²), Q is the charge, and r is the distance from the charge.

We are provided with:

Electric field, E = 180,000 N/C

Distance, r = 2.8 cm = 0.028 m

Rearranging the formula to solve for Q gives:

Q = Er²/k

Substituting in the given values:

Q = (180,000 N/C) × (0.028 m)² / (8.99 × 10⁹ N·m²/C²)

Q = (180,000) × (0.000784) / (8.99 × 10⁹)

Q ≈  1.57 × 10⁻⁵ C

The object's charge q is approximately  1.57 × 10⁻⁵ C.

Explanation:

It is given that,

Number of turns, N = 200

Area of cross section, A = 8.5 cm²

Magnetic field is directed out of the paper and is, B = 0.06 T

The magnetic field is  out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

[tex]\epsilon=-N\dfrac{d\phi}{dt}[/tex]

Since, [tex]\epsilon=IR[/tex]

I is the induced current

[tex]I=-\dfrac{N}{R}\dfrac{d\phi}{dt}[/tex]

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.

Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.

Answer:

424088766.068 m

Explanation:

Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R

R = Radius of earth = 6371000 m (mean radius)

In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit

Circumference of a circle = 2×π×r

⇒Distance travelled in 5.89 hours = 2×π×2.6 R

⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000

⇒Distance travelled in 5.89 hours = 104078451.3393m

Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m

∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m

Final answer:

The satellite orbits Earth at approximately 2.60 Earth radii and travels a total distance of about 4.23×10⁸ meters in one day.

Explanation:

The question pertains to calculating the total distance traveled by a satellite orbiting Earth at a given distance over the duration of one day. To solve this, we need to determine the circumference of the satellite's circular orbit and then calculate how many orbits it completes in a day.

To determine the satellite's orbit circumference, we use the formula for the circumference of a circle, C = 2πr, where r represents the orbit radius. Given that the satellite orbits at a distance of about 2.60 Earth radii, we can express this distance in meters by multiplying the Earth's radius, 6.37×10⁶ meters, by 2.60, yielding a radius of approximately 1.6562×10⁷ meters. Therefore, the satellite's orbit circumference is approximately 1.04×10⁸ meters.

Since the satellite completes one orbit in 5.89 hours, we can now calculate how many orbits it completes in a day (24 hours) by dividing 24 by 5.89, which is approximately 4.07 orbits per day. The total distance traveled in one day is the circumference multiplied by the number of orbits, resulting in a distance of approximately 4.23×10⁸ meters per day.

Answer:

Kinetic energy, [tex]E_k=61.74\ J[/tex]

Explanation:

It is given that,

Mass of ball, m = 0.9 kg

It is dropped form a height of 7 m, h = 7 m

When the ball is at certain height it will have only potential energy. As it hits the ground, its potential energy gets converted to kinetic energy as per the law of conservation of energy.

So, [tex]E_k=PE=mgh[/tex]

[tex]E_k=0.9\ kg\times 9.8\ m/s^2\times 7\ m[/tex]

[tex]E_k=61.74\ J[/tex]

So, the kinetic energy when it hits the ground is 61.74 Joules.    

The kinetic energy of a 0.9 kg ball dropped from a height of 7 m is calculated using the conservation of mechanical energy, resulting in 61.74 Joules when it hits the ground.

The kinetic energy of a 0.9 kg ball when it hits the ground after being dropped from a height of 7 m. The principle used to determine this is the conservation of mechanical energy, which states that the sum of the potential and kinetic energies in a system remains constant if only conservative forces are doing work.

When the ball is at the height of 7 m, it posses gravitational potential energy (GPE) which can be calculated using the formula GPE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the ball is dropped.

Since the ball is dropped from rest, its initial kinetic energy is 0. When the ball reaches the ground, its potential energy is converted into kinetic energy.

Therefore, the kinetic energy (KE) of the ball when it hits the ground can be calculated using the potential energy formula, assuming no energy is lost:
KE = mgh = 0.9 kg x 9.8 m/s² x 7 m = 61.74 Joules.

This means that the ball will have a kinetic energy of 61.74 Joules when it impacts the ground.

Answer:

The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]

Explanation:

Given that,

mass of bar = 150 g

Length l = 36 cm

We need to calculate the moment of inertia of the bar

Using formula of moment inertia

[tex]I=\dfrac{1}{12}Ml^2[/tex]

Where,

M = mass of the bar

L = length of the bar

Put the value into the formula

[tex]I=\dfrac{1}{12}\times150\times10^-3\times36\times10^{-2}[/tex]

[tex]I=45\times10^{-4}\ kg-m^2[/tex]

Hence, The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]

The moment of inertia of a bar with a mass of 150g and length of 36cm, rotating about its center, is approximately 0.00162 kg·m². The calculation uses the formula I = (1/12) * M * L². First, convert the mass and length to SI units and then substitute them into the formula.

To find the moment of inertia of a uniform bar with a mass of 150g and a length of 36cm, we can use the formula for a rod rotating about its center:

I = (1/12) * M * L²

I = (1/12) * 0.15kg * (0.36m)²

I = (1/12) * 0.15kg * 0.1296m²

I ≈ 0.00162 kg·m²

Therefore, the moment of inertia of the bar rotating about its center is approximately 0.00162 kg·m².

Final answer:

Using the Doppler Effect formulae for electromagnetic waves, the difference in frequency (Dopler shift) experienced by radar signals upon hitting and returning from a moving vehicle allows a radar gun to calculate the vehicle's speed. The principle involves the use of shift in frequency and the speed of light to measure the speed at which the vehicle is moving.

Explanation:

The question involves the Doppler Effect in physics, specifically its application in a radar speed trap. To find the frequency shift, we must use the Doppler Effect formulae for electromagnetic waves. However, it seems there might have been a mix-up with the frequencies provided in several example problems. Since those seem to be examples rather than the actual frequencies we are working with, let's focus on finding the principle behind calculating the speed of the vehicle based on a known frequency shift of a radar emission and its return signal.

The frequency shift (Δf) in the Doppler Effect for electromagnetic waves such as radar can be calculated by the formula: Δf = (2 * f * v) / c, where f is the original frequency emitted by the radar gun, v is the speed of the vehicle, and c is the speed of light. The factor of 2 is because the radar signal experiences a frequency shift once when it hits the moving vehicle and another shift when the echo returns. The radar gun's internal processors calculate the difference in frequency (the Dopler shift) to find the speed of the vehicle. For accurate measurement, the radar unit must be able to discern even small frequency shifts to effectively differentiate speeds with fine resolution.

Answer:

The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Explanation:

Given that,

Initial speed = 10.0 m/s

Acceleration = 3.00 m/s^2

Distance [tex]d = \dfrac{1}{4}\times1.609[/tex]

[tex]d = 0.40225\ km=0.40225\times10^{3}\ m[/tex]

We need to calculate the speed of the car,

Using equation of motion

[tex]v^2-u^2=2as[/tex]

Where, u = initial velocity

v = final velocity

a = acceleration

Put the value in the equation

[tex]v^2=(10.0)^2+2\times3.00\times0.40225\times10^{3}[/tex]

[tex]v^2=2513.5[/tex]

[tex]v=50.13\ m/s[/tex]

(B). We need to calculate the stopping distance of the car,

Using equation of motion again

[tex]v^2=u^2+2as[/tex]

Here,initial velocity = 50.13 m/s

Final velocity = 0

Acceleration = -4.50 m/s²

Put the value in the equation

[tex]0=(50.13)^2-2\times4.50\times s[/tex]

[tex]s=\dfrac{(50.13)^2}{2\times4.50}[/tex]

[tex]s=279.22\ m[/tex]

Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Answer:

The beat frequency is 33.33 Hz.

Explanation:

Given that,

Length of first pipe =60 cm

Length of other pipe = 68 cm

Speed of sound = 340 m/s

We need to calculate the frequency

We know that,

When they operate at fundamental frequency then the length is given by,

[tex]L=\dfrac{\lambda}{2}[/tex]

The wavelength is given by

[tex]\lambda=2L[/tex]

For first organ pipe,

Using formula of frequency

[tex]f=\dfrac{v}{\lambda}[/tex]

[tex]f_{1}=\dfrac{v}{2L_{1}}[/tex]...(I)

Put the value into the formula

[tex]f_{1}=\dfrac{340}{2\times60\times10^{-2}}[/tex]

[tex]f_{1}=283.33\ Hz[/tex]

For second organ pipe,

[tex]f_{2}=\dfrac{v}{2L_{2}}[/tex]...(II)

Put the value in the equation (II)

[tex]f_{2}=\dfrac{340}{2\times68\times10^{-2}}[/tex]

[tex]f_{2}=250\ Hz[/tex]

Therefore the beat frequency

[tex]\Delta f=f_{1}-f_{2}[/tex]

[tex]\Delta f=283.33-250[/tex]

[tex]\Delta f=33.33\ Hz[/tex]

Hence,  The beat frequency is 33.33 Hz.

The beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, where one pipe is 60 cm long and the other is 68 cm long, is 33.33 Hz.

To find the beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, we need to calculate the frequencies of each pipe. The formula for frequency is f = v/λ, where v is the speed of sound and λ is the wavelength. Since the pipes are open at both ends, the wavelength is twice the length of the pipe. So, the frequency of the 60 cm pipe would be f₁ = v/(2L₁), and the frequency of the 68 cm pipe would be f₂ = v/(2L₂). Now, to calculate the beat frequency, we subtract the frequencies: beat frequency = |f₁ - f₂|.

Using the given speed of sound (340 m/s), we can substitute the lengths of the pipes into the frequency formula to find the frequencies:

f₁ = 340/(2 x 0.6) Hz = 283.33 Hz

f₂ = 340/(2 x 0.68) Hz = 250 Hz

Now, we can calculate the beat frequency:

beat frequency = |283.33 - 250| Hz = 33.33 Hz

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Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

[tex]T + mg =\frac{mv^{2}}{r}[/tex]

Inserting the values

[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]

T = 31.1 N

Answer: 41N

Explanation :

T= mv^2/R + mgcos θ

At the highest point on the circle θ=0

Cos 0 = 1

T= mv^2/R + mg

m = 0.5kg

Velocity at the highest point (amplitude)= 12m/s

T = 0.5× 12^2/2 + 0.5×10

0.5×144/2 +5

T = 0.5×72 + 5

T = 36+5

T = 41N

The area of a square is given by:

A = s²

A is the square's area

s is the length of one of the square's sides

Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:

d[A]/dt = d[s²]/dt

The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt

dA/dt = 2s(ds/dt)

A) Given values:

s = 14m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(14)(3)

dA/dt = 84m²/s

B) Given values:

s = 25m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(25)(3)

dA/dt = 150m²/s

When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.

When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.

The given parameters;

The area of the square is calculated as;

A = L²

The change in the area is calculated as;

[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]

When the side of the square is 14 m, the rate at which the area is changing is calculated as;

[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]

When the side of the square is 25 m, the rate at which the area is changing is calculated as;

[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]

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Answer:

D = 230.2 Km

Explanation:

let distance between seismograph and focus of quake is D

From time distance formula we  can calculate the time taken by the S wave

[tex]T_1 =\frac{D}{4.5}[/tex]

From time distance formula we  can calculate the time taken by the P wave

[tex]T_2 =\frac{D}{6.90}[/tex]

It is given in equation both waves are seperated from each other by 17.8 sec

so we have

[tex]T_1 - T_2 = 17.8[/tex]sec

Putting both time value to get distance value

[tex]\frac{D}{4.5} - \frac{D}{6.90} = 17.8[/tex]

D = 230.2 Km