A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the particle's motion?Check all that apply.Increasing its chargeIncreasing its massIncreasing the field strengthIncreasing its speed

Answer:

-6.2 m/s (downward)

Explanation:

The velocity of an object thrown vertically upward is given by:

[tex]v= u + at[/tex]

where:

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this problem,

u = 27.1 m/s

t = 3.4 s

So, the velocity after 3.4 s is

[tex]v=27.1 m/s + (-9.8 m/s^2)(3.4 s)=-6.2 m/s[/tex]

and the negative sign means the velocity points downward.

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

[tex]\frac{V_1}{N_1}=\frac{V_2}{N_2}[/tex]

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

[tex]V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V[/tex]

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

Answer:

EMF = 187.5 volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in the flux will induce EMF

so we can say

[tex]EMF = -\frac{d\phi}{dt}[/tex]

[tex]EMF = \frac{\phi_i - \phi_f}{\Delta t}[/tex]

now we will have

[tex]EMF = \frac{NBA - 0}{\Delta t}[/tex]

[tex]EMF = \frac{50(1.5)(0.25)}{0.10}[/tex]

[tex]EMF = 187.5 Volts[/tex]

So the induced EMF in the coil will be 187.5 Volts

Answer:

Minimum voltage, V = 0.682 mV          

Explanation:

It is given that,

Current passing through the human heart, [tex]I=2.5\times 10^{-6}\ A[/tex]

A patient is undergoing open-heart surgery, and the patient’s heart has a constant resistance, R = 273 ohms

We have to find the minimum voltage across the heart that could pose a danger to the patient. It can be calculated using Ohm's law as :

V = IR

[tex]V=2.5\times 10^{-6}\ A\times 273\ \Omega[/tex]

V = 0.0006825 volts

or

V = 0.682 mV

Hence, the minimum voltage across the heart that could pose a danger to the patient is 0.682 mV    

Answer:

A helium filled balloon floats forwards in a accelerating car because of the pressure difference between the front and the back of the car. When the car is accelerating, the air moves relitive to the car and the consequence is that the pressure in the back is slightly higher than in the front; which results in net force in forward direction.

hit that heart please leave brainliest and let me know if want me to answer or explain it in a different way :)

The helium balloon tilts forward in a car that is accelerating because the air inside the car is thrust backwards, creating more air pressure at the back and causing the balloon to float towards the lower pressure at the front. This behavior is unlike a pendulum that tilts backward due to the force of gravity.

The tilting of the helium balloon in a car that is accelerating forward can be explained using concepts from physics, particularly related to Newton's laws, the concept of a pendulum in equilibrium, and the buoyant Archimedean force.  

When a car accelerates, everything inside the car, including the air, is thrust backwards. This creates a higher air pressure at the back of the car than at the front. The helium balloon will float towards the lower pressure, which is at the front of the car, so the balloon will appear to tilt forward. This contrasts with the backward tilt of a pendulum, as a pendulum is influenced primarily by the force of gravity, which acts downward.

In terms of calculating the angle of tilt, it would be necessary to know the buoyant force applied to the balloon (which depends on the pressure gradient in the air), the acceleration of the car, and the mass of the balloon. However, without numerical values for these variables, a specific angle of tilt cannot be determined in this context.

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Answer:

902 ft

Explanation:

First convert mi/h to ft/s.

53.0 mi/h × (5280 ft / mi) × (1 h / 3600 s) = 77.7 ft/s

Sum of the forces on the car in the y direction:

∑F = ma

N - W = 0

N = mg

Sum of the forces on the car in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substituting;

-mgμ = ma

-gμ = a

Acceleration is constant, so:

v² = v₀² + 2a(x - x₀)

(0 ft/s)² = (77.7 ft/s)² + 2(-32.2 ft/s² × 0.104)(x - 0)

x = 902 ft

The minimum stopping distance is 902 ft.

Answer:

[tex]a_c = 1.1 m/s^2[/tex]

Explanation:

As we know that net force on the Satellite due to gravity will provide it centripetal force

so we can say here

[tex]F_g = F_c[/tex]

[tex]\frac{GMm}{r^2} = ma_c[/tex]

now we will have

acceleration given by the equation

[tex]a_c = \frac{GM}{r^2}[/tex]

now we have

r = R + 2R = 3R

[tex]a_c = \frac{GM}{9R^2}[/tex]

also we know that acceleration due to gravity on the surface of earth is given as

[tex]g = \frac{GM}{R^2} = 9.8 m/s^2[/tex]

so the acceleration of satellite is given as

[tex]a_c = \frac{9.8}{9} = 1.1 m/s^2[/tex]

Final answer:

The acceleration of a satellite in orbit at an altitude of twice Earth's radius is best approximated using the formula for gravitational acceleration, g = GME / r². Plugging in the values and simplifying, the acceleration is approximately 2.45 m/s², with option B (2.5 m/s²) being the closest approximation.

Explanation:

Calculating the Acceleration of a Satellite in Orbit

In order to best approximate the acceleration a satellite experiences in orbit at an altitude of twice the Earth's radius, we must apply Newton's law of universal gravitation and the formula for gravitational acceleration, g = GME / r². Given that the satellite weighs 104 N at ground control which reflects the gravitational force at Earth's surface, we use this formula to find its gravitational acceleration in orbit. The relevant equation provided, Fpa = GME2thm/r, appears to be a typo, but our main equation for gravitational acceleration does not require mass, as it cancels out during the calculation:

g = GME / r²

Since we're given that the altitude is twice the Earth's radius, the distance r from Earth's center to the satellite in orbit is 3RE (1RE for Earth's radius and 2RE for the altitude above Earth). Considering this, we can calculate:

g = (6.674 x 10-11 m³/kg s²) (5.972 x 1024 kg) / (3 x 6.371 x 106 m)²

After simplifying, we get an acceleration of approximately 2.45 m/s². Therefore, option B, 2.5 m/s², best approximates the acceleration the satellite experiences in orbit at this altitude.

Answer:[tex]5.11(10)^{13}miles[/tex]

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.  

This unit is equivalent to [tex]5.879(10)^{12}miles[/tex], which mathematically is expressed as:

[tex]1Ly=5.879(10)^{12}miles[/tex]

Doing the conversion:

[tex]8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles[/tex]  This is the distance from Earth to Sirius in miles.

Final answer:

The distance from Earth to Sirius is approximately 50 trillion miles.

Explanation:

To calculate the distance from Earth to Sirius in miles, we need to convert the light year measurement to miles. We know that 1 light year is the distance light travels in one year, which we can calculate by multiplying the speed of light by the number of seconds in a year. Here's the step-by-step calculation:

Speed of light = 3.00 × 10^8 m/s

Number of seconds in a year = 365 days * 24 hours * 60 minutes * 60 seconds = 31,536,000 seconds

Distance in meters = speed of light * number of seconds in a year = 3.00 × 10^8 m/s * 31,536,000 seconds

Distance in miles = distance in meters / (5280 ft/mi * 12 in/ft)

Plugging in the values into the equation, we get:

Distance in miles = (3.00 × 10^8 m/s * 31,536,000 seconds) / (5280 ft/mi * 12 in/ft)

Simplifying the equation, the distance from Earth to Sirius is approximately 50 trillion miles.

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

Final answer:

The magnitude of the net force acting on the kayak, which accelerates from 0 to 0.61 m/s over a distance of 0.39 m with a mass of 74 kg, is approximately 35.25 Newtons (N).

Explanation:

A person in a kayak accelerates from 0 to 0.61 m/s over a distance of 0.39 meters. To find the magnitude of the net force acting on the kayak, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. We can calculate this as:

To find the net force, we need to first calculate the change in kinetic energy, which is the kinetic energy at 0.61 m/s minus the initial kinetic energy at 0 m/s:

Since work done is equal to force times distance and to the change in kinetic energy, we can express the net force as:

Therefore, the magnitude of the net force acting on the kayak is approximately 35.25 Newtons (N).

Answer:

The potential energy change is 0.596 J.

Explanation:

Given that,

Potential difference =14.9 V

Charge q =0.0400 C

We need to calculate the change potential energy

The potential energy change is the product of the charge and potential difference.

Using formula of change potential energy

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

[tex]\Delta U=0.0400\times14.9[/tex]

[tex]\Delta U=0.596\ J[/tex]

Hence, The potential energy change is 0.596 J.

Answer:

0.596 J.

Explanation:

acellus

The specific heat of the gas is 0.381 J/g°C.

The specific heat of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. To find the specific heat of the gas, we can use the equation:

Q = mass * specific heat * change in temperature

In this case, the mass of the gas is 80.0g, the change in temperature is 225°C - 25°C = 200°C, and the energy absorbed by the gas is 6085J. Plugging these values into the equation, we can solve for the specific heat:

6085J = 80.0g * specific heat * 200°C

Specific heat = 6085J / (80.0g * 200°C) = 0.381 J/g°C

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The specific heat of the given gas can be calculated using the formula c = Q / (m * ΔT). By substitifying Q = 6431J (sum of work done by the system and increase in internal energy), m = 80g (mass of gas), and ΔT = 200°C (change in temperature), we find that the specific heat of the gas is roughly 0.40 J/g°C

The specific heat of a substance is the heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For the given gas, we first need to calculate the heat absorbed which is the sum of the work done by the system and the increase in its internal energy, which equals 346J + 6085J = 6431J. The following information is given, mass m = 80.0g and the change in temperature ΔT = 200°C (225 °C - 25°C).

The formula for specific heat is:
Q = m*c*ΔT
where:
Q is the heat energy absorbed (or released),
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
Solving this equation for the specific heat (c), we find:
c = Q / (m * ΔT).

So, substituting the given values into this formula, we have:
c = 6431J /  (80.0g * 200°C = 0.40 J/g °C which will be the specific heat of the gas.

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Answer:

V=-5304.6V

Explanation:

(I set the charge density unit to microcoulombs per metre cubed)

Answer:

In the picture.

Final answer:

To find the electric potential 5 m from the center of the uniformly charged sphere, the sphere is treated as a point charge resulting from its total charge due to the volume charge density. The potential is calculated using Coulomb's law for a point charge, yielding a value of -0.399 MV.

Explanation:

To calculate the electric potential a distance 5 m from the center of a uniformly charged solid sphere with radius R = 0.4 m and charge density [tex]\\rho = -11 \mu C/m^3\[/tex], we use the concept of electric potential due to a continuous charge distribution. Since the point where we need to find the potential is outside the sphere, we can treat the sphere as a point charge located at its center with total charge Q.

The total charge Q can be found by integrating the charge density over the volume of the sphere:

[tex]Q = \int \rho \, dV = \rho \frac{4}{3}\pi R^3[/tex]

Plugging the values in, we have:

[tex]Q = -11 \times 10^{-6} C/m^3 \times \frac{4}{3}\pi \times (0.4 m)^3\\Q = -2.21 \times 10^{-7} C[/tex]

Now, the electric potential V at a distance r from a point charge Q is given by:

[tex]V = \frac{kQ}{r}[/tex]

where k is Coulomb's constant, k = 8.99 \times 10^9 N m^2/C^2 and r is the distance from the center of the sphere, which is 5 m in this case. Thus:

[tex]V = \frac{8.99 \times 10^9 N m^2/C^2 \times (-2.21 \times 10^{-7} C)}{5 m}[/tex]

V = -0.399 MV

Answer:

2.45 N/m

Explanation:

We can find the spring constant by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force applied to the spring

k is the spring constant

x is the stretching of the spring

Here, the force applied is the weight of the mass hanged on the spring. The mass is

m = 10 g = 0.010 kg

So the weight is

[tex]F=mg=(0.010 kg)(9.8 m/s^2)=0.098 N[/tex]

while the stretching is

x = 4 cm = 0.04 m

So the spring constant is

[tex]k=\frac{F}{x}=\frac{0.098 N}{0.04 m}=2.45 N/m[/tex]

When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

An object is said to be in a condition of rotational equilibrium if there is no net torque operating on it. In other words, the object is keeping a constant angular velocity while not suffering any rotational acceleration. The object has a constant angular momentum when the net torque is zero.

If the net torque is zero, the item is being pulled in a direction that is equal to the sum of all the clockwise and anticlockwise torques pulling on it. To put it another way, any propensity the object has to spin in one direction is counterbalanced by an equal tendency to rotate in the opposing direction.

Hence, When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

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Final answer:

When the net torque is zero, it means that the clockwise and counterclockwise torques are balanced.

Explanation:

When the net torque is zero, it implies that the clockwise and counterclockwise torques balance each other out. In other words, the total torque in the clockwise direction is equal in magnitude but opposite in sign to the total torque in the counterclockwise direction.

Answer:

The velocity after impact 24.96 m/s

Explanation:

Given that,

Mass of falcon = 1.55 kg

Mass of dove = 0.395 kg

Velocity of falcon = 29.5 m/s

Velocity of dove = 7.15 m/s

We need to calculate the velocity after impact

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Where,

u = initial velocity

v = final velocity

[tex]m_{1}[/tex] = mass of falcon

[tex]m_{2}[/tex] = mass of dove

Put the value in the equation (I)

[tex]1.55\times29.5+0.395\times7.15=(1.55+0.395)v[/tex]

[tex]v=\dfrac{1.55\times29.5+0.395\times7.15}{(1.55+0.395)}[/tex]

[tex]v=24.96\ m/s[/tex]

Hence, The velocity after impact 24.96 m/s

Answer:

P = 2234400 Watt

Explanation:

Power generation is given as rate of work done

Here the turbine generator is located at 120 m below the free surface

So here rate of work done is given as rate of potential energy

[tex]Power = \frac{dw}{dt}[/tex]

[tex]Power = \frac{mgh}{t}[/tex]

so we have

[tex]Power = (\frac{dm}{dt})gh[/tex]

now we have

[tex]Power = 1900(9.8)(120) = 2234400 Watt[/tex]

so the power generation potential will be 2234400 Watt

Explanation:

For pendulum 1 :

Mass, m₁ = 0.1 kg

Length, l₁ = 5 m

For pendulum 2 :

Mass, m₂ = 0.9 kg

Length, l₂ = 1 m

The time period the simple pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]..........(1)

And we know that frequency f is given as :

[tex]f=\dfrac{1}{T}[/tex]

[tex]f=\dfrac{1}{2\pi\sqrt{\dfrac{L}{g}}}[/tex]

From equation (1) it is clear that the time period and frequency depend on the length of the bob and acceleration due to gravity only. It is independent of its mass.

Answer:

The distance is 0.86 m.

Explanation:

Given that,

Mass = 39 kg

Initial kinetic energy, [tex]K.E_{i} = 160\ J[/tex]

Final kinetic energy, [tex]K.E_{f} = 490\ J[/tex]

We need to calculate the work done

According to work energy theorem

[tex]W = \Delta K.E[/tex]

[tex]W=K.E_{f}-K.E_{i}[/tex]...(I)

Work done is the product of the force and displacement.

[tex]W = mgh[/tex]....(II)

From equation (I) and (II)

[tex]K.E_{f}-K.E_{i}=mgh[/tex]

[tex]490-160=39\times9.8\times h[/tex]

[tex]h = 0.86\ m[/tex]

Hence, The distance is 0.86 m.

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

[tex]F \Delta t = m (v-u)[/tex]

where

F is the average force

[tex]\Delta t[/tex] is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

[tex]\Delta t = 60.0 ms = 0.06 s[/tex]

Solving for F,

[tex]F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N[/tex]

And since we are interested in the magnitude only,

F = 106.7 N

The magnitude of the average force applied to the ball was about 107 N

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial velocity of the ball = u = -20.0 m/s

final velocity of the ball = v = 12.0 m/s

contact time = t = 60.0 ms = 0.06 s

mass of the ball = m = 0.200 kg

Asked:

average force applied to the ball = F = ?

Solution:

We will use this following formula to solve this problem:

[tex]\Sigma F = ma[/tex]

[tex]F = m ( v - u ) \div t[/tex]

[tex]F = 0.200 ( 12 - (-20) ) \div 0.06[/tex]

[tex]F = 0.200 ( 32 ) \div 0.06[/tex]

[tex]F = 6.4 \div 0.06[/tex]

[tex]F = 106 \frac{2}{3} \texttt{ N}[/tex]

[tex]F \approx 107 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Explanation:

Impulse is change in momentum:

I = Δp

I = (1400 kg) (27 m/s) - (1400 kg) (0 m/s)

I = 37800 kg m/s

Impulse is also average force times time:

I = F ΔT

37800 kg m/s = F (7 s)

F = 5400 N

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

[tex]y = h + \frac{1}{2}at^2[/tex]

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

[tex]0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s[/tex]

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

[tex]v_x = 23.4 m/s[/tex]

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

[tex]d=v_x t = (23.4 m/s)(3.35 s)=78.4 m[/tex]

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

[tex]v_x = 23.4 m/s[/tex]

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

[tex]v_y = u_y +at[/tex]

where

[tex]u_y = 0[/tex] is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

[tex]v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s[/tex]

and the negative sign means it points downward.

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

For the motion of horizontal and vertical direction we use following equation in projectile motion.

[tex]y=y_o+\dfrac{1}{2}gt^2[/tex]

Here, (g) is the gravity and (t) is time.

Here, in the given problem, the projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall.

As the height of the projectile before the shot is 55 meters and the the height of the projectile is 0 meter when it reaches to the ground. Thus from the above equation,

[tex]0=55+\dfrac{1}{2}\times9.81\times(t)^2\\t=3.35\rm s[/tex]

Thus the time required for the projectile to reach the ground below is 3.35 seconds.

The distance traveled by the object is the product of velocity and time taken. Thus, distance from the base of the building that the projectile lands is,

[tex]d=23.4\times3.35\\d=78.4\rm m[/tex]

Horizontal component of velocity is equal to the velocity of projectile shot just before it reaches to the ground. Thus, horizontal component of velocity is,

[tex]v_x=23.4\rm m/s[/tex]

Now the time taken is 3.35 seconds. Thus the vertical component of velocity can be found with the formula of first equation of motion with  0 initial velocity as,

[tex]v_y=0+(-9.81)\times(3.35)\\v_y=-32.8\rm m/s[/tex]

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

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4)none of the above.

A transformer is an electrical device  that allows to increase or decrease from one voltage level to another in an AC circuit while conserving the power.  

It is basically composed of two coils magnetically coupled in a core of ferromagnetic material increasing its permeability and effectiveness.

Answer: None of the above

Explanation:

Answer:

19 cm

Explanation:

inner radius, a = 16 cm, C = 113 p F = 113 x 10^-12 F

Let b cm be the radius of outer sphere.

The formula of capacitance of spherical capacitor is given by

C = 4π∈0 a b / (b - a)

a b / ( b - a) = C / 4π∈0

a b / ( b - a ) = 113 x 10^-12 x 9 x 10^9

a b / ( b - a ) = 1.017

16 x 10^-2 x b x 10^-2 = 1.017( b - 16) x 10^-2

0.16 b = 1.017 b - 16.272

0.857 b = 16.272

b = 19 cm

Answer: 2.52 Kilometers

Explanation:

We know that the formula to calculate speed is given by :-

[tex]\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}\\\\\Rightarrow\text{Distance}=\text{Speed}\times\text{Time}[/tex]

Given: The speed of sound during the storm : [tex]V=3350.0m/s[/tex]

The time taken by sound to travel : [tex]t= 7.20\text{ seconds}[/tex]

Then , the distance traveled by sound :-

[tex]\text{Distance}=350\times7.2=2520\text{ meters}=2.52\text{ kilometers}[/tex]

To find the distance of a lightning strike heard 7.20 seconds after the flash, multiply the sound's speed (350.0 m/s) by the time delay (7.20 s), resulting in a distance of 2520 meters.

We use the speed of sound. Since sound traveled at 350.0 m/s during the storm, we multiply the time it took to hear the thunder by the speed of sound to calculate the distance:

Distance = Speed of Sound × Time

Substituting in the given values:

Distance = 350.0 m/s × 7.20 s = 2520 meters

Therefore, the lightning struck approximately 2520 meters away.

(a) [tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]

In the first part of the motion, the car accelerates at rate [tex]a_1[/tex], so the final velocity after a time t is:

[tex]v = u +a_1t[/tex]

Since it starts from rest,

u = 0

So the previous equation is

[tex]v= a_1 t[/tex]

So the time taken for this part of the motion is

[tex]t_1=\frac{v}{a_1}[/tex] (1)

In the second part of the motion, the car decelerates at rate [tex]a_2[/tex], until it reaches a final velocity of v2 = 0. The equation for the velocity is now

[tex]v_2 = v - a_2 t[/tex]

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

[tex]t_2=\frac{v}{a_2}[/tex] (2)

So the total time taken for the trip is

[tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]

(b) [tex]d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]

In the first part of the motion, the distance travelled by the car is

[tex]d_1 = u t_1 + \frac{1}{2}a_1 t_1^2[/tex]

Substituting u = 0 and [tex]t_1=\frac{v}{a_1}[/tex] (1), we find

[tex]d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}[/tex]

In the second part of the motion, the distance travelled is

[tex]d_2 = v t_2 - \frac{1}{2}a_2 t_2^2[/tex]

Substituting [tex]t_2=\frac{v}{a_2}[/tex] (2), we find

[tex]d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}[/tex]

So the total distance travelled is

[tex]d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]

The total time elapsed from start to stop for the electric vehicle is the sum of the time taken to accelerate and decelerate, calculated as T = v / a1 + v / a2. The total distance moved is the sum of the distances during acceleration and deceleration given by S = 0.5 * a1 * (v / a1)^2 + 0.5 * a2 * (v / a2)^2.

The question inquires about the time elapsed and the distance traveled by an electric vehicle which accelerates from rest until it reaches a certain velocity, and then decelerates to a stop. To answer part (a), we need to calculate the time taken for both acceleration and deceleration phases. For acceleration, we use the formula t1 = v / a1, and for deceleration t2 = v / a2. The total time elapsed T is then the sum of t1 and t2.

For part (b), the total distance covered S is the sum of the distances covered during acceleration and deceleration. The distance covered during acceleration s1 can be found using the equation s1 = 0.5 * a1 * t1^2, and the distance covered during deceleration s2 can be found with s2 = 0.5 * a2 * t2^2.

Hence, for an electric vehicle that accelerates to a speed v at a rate a1 and then decelerates at a rate a2 to a stop:

The location of the coin's final image is at -6.00 cm from the diverging lens, and the magnification of the image is 0.50, indicating that the image is half the size of the object.

To find the location and magnification of the coin's final image, we need to use the lens equation in combination with the mirror equation. First, we will calculate the position of the coin's image formed by the converging lens:

So, the location of the coin's final image is at -6.00 cm from the diverging lens. To calculate the magnification, we can use the formula m = -di/do, where di is the image distance and do is the object distance. Substituting the values, we get m = -(-6.00 cm)/(-12.0 cm) = 0.50. Therefore, the magnification of the coin's final image is 0.50, indicating that the image is half the size of the object.

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Explanation:

It is given that,

Momentum of an object, p = 23.3 kg-m/s

Kinetic energy, E = 262 J

(a) Momentum is given by, p = mv

23.3 = mv...........(1)

Kinetic energy is given by, [tex]E=\dfrac{1}{2}mv^2[/tex]

m = mass of the object

v = speed of the object

[tex]E=\dfrac{1}{2}\times (mv)\times v[/tex]

[tex]262=\dfrac{1}{2}\times 23.3\times v[/tex]

v = 22.48 m/s

(2) Momentum, p = mv

[tex]m=\dfrac{p}{v}[/tex]

[tex]m=\dfrac{23.3\ kg-m/s}{22.48\ m/s}[/tex]

m = 1.03 Kg

Hence, this is the required solution.

An object with momentum 23.3 kg.m/s and kinetic energy of 262J is traveling at a speed of approximately 30.21 m/s and its mass is approximately 0.771 kg based on the physics principles of kinetic energy and momentum.

The question involves the physics concepts of momentum and kinetic energy. We are given the momentum (p) of 23.3 kg.m/s and the kinetic energy (KE) of 262 J of an object.

(a) The formula for kinetic energy is KE = 0.5 * m * v^2, where m is the mass of the object and v is its velocity. We can rearrange to find v = sqrt((2*KE) / m). The mass can be obtained from the momentum formula, p = m * v, hence m = p / v. Substituting the second equation into the first gives v = sqrt((2 * KE * v) / p), which simplifies to v = sqrt((2 * 262 J) / 23.3 kg.m/s) = 30.21 m/s.

(b) With the velocity calculated in (a), the mass of the object can now be found by rearranging the momentum formula to m = p / v = 23.3 kg.m/s / 30.21 m/s = 0.771 kg.

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Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{4.86\ m/s-0}{2.43\ s}[/tex]

a₁ = 2 m/s²

(b) Acceleration of the car,

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}[/tex]

a₂ = 4.97 m/s²

(c) Distance covered by the car,

[tex]d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2[/tex]

[tex]d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2[/tex]

d₁ = 5.904 m

Distance covered by the jogger,

[tex]d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2[/tex]

[tex]d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2[/tex]

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

Answer:

Fringe width = 21 mm

Explanation:

Fringe width is given by the formula

[tex]\beta = \frac{\Lambda L}{d}[/tex]

here we know that

[tex]\Lambda = 564 nm[/tex]

L = 4.0 m

d = 0.108 mm

now from above formula we will have

[tex]\beta = \frac{(564 \times 10^{-9})(4.0 m)}{0.108\times 10^{-3}}[/tex]

[tex]\beta = 0.021 meter[/tex]

so fringe width on the wall will be 21 mm

Final answer:

In Young's double slit experiment with given parameters, the fringe separation between the fourth order and central fringe is calculated to be 83.6 mm.

Explanation:

The question involves calculating the fringe separation in Young's double slit experiment using a green laser with a wavelength of 564 nm, a slit separation of 0.108 mm, and a distance to the screen of 4.0 m. To find the separation between the fourth order and the central fringe, we use the formula for fringe separation in a double slit experiment, which is Δy = λL/d, where Δy is the fringe separation, λ is the wavelength of the light, L is the distance from the slits to the screen, and d is the separation between the slits. Plugging in the values, we get Δy = (564 × 10^-9 m)(4 m) / (0.108 × 10^-3 m) = 0.0209 m or 20.9 mm for the separation between each fringe. However, since we need the separation between the fourth order and the central fringe, we simply multiply by the order number, giving us 4 × 20.9 mm = 83.6 mm.