A 1100- kg car collides with a 1300- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 20.0 km/h in a direction of 30 o with respect to the positive x axis. The heavier car moves at 23 km/h at -46 o with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)

Answer:

a) W = 139.4 kJ

b) W = 200J

Explanation:

a) given;

Force F = 410kN

distance d = 34cm = 0.34m

Workdone = force × distance = Fd

Substituting the values;

Workdone = 410kN × 0.34m = 139.4 kJ

b) given;

Force F = 4000N

distance d = 5cm = 0.05m

Workdone = force × distance = Fd

Substituting the values;

Workdone = 4000N × 0.05m = 200J

Answer:

4.74*10^-7 m

Explanation:

TO find the wavelength of the photon you calculate the energy of the photon emitted in a harmonic oscillator:

[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]

m: initial state = 3

n: final state = 1

k = 3.6N/m

By replacing the values of m for the electron, m,n and ħ you obtain:

[tex]E_{m-n}=(1.055*10^{-34}Js)\sqrt\frac{3.6N/m}{9.11*10^{-31}kg}}(3-1)=4.19*10^{-19}J[/tex]

Furthermore, this energy is equivalent to the expression:

[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]

By replacing you obtain:

[tex]\lambda=\frac{2\pi (1.055*10^{-34}Js)(3*10^8m/s)}{4.19*10^{-19}J}=4.74*10^{-7}m[/tex]

hence, the wavelength of the photon is 4.74*10^-7 m

The wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m - [tex]4.74\times10^-7 m[/tex]

The energy of the photon emitted in a harmonic oscillator:

[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]

m: initial state = 3

n: final state = 1

k = 3.6N/m

By replacing the values of m for the electron, m, n and ħ you obtain:

[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]

Thus, the wavelength of the photon is [tex]4.74\times10^-7 m[/tex]

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Answer:

Find the given attachment

Answer:

471.4 nm

Explanation:

Please kindly see attachment for the step by step and very detailed solution to the given problem.

The attached file gave an explicit solution

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

From the question we are told that

   The center of gravity of the plane to its nose  is  [tex]z = 2.58 m[/tex]

    The distance of the front wheel of the plane to  its nose is [tex]l = 0.800\ m[/tex]

     The distance of the main wheel of the plane to its nose is [tex]e = 3.02 \ m[/tex]

At equilibrium  the Torque about the nose of the airplane is mathematically represented as

          [tex]mg (z- l) - G_B *(e - l) = 0[/tex]

Where m is the mass of the airplane

          [tex]G_B[/tex] is the weight of the airplane supported by the main wheel  

       So  

             [tex]G_B =\frac{mg (z-l)}{(e - l)}[/tex]

Substituting values

            [tex]G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}[/tex]

           [tex]G_B = 0.8018 mg[/tex]

Now the weight supported at the frontal wheel is mathematically evaluated as

           [tex]G_F = mg - G_B[/tex]

Substituting values      

       [tex]G_F = mg - 0.8018mg[/tex]    

      [tex]G_F = (1 - 0.8018) mg[/tex]      

     [tex]G_F = 0.1982 mg[/tex]    

Now the weight of the airplane is  =  mg

Thus percentage of this weight supported by the front wheel is  [tex]A = 0. 1982 *100 =[/tex] 19.82 %

Answer:

Explanation:

The ball was moving with velocity of 20 m /s earlier in horizontal direction . Due to kicking, additional V velocity was added to it at 40° because he kicked it at this angle but the ball travelled in the direction of resultant which was making an angle of 30° with the horizontal  .

From the relation of inclination of resultant

Tan θ = V sinα / (u + V cosα) where α is angle between u and V , θ is inclination of resultant

Tan30 = [tex]\frac{Vsin40}{20+ Vcos40}[/tex]

[tex]\frac{1}{\sqrt{3 } } =\frac{V\times .64}{20+ V\times.766}[/tex]

20 + .766 V = 1.11 V

20 = .344 V

V = 58 m /s

To know the force , we shall apply concept of impulse

F x t = mv  , F is force for time t creating a change of momentum mv

F x .1 = .4 x 58

F = 232 N

Answer:[tex]d+\frac{L}{2}[/tex]

Explanation:

Given

Length of skateboard is L

distance of skateboard from the wall is d

Suppose mass of skateboard is M

so mass of Professor is M

When Professor moves towards wall skateboard started moving away from wall.

If the professor moves L distance on the skateboard

Therefore relative displacement of the skateboard is

[tex]=\frac{ML}{M+M}[/tex]

[tex]=\frac{L}{2}[/tex]

Therefore professor is at a distance of [tex]d+\frac{L}{2}[/tex] from wall

Final answer:

The professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.

Explanation:

To solve this problem, we can apply the principle of conservation of momentum. The initial momentum of the system, consisting of the professor and the skateboard, is equal to the final momentum of the system. Initially, both the professor and the skateboard are at rest, so the initial momentum is zero.

When the professor walks towards the wall, he exerts a force on the ground, causing an equal and opposite force on him (according to Newton's third law). This force propels the skateboard forward, resulting in a change in momentum of the professor-skateboard system. The total momentum is conserved during this process.

When the professor reaches the opposite end of the skateboard, he comes to a stop. At this point, the skateboard would have moved a certain distance, which we'll call x. If the professor and the board have the same mass, the professor would have moved a distance equal to (d + x) from the wall, while the board would have moved a distance equal to x from the wall.

From the conservation of momentum, we can write:

(0) + (m)(v) = (m)(v) + (M)(V)

Here, m represents the mass of the professor, v represents his initial velocity, M represents the mass of the skateboard, and V represents its final velocity.

Since the professor starts from rest, his initial velocity v is zero. The final velocity V of the skateboard can be calculated using the equation:

(m)(v) = (M)(V)

From the given information, we know that the professor's mass is equal to the skateboard's mass, so m = M. Plugging this into the equation, we get:

(M)(0) = (M)(V)

This simplifies to:

0 = V

Since the final velocity V of the skateboard is zero, we can conclude that the professor comes to a stop at the opposite end of the skateboard. Therefore, he would be a distance x from the wall. To find the value of x, we need to analyze the forces and motion of the system.

When the professor exerts a force on the ground, causing an equal and opposite force on him, the system experiences a net force. According to Newton's second law, the net force on the system is equal to the product of the mass of the system and its acceleration.

Let's define the positive direction as the direction the professor is moving towards the wall. The net force on the system in the positive direction is:

F_net = F_applied - F_opposing

Where F_applied is the force exerted by the professor on the ground, and F_opposing is the sum of all the opposing forces, such as friction.

Since the skateboard rolls with negligible friction, the opposing forces can be considered negligible. Therefore, we have:

F_net = F_applied

From Newton's second law, we can write:

F_net = (M + m)a

Where a is the acceleration of the system.

Plugging in the given information, we have:

150 N = (2M)a

Solving for a gives:

a = 75 N / M

Now, let's consider the motion of the skateboard. Since there is no friction, the only horizontal force acting on the skateboard is the force exerted by the professor on the ground. This force causes the skateboard to accelerate in the positive direction.

The distance x is related to the acceleration a and the time taken t to reach the opposite end of the skateboard. We can use the kinematic equation:

x = (1/2)at^2

Since the professor is initially at rest and comes to a stop at the opposite end, his final velocity vf is zero. We can use the equation:

vf = vi + at

Where vi is the initial velocity of the professor, and a is the acceleration.

Since the professor is initially at rest, his initial velocity vi is zero. Plugging in the known values, we have:

0 = 0 + (75 N / M)t

This simplifies to:

t = 0

Since the time t is zero in this case, the professor reaches the opposite end of the skateboard instantaneously. Therefore, the distance x is also zero.

In conclusion, the professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.

Answer:

[tex]L=2.48*10^{-3} m[/tex]

Explanation:

The period equation for a pendulum is given by:

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]

and we know that T = 1/f, where f is the frequency, so we will have:

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

Now, we just need to solve this equation for L.

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

[tex]L=\frac{g}{(2\pi f)^{2}}[/tex]

[tex]L=\frac{9.78}{(2\pi*10)^{2}}[/tex]

[tex]L=2.48*10^{-3} m[/tex]

I hope it helps you!

Answer:

The intensity of the light transmitted through the third filter is  [tex]I_3 = \frac{I_o}{8}[/tex]

Explanation:

From the question we are told

   The intensity of the unpolarised light [tex]I_o[/tex]

   The angle between the first and second polarizer is  [tex]\theta _1 = 45^o[/tex]

     The angle between the first and third  polarizer is  [tex]\theta _2 = 90^o[/tex]

Generally the intensity of light emerging from the first polarizer is mathematically represented as

           [tex]I_1 = \frac{I_o}{2}[/tex]

According to Malus law the intensity of light emerging from the second polarizer is mathematically represented as

         [tex]I_2 = I_1 cos^2 (\theta_1)[/tex]

Substituting for [tex]I_1[/tex] and [tex]\theta _1[/tex]

          [tex]I_2 = \frac{I_o}{2} cos^2 (45)[/tex]

          [tex]I_2 = \frac{I_o}{4 }[/tex]

According to Malus law the intensity of light emerging from the third polarizer is mathematically represented as

         [tex]I_3 = I_2 cos ^2 (\theta_2 - \theta_1)[/tex]

Substituting for [tex]I_2[/tex] and [tex]\theta _1 \ and \ \theta _2[/tex]

         [tex]I_3 = \frac{I_o}{4} cos ^2 (90 - 45)[/tex]

         [tex]I_3 = \frac{I_o}{8}[/tex]

Answer:

Maximum radius = 5.1m

Explanation:

For us to get the radius of the circle of light, we have to first calculate the critical angle which is the angle of incidence above which total internal reflection occurs, i.e. the angle of incidence when θ2 = 90° .

At the point where the total internal reflection occurs, the light ray doesn't pass through the interface, thus, this point is on the edge of the circle of light.

From Snell's law, we have:

n_water * sin (θ1) = n_air * sin(θ2)

Thus;

sin(θ1) = n_air/(n_water * sin(θ2))

Since the critical angle is the value of θ1 when θ2 = 90° and that sin(90°) =1, thus;

sin(θ1) = (n_air/n_water) * sin(90°) = n_air/n_water

We are told the Refractive index of water is 1.33. meanwhile the Refractive index of air is not given but it has a constant value of 1.

Thus, we can determine θ1:

sin(θ1) = (nair/nwater) = 1/1.33

sin(θ1) = 1/1.33

(θ1) = sin^(-1)(1/1.33)

(θ1) = 48.75°

The question when looked at critically, depicts a right triangle with vertices including the light and the extremity of the circle, and we know one of its angles(θ1 = 48.75°) and one of its sides(4.5 m).

Thus, from trigonometric ratio, we can determine the radius as;

r/4.5 = tan(θ1)

r = 4.5tan(48.75°) = 5.1 m

Answer:

2.09 m/s

Explanation:

As the  spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.

Therefore, for the period of a full oscillation of the system

T= 2t => 2(0.6)=> 1.2 s

As the frequency is the reciprocal of the period, we have

f= 1/T => 1/1.2

f= 0.833 Hz

The angular frequency'ω' is given by,

ω= 2πf => 2π x 0.833=>5.23 rad/s

For the maximum velocity of the object  in a spring-mass system:

V[tex](_{max} )[/tex]= Aω

where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)

V[tex](_{max} )[/tex]= 0.4 x 5.23 =>2.09 m/s

Answer:

a) [tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex], b) [tex]\mu_{s} = 0.028[/tex], c) [tex]\mu_{s} = 0.036[/tex]

Explanation:

a) The linear acceleration of the watermelon seed is:

[tex]a_{r} = \omega^{2}\cdot r[/tex]

[tex]a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)[/tex]

[tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex]

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

[tex]\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a[/tex]

[tex]a = \mu_{s}\cdot g[/tex]

[tex]\mu_{s} = \frac{a}{g}[/tex]

[tex]\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\mu_{s} = 0.028[/tex]

c) Angular acceleration experimented by the turntable is:

[tex]\alpha = \frac{\omega-\omega_{o}}{\Delta t}[/tex]

[tex]\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}[/tex]

[tex]\alpha = 9.6\,\frac{rad}{s^{2}}[/tex]

The tangential acceleration experimented by the watermelon seed is:

[tex]a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)[/tex]

[tex]a_{t} = 0.221\,\frac{m}{s^{2}}[/tex]

The linear acceleration experimented by the watermelon seed is:

[tex]a = \sqrt{a_{t}^{2}+a_{r}^{2}}[/tex]

[tex]a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}[/tex]

[tex]a = 0.353\,\frac{m}{s^{2}}[/tex]

The minimum coefficient of static friction is:

[tex]\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\mu_{s} = 0.036[/tex]

Answer:

When doing ballistic stretching, it is using motion to bounce and stretch your body past its natural range of motion. In doing so you can harm yourself if you don't have a professional to help you as you might tear, damage, or pop your tendons, ligaments, or joints.

Explanation:

100% On Edge for 2021

The ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.

The stretching activity that utilizes the momentum of body to achieve greater range of motion and flexibility, is known as Ballistic stretching. It is one of the intense stretching method that involves the bouncing movements to force the body beyond the normal range of motion.

This can be harmful if an athlete do not have a professional trainer to train for the cause. This may cause tear, damage of tendons, ligaments, or joints.

Following are the  ways to perform a perform a proper stretch:

Thus, we can conclude that the ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.

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Answer:A generator is the appropriate device to be used.

Explanation:

A generator is a device which transforms mechanical energy to electrical energy.

The different parts of a generator includes:

- Engine, alternator, fuel system, voltage regulator, lubricating system, cooling and exhaust system, Battery charger and control panel.

The generator works on the principle of electromagnetic induction which states that

that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.

Therefore, the use of a generator is the best device for the campers to provide electrical power source for their heaters and light in the cabin.

A generator would be the appropriate device to power a heater and lights in a cabin with no electrical power, as it can convert mechanical energy into electrical energy required to operate these devices.

To provide power for a heater and lights in a cabin with no electrical power, the appropriate device among the options provided is a generator. A generator converts mechanical energy into electrical energy, which can then be used to power electrical devices such as heaters and lights. It does not rely on external power sources, making it suitable for remote locations like a cabin.

Other devices mentioned, like a resistor, are components typically used within electronic circuits to control current flow or convert electrical energy into heat. An insulator is a material that prevents the transfer of electricity, and a voltmeter is a tool used to measure voltage across points in a circuit, rather than to provide power. None of these would be suitable to power a heater and lights in the absence of an existing power source.

Answer:

Explanation:

flux through the coil = NBA

Change in flux = NBA - 0 = NBA

rate of change of flux = NBA / Δt

emf induced = NBA / Δt

current i = emf / resistance

= NBA / (RΔt)

Charge flowing through the search coil

=  NBA Δt/ (RΔt)

Q = NBA/R

Answer:The object characteristics

Explanation:the objects characteristics

Answer:

C- The objects characteristics

Explanation:

I just did it

Answer:

The speed is  [tex]v = 4.425 m/s[/tex]

Explanation:

From the question we are told that

     The spring constant is  [tex]k = 75 \ N /m[/tex]

      The mass of the foam dart is [tex]m = 5 g = \frac{5}{100} = 0.05 \ kg[/tex]

      The compression distance is  [tex]d = 10 cm = 0.1 m[/tex]

       The height which the gun raised the dart is  [tex]h = 5 m[/tex]

        The change in height is  [tex]\Delta h = 2 m[/tex]

        The new height is [tex]h_2 = 5 -2 = 3 m[/tex]

Generally from the law of conservation of energy

            [tex]E_s = KE[/tex]

Where [tex]E_s[/tex] is the energy stored in spring and it is mathematically represented as

            [tex]E_s = \frac{1}{2} k d^2[/tex]

  KE is the kinetic energy possessed by the dart when it is being shut and this is mathematically represented as

              [tex]KE = \frac{1}{2} mv^2_r[/tex]

So

          [tex]\frac{1}{2} k d^2 = \frac{1}{2} mv^2_r[/tex]

Substituting values

          [tex]0.5 * 75 * 0.1 = 0.5 * 0.0005 * v^2_r[/tex]

=>     [tex]v_r = \sqrt{\frac{0.5 * 75 * 0.1}{0.5 * 0.0005 } }[/tex]

        [tex]v_r = 12.25 m/s[/tex]

When the dart is at  the maximum height the

     let it acceleration due air resistance be z

So by equation of motion

          [tex]v^2 = u^2 - 2ah[/tex]

Where v is the velocity at maximum height which is equal to zero

    and  u is it initial velocity before reaching maximum height which we calculated as [tex]v_r = 12.25 m/s[/tex]

       and a is the acceleration due to gravity + the acceleration due to air resistance

     So

          a =  z+g

             =  9.8 + z

=>    [tex]v^2 = u^2 - 2(9.8 +z)h[/tex]

Substituting values

          [tex]0 = 12.25^2 - 2(9.8 +z)h[/tex]

Making z the subject

          [tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]

         [tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]

         [tex]z = 5 m/s[/tex]

When the dart is moving downward we can mathematically represent the motion as

        [tex]v^2 = u^2 + 2ah[/tex]

Since the motion is downward and air resistance is upward we have that

         a =  g - z

and the the initial velocity u becomes the velocity at maximum height

i.e u = 0

     And v is the velocity the dart has when it is moving downward

               So

                         [tex]v^2 = 0 + 2 * (g -z )h[/tex]

Substituting values

                        [tex]v = \sqrt{0+ 2 (10 - 5) * 2}[/tex]

                        [tex]v = 4.425 m/s[/tex]

Answer:the object with the highest mass and with the highest acceleration

Explanation:

Force is directly proportional to mass

The higher the mass, the higher the force

Answer:

T = 9.19 seconds

Explanation:

It is given that,

Length of the pendulum, l = 21 m

We need to find its time period. We know that the time period of the pendulum is given by :

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

g is acceleration due to gravity

So,

[tex]T=2\pi \sqrt{\dfrac{21}{9.8}}\\\\T=9.19\ s[/tex]

So, the period of the pendulum is 9.19 seconds.

Final answer:

The period of the Smithsonian's 21.0 m long pendulum is approximately 9.2 seconds, which is calculated using the formula for a simple pendulum's period T = 2π√(L/g).

Explanation:

The period of a pendulum primarily depends on its length and the acceleration due to gravity. Using the formula for the period of a simple pendulum T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), we can calculate the period of the Smithsonian's pendulum.

Given that the length L is 21.0 m and by taking the standard value for g, 9.81 m/s², we plug these values into the formula to find the period:

T = 2π√(21.0 m / 9.81 m/s²)

Calculating this we get:

T ≈ 2π√(2.14 s²)

T ≈ 2π√(2.14 s²)

T ≈ 2π(1.46 s)

T ≈ 9.2 s

This means the Smithsonian's pendulum would have taken approximately 9.2 seconds to complete one full swing, back and forth. This is the time it would take to go from one side to the other and back again, corresponding to one complete oscillation.

The orbital speed of Dactyl around Ida is approximately 161.6 m/s.

The orbital speed of an object can be calculated using the formula v = √ (G * (M + m) / r), where v is the orbital speed, G is the gravitational constant (6.672x10-11 N-m2/kg2), M is the mass of the primary object (Ida), m is the mass of the secondary object (Dactyl), and r is the distance between the center of the two objects.

Plugging in the given values, we have:

v = √ ((6.672 x 10-11 N-m2/kg2) * ((4.4 x 1016 kg) + (2.6 x 1012 kg)) / (95,000 m))

Simplifying the equation, we get:

v = 161.6 m/s

Therefore, the orbital speed of Dactyl around Ida is approximately 161.6 m/s.

Final answer:

To calculate the orbital speed of Dactyl in a circular orbit around Ida, you can use the formula for orbital speed.

Explanation:

Orbital speed of Dactyl:

The orbital speed of a satellite in a circular orbit can be calculated using the formula: v = √(G * M / r), where v is the orbital speed, G is the gravitational constant, M is the mass of the central body, and r is the distance from the center of the central body.

Substitute the given values: G = 6.67²x10-11 N-m²/kg², M = 4.4 x 10¹⁶ kg, and r = 95 km (converted to meters).

Answer: the first box is 'red' and the second box is 'supergiants'

Explanation: just did it edg. And it was correct

A star is observed to have a temperature of 3000 K and then luminosity of 105. The color of the star is red.

A spectrum is a collection of all visible light. The region of the electromagnetic spectrum that is visible to the human eye is the light that we see, which includes the rainbow's hues.

All electromagnetic energy emits some kind of radiation, whether it is in the form of visible light or another type, and it also radiates heat. Other stars emit heat and light, just like our sun does. Numerous star measurements have revealed a strong correlation between star temperature and star hue.

Given parameters:

Temperature of the star: T = 3000 K.

luminosity of the star = 105.

For this temperature of the star,  the color of the star is red.

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Answer:

Explanation:

a ) Magnetic force on proton

= B q v , B is magnetic field , q is charge with velocity v

= 2.85 x 10⁻³ x 1.6 x 10⁻¹⁹ x 2840

= 12.95 x 10⁻¹⁹ N

Its direction will be towards positive z - direction according to Fleming's left hand rule.

force on proton due to electric field = charge x electric field.

= 1.6 x 10⁻¹⁹ x 5.38

= 8.608 x 10⁻¹⁹ N

this force will be along the field ie in positive z direction so both the forces are acting in the same direction so they will add up.

total force =  (12.95 + 8.608)x 10⁻¹⁹

= 21.558 x 10⁻¹⁹ N .

b ) in this case , both the forces are acting in opposite direction . net force

=  (12.95 - 8.608)x 10⁻¹⁹

= 4.342 x 10⁻¹⁹ N

c ) In this case both the forces are acting perpendicular to each other

resultant = √(12.95² + 8.608²) x 10⁻¹⁹ N

= 15.54 x 10⁻¹⁹ N .

Answer:

Maximum Radius = 2.89m

Explanation:

The maximum radius will be determined by the angle of incidence which is equal to the critical angle. Now, any angle larger than that will make the light to be totally internally reflected. Hence, we can figure out that angle from Snell’s law where the refracted angle is 90°, and then use the tangent function.

From Snell's law;

n_air*sin90° = n_water*sin(θ _c)

Where;

θ_c is the critical angle

Refractive index of water; n_water = 1.333

Refractive index of air;n_air = 1

Thus;

1*1 = 1.33sinθ_c

sinθ_c = 1/1.33

θ_c = sin^(-1)0.7519

θ_c = 48.76°

Like I said earlier, we'll use tangent to find the radius.

Thus;

tanθ_c = d/R

From the question, d = 3.3m

Thus;

3.3/tan48.76 = R

So, R = 2.89m

Answer:

The acceleration of bicyclist is greater than that of the runner.

Explanation:

It is given that,

Initial speed of both bicyclist and a runner is 0 as they both are waiting at a red light,

When the light turns green they start to speed up.

Final speed of the bicyclist is 20 mph in 5 seconds

The runner gets to a final speed of 11 mph in 3 seconds.

20 mph = 8.94 m/s

11 mph = 4.91 m/s

Acceleration of bicyclist is :

[tex]a_b=\dfrac{v}{t}\\\\a_b=\dfrac{8.94\ m/s}{5\ s}\\\\a_b=1.78\ m/s^2[/tex]

Acceleration of runner is :

[tex]a_r=\dfrac{v}{t}\\\\a_r=\dfrac{4.91\ m/s}{3\ s}\\\\a_r=1.63\ m/s^2[/tex]

It is clear that the acceleration of bicyclist is greater than that of the runner.

The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².

Hence, the bicyclist has the greater acceleration.

Given the data in the question;

Since the runner and the bicyclist where initially at rest;

Acceleration is simply the rate at which velocity changes with respect to time. Formula for acceleration can be derived from the First Equation of Motion;

[tex]v = u + at\\\\at = v - u\\\\a = \frac{v - u}{t}[/tex]

Where a is acceleration, v is final velocity, u is initial velocity and t is time elapsed.

Now, to determine who has the greater acceleration, we substitute our given values into the expression above.

For the bicyclist;

[tex]a_b = \frac{v -u}{t} \\\\a_b = \frac{ 8.9408m/s - 0}{5s}\\ \\a_b = \frac{8.9408m/s}{5s}\\ \\a_b = 1.788m/s^2[/tex]

For the runner;

[tex]a_r = \frac{v -u}{t} \\\\a_r = \frac{ 4.91744m/s - 0}{3s}\\ \\a_r = \frac{4.91744m/s}{3s}\\ \\a_r = 1.639m/s^2[/tex]

The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².

Therefore, the bicyclist has the greater acceleration.

Learn more about Equation of Motion: https://brainly.com/question/18486505

Answer:

Explanation:

mass of particle m = 2 x 10⁻⁶ kg

charge q = 1 x 10⁻⁹ C

electric field E = 2000 N/C

force on charge = E q

= 2000 x  1 x 10⁻⁹

acceleration = force / mass

= 2000x10⁻⁹ / 2 x 10⁻⁶

= 1 m /s²

initil velocity u = 6 m /s

distance s = 4 x 10⁻²

time = t

s = ut + .5  t²

4 x 10⁻² = 6t + .5 x 1 x t²

t² + 12t - .08 = 0

= .0067 s .

= 6.7 ms .

Answer:

the length of the sides s is  [tex]s = 1.998 \ cm[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 60 \ turn[/tex]  

      The magnetic field is [tex]B = 1.20 \ T[/tex]

     The angle the loop makes with the x-axis [tex]\theta = 15 ^o[/tex]

      The current flowing through the loop is [tex]I = 2.50 A[/tex]

       The magnitude of the torque is [tex]\tau = 0.0186 \ N[/tex]

        the length of the sides of the square is  [tex]s[/tex]

Generally, we can represent the torque magnitude as

            [tex]\tau = N I A B sin \theta[/tex]

Where A is the area of the square which is mathematically represented as

        [tex]A = s^2[/tex]

Substituting this into the formula for torque

        [tex]\tau = N I s^2 B sin \theta[/tex]

making s the subject

         [tex]s = \sqrt{\frac{\tau }{NIB sin \theta } }[/tex]

    Substituting values

         [tex]s = \sqrt{\frac{0.0186}{(60) * (2.50) * (1.20) * (sin (15))} }[/tex]

         [tex]s = 0.01998 m[/tex]

Converting to centimeters

      [tex]s = 0.01998 * 100[/tex]

      [tex]s = 1.998 \ cm[/tex]

Answer:

2 cm

Explanation:

To fins the lengths of the sides of the loop you use the following formula for the calculation of the torque experienced by the loop in a magnetic field:

[tex]\tau=NiABsin\theta[/tex]

N: turns of the loop = 60

i: current in the loop = 2.50A

A: area of the loop = s*s

B: magnitude of the magnetic field = 1.20T

θ: angle between the plane of the lop and the direction of B = 15°

BY replacing the values of the torque and the other parameters in (1) you  can obtain the area of the loop:

[tex]A=\frac{\tau}{NiBsin\tetha}=\frac{0.0186Nm}{(60)(2.5A)(1.2T)sin15\°}=3.99*10^{-4}m^2[/tex]

but the area is s*s:

[tex]A=\sqrt{s}=\sqrt{3.99*10^{-4}m^2}=0.019m\approx2cm[/tex]

hence, the sides of the square loop have a length of 2cm

Answer:

Check the explanation

Explanation:

The potential energy (is the energy by virtue of a particular object's location relative to that of other objects. This term is most of the time linked or associated with restoring forces such as a spring or the force of gravity.) seems to be U = mgy [tex](1/2)(k k')(x^2 y^2)[/tex]. In fact, the mgy term has disappeared from the development.

Kindly check the attached image below for the step by step explanation to the question above.

Answer:

it is kenetic

Explanation: its in motion:D

Answer:

5.98 km

Explanation:

This question can be easily solved by using the trigonometric properties of a right angled triangle.

See attachment for pictorial explanation

To get x we have

Sinθ = opp / hyp

Sin25 = x / 4.5

x = 4.5 sin 25

x = 4.5 * 0.423

x = 1.9 km

To get y, we have

Cosθ = adj / hyp

Cos25 = y / 4.5

y = 4.5 cos 25

y = 4.5 * 0.906

y = 4.08 km

x + y = 1.9 + 4.09 = 5.98 km

Thus, the minimum distance required is 5.98 km

Final answer:

To find the minimum distance to the helicopter, calculate the eastward and northward components using trigonometry and sum them.

Explanation:

The minimum distance you must travel to reach the helicopter is determined by decomposing the direct diagonal path into two perpendicular paths that correspond to the grid of streets running east-west and north-south. Since the direction to the helicopter is 25° north of east, you can use trigonometry to find the lengths of the east and north legs of your journey. Using the cosine function for the eastward distance (cos(25°) × 4.50 km) and the sine function for the northward distance (sin(25°) × 4.50 km), you can calculate the exact distances you need to travel east and north:

Eastward distance = cos(25°) × 4.50 km

Northward distance = sin(25°) × 4.50 km

Sum these two distances to get the total minimum distance you need to travel.

Answer:

c. the magnitude of the ratio of the current to the time rate of change of the current

Explanation:

In a LR circuit where battery is connected , the expression for decay of current is given by

[tex]i = i_0e^{-\frac{t}{\lambda}[/tex]     where i is instantaneous current at time t , i₀ is maximum current , λ is constant .

differentiating on both sides with respect to t

di / dt = [tex]- \frac{i_0}{\lambda} e^{-\frac{t}{\lambda}[/tex]

[tex]\lambda = - \frac{i}{\frac{di}{dt} }[/tex]

So time constant is equal to magnitude of ratio of current to  time rate of change of current .

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.