Yea, gonna need some help. Thanks

Answer:

v = speed

c = speed of light

using the equation

L = Lc Sqrt( 1 - (v/c)2)

9 = 14 Sqrt( 1 - (v/c)2)

v/c = 0.765

v = 0.765 c

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

pls mark me as a brainlist

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

Answer:

B) the medium

Explanation:

B. Depends on the medium.

Answer:

To cover a larger distance we use km(kilometre) , hm (hectometre) , and dac(decametre). These are called the multiples of Metre. For the distance is smaller , we use Dm (decimetre , cm(centimetre) and mm (millimetre) . These are called the submultiples of

Answer:

[tex]\theta=34 \textdegree[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=55kg[/tex]

Angle [tex]\theta =28.0[/tex]

Coefficient of static friction [tex]\alpha =0.680[/tex]

Generally, the equation for Newtons second Law is mathematically given by

For

[tex]\sum_y=0[/tex]

[tex]N=mgcos \theta[/tex]

for

[tex]\sum_x=0[/tex]

[tex]F_{s}=mgsin\theta[/tex]

Where

[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]

[tex]F_{s}=0.68*55*9.8*cos 28[/tex]

[tex]F_{s}=323.62N[/tex]

Therefore

[tex]\alpha mgcos \theta=mg sin \theta[/tex]

[tex]\theta=tan^{-1}(0.68)[/tex]

[tex]\theta=34 \textdegree[/tex]

(a) The static frictional force which holds the box in place is 323.62 N.

(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.

The net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.

∑F = 0

The static frictional force is calculated as follows;

Fs = μFncosθ

Fs = 0.68 x (55 x 9.8) x cos28

Fs = 323.62 N

Fn(sinθ) - μFn(cosθ) = 0

mg(sinθ) - μmg(cosθ) = 0

μmg(cosθ) = mg(sinθ)

μ(cosθ) = (sinθ)

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(μ)

θ = tan⁻¹(0.68)

θ = 34.2⁰

Learn more about net force of inclined here: https://brainly.com/question/25784024

Answer:

[tex]P_d=6203.223062W/mm^2[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=45kV[/tex]

Current [tex]I=50mAmp[/tex]

Diameter  [tex]d=0.50mm[/tex]

Heat transfer factor [tex]\mu= 0.87.[/tex]

Generally the equation for  Power developed is mathematically given by

[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]

[tex]P=2.250[/tex]

Therefore

Power in area

[tex]P_a=1400*0.87[/tex]

[tex]P_a=1218watt[/tex]

Power Density

[tex]P_d=\frac{P_a}{Area}[/tex]

[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]

[tex]P_d=6203.223062W/mm^2[/tex]

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

Answer:

x = A sin (wt + theta)        where w = angular frequency - basic SHM equation

w = 8 pi = 2 pi f

f = 4         basic frequency

N = 8     number of times thru origin

Each cycle the particle will pass thru the origin +x and -x    twice

Answer:

The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.

Explanation:

brainliest plz . . .

Answer:

wavelength= 1.05 × 10^ -46 m

Explanation:

the formula : λ= hc/E

where; "h" = Planck's constant [6.626 × 10^ -34]

c= speed of light [3.0 × 10^ 8]

you first have to convert the energy of the photon to Joules by dividing the constant by 1000

2.09 × 10^ -18 / 1000 = 2.09 × 10^ -21

then you replace you data into the equation

λ= 6.626 × 10^ -34 × 3.0 × 10^ 8 / 2.09 × 10 ^ -21

first multiply the Planck's constant and the speed of light then divide it by the energy which is in "Joules"

:. λ = 1.05 × 10^ -46

hope this helps

Answer:

Work done = 2289.084 Joules

Explanation:

Given the following data;

Mass = 102 Kg

Height = 2.29

Time = 1.32 seconds

We know that acceleration due to gravity, g = 9.8 m/s²

a. To find the work done by the person;

Here, work would be done in the form of gravitational potential energy.

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

Work done = 102 * 2.29 * 9.8

Work done = 2289.084 Joules

Answer:

The amplitude of vibration of the spring is "24 cm"

The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.

From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.

Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.

Amplitude = 24 cm

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

[tex]g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}[/tex]

g = 0.0114 m/s²

Explanation:

Given:

[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]

[tex]V = 12 V[/tex]

[tex]I = 1.2 A[/tex]

Recall that power P is given by

[tex]P = VI[/tex]

so the amount of energy dissipated [tex]\Delta E[/tex] is given by

[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]

[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]

Answer:

15.88

is the correct answer

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

Answer:

Electric field at A = 9.28 x 10¹² N/C

Explanation:

Given:

K = 8.99 x 10¹² N/C

Missing information:

Length = 11 cm = 11 x 10⁻² m

q = 12.5 C

Find:

Electric field at A

Computation:

Electric field = Kq / r²

Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²

Electric field at A = 9.28 x 10¹² N/C

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

Answer:

it will drop simultaneously

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

[tex]x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}[/tex] (1)

Where:

[tex]x_{o}[/tex] - Initial x-position, in meters.

[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.

[tex]t[/tex] - Time, in seconds.

[tex]a_{x}[/tex] - x-acceleration, in meters per second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]

[tex]x(t) = 0.324\,m[/tex]

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

[tex]y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}[/tex] (2)

Where:

[tex]y_{o}[/tex] - Initial y-position, in meters.

[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.

[tex]t[/tex] - Time, in seconds.

[tex]a_{y}[/tex] - y-acceleration, in meters per second.

If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:

[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]

[tex]y(t) = -2.16\,m[/tex]

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:

[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)

If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:

[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]

[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]

The y-velocity of the skateboard is -3.6 meters per second.

Answer:

always runs slower than normal.

Explanation:

The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.

According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.

In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.