What is the difference between a wave and energy?

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

Learn more about speed: https://brainly.com/question/7359669

Answer:

Explanation:

equation of wave is given by the following equation

y = (2.6 cm) sin[1.8 - (5.8 s-1)t].

Comparing it with standard form of wave

y = A sin ( ωt - kx )

we get

ω = 5.8

2πn = 5.8

n = .92 per second

kx = 1.8

k x 6 = 1.8

k = 0.3

[tex]\frac{2\pi}{\lambda}[/tex] = 0.3

λ = 20.9 cm

Answer:I think it’s self monitoring sorry if wrong

Explanation:

Answer:

It self monitoring

Explanation:

I took the test

Answer:

6N

Explanation:

Given parameters:

Mass of object  = 6kg

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time  = 30s

Unknown:

Net force acting on the object  = ?

Solution:

From Newton's second law of motion:

   Force  = mass x acceleration

Acceleration is the rate of change of velocity with time

  Acceleration  = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

  Force  = mass x  [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

So;

 Force  = 6 x [tex]\frac{25 - 5}{30}[/tex]    = 6N

Answer:

[tex]E=6.8Kv/m[/tex]

Explanation:

From the question we are told that

Distance b/w plate [tex]d=10cm=>0.1m[/tex]

P_1 Potential at 7.35 [tex]V=533v[/tex]

Generally the equation for electric field at a distance is mathematically given as

[tex]E=\frac{v}{d}[/tex]

[tex]E=\frac{533}{7.85*10^-^2}[/tex]

[tex]E=6789.808917[/tex]

[tex]E=6.8*10^3[/tex]

[tex]E=6.8Kv/m[/tex]

Answer:

1.9kHz

Explanation:

Given data

wavelength [tex]\lambda= 0.32m[/tex]

velocity [tex]v= 622 m/s[/tex]

We know that

[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]

substitute

[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]

Hence the frequency is 1.9kHz

Answer:

971.2

Explanation:

It was right on acellus :)

Answer:

ieces A and B must also have the same type of charges

Explanation:

In electrostatics, charges of the same sign repel and charges of different signs attract.

If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.

Also part A and C repel each other, therefore they have the same type of charge.

If we use the transitive property of mathematics, pieces A and B must also have the same type of charges

Answer:

15km/h

Explanation:

→ Speed = Distance ÷ Time

30 ÷ 2 = 15km/h

Answer:

D- Repulsion

Explanation:

A positively charged object will exert a repulsive force upon a second positively charged object.

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

Explanation:

kinetic energy = 14.1 MJ = 14.1 x 10⁶ J

Let radius of flywheel be r .

volume of flywheel = π r² x t where t is thickness

= 3.14 x r² x .113 m³

= .04 r² m³

mass = volume x density

= .04 r² x 7800 = 312.73 r²kg

moment of inertia I = 1 / 2 mass x radius²

= .5 x 312.73 r² x r²

= 156.37 r⁴ kg m²

angular velocity ω = 2π x 93/60

= 9.734 rad /s

kinetic energy = 1/2 Iω² where ω is angular velocity

= .5 x 156.37 r⁴ x 9.734²

= 7408.08 r⁴

Given

7408.08 r⁴ =  14.1 x 10⁶

r⁴ = .19 x 10⁴

r = .66 x 10

= 6.60 m .

Diameter = 13.2 m

b )

centripetal acceleration of a point on its rim = ω² r

= 9.734² x 6.6

= 625.35 m /s²

Answer:

the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Explanation:

from coulumbs law, The force that is acting over both charge can be computed as

F=( kq1q2)/r^2..............eqn(1)

Where

F=electrostatic force= 3.89 x 10-21 N

k= column constant= 9 x 10^9 Nm^2/C^2

q1 and q2= magnitude of the charges

r= distance between the charges= 1.08 x 10-3 m.

Since both charges are experiencing the same force, eqn(1) can be written as

F=( kq^2)/r^2.

We can make q subject of the formula

q= √(Fr^2)/k

= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9

q= 71.043×10^-20 C

Hence, the charge is 71.043×10^-20 C

From quantization law, the number of electron can be computed as

N=q/e

Where

N= number of electron

q= charges

=1.6×10^-19C

N=71.043×10^-20/1.6×10^-19

=4.44

Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:

[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)

Where:

[tex]m[/tex] - Mass of the sphere, measured in kilograms.

[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.

[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.

[tex]\Delta t[/tex] - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)

Where:

[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.

[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.

[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]

If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:

[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]

[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,

Fs = m a = W a / g

(a = centripetal acceleration, m = mass, g = acceleration due to gravity)

We have

a = v ² / R

(v = tangential speed, R = radius of the curve)

so that

Fs = W v ² / (g R)

Solving for v gives

v = √(Fs g R / W)

Perpendicular to the road, the car is in equilibrium, so Newton's second law gives

N - W = 0

(N = normal force, W = weight)

so that

N = W

We're given a coefficient of static friction µ = 0.4, so

Fs = µ N = 0.4 W

Substitute this into the equation for v. The factors of W cancel, so we get

v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

Answer:

   [tex]g=4.38*10^{7} m/s^2[/tex]

Explanation:

To solve this, we need to know the mass of the sun, and the radius of the moon

[tex]M_{s} = 1.989*10^{30}kg \\R_{m} = 1737400m[/tex]

Now we can plug our values into our equation:

[tex]g=G*\frac{M_{E} }{r^{2} }[/tex]

This gives us:

[tex]g=6.67*10^{-11}*\frac{1.989*10^{30}}{1737400^{2}}[/tex]

This equals:

[tex]g=4.38*10^{7} m/s^2[/tex]

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

Answer:

form 1 question??????????

Answer:

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

Explanation:

The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.

Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

Answer:

It means when you look into the lens your vision magnifies by x1

Explanation:

Answer:

it will option option A hope it helps

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Answer:

force = mass * acceleration

therefore the SI unit is kg*m/s2 or newton's

it's a vector quantity because it has both direction(acceleration) and size (mass)

Answer:

[tex]v=0.8m/s[/tex]

Explanation:

From the question we are told that

Distance [tex]d=1.5m/sm/s[/tex]

Time  [tex]t_1=60s[/tex]  

Time  [tex]t_2=90s[/tex]  

Generally the  the equation for the distance traveled is mathematically given as

[tex]d=vt[/tex]

[tex]d=1.5*90[/tex]

[tex]d=138m[/tex]

Generally equation for speed of side walk is mathematically given as

[tex]d=(v+u)t[/tex]

[tex]v=\frac{d}{t}-u[/tex]

[tex]v=\frac{138}{60}-1.5[/tex]

[tex]v=0.8m/s[/tex]

Answer:

B

Explanation:

Answer:

Force = 2240 Newton.

Explanation:

Given the following data;

Mass A= 65kg

Mass B = 215kg

Acceleration = 8m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

First of all, we would have to find the total mass.

Total mass = Mass A + Mass B

Total mass = 65 + 215

Total mass = 280kg

Substituting into the equation, we have

[tex] Force = 280 * 8 [/tex]

Force = 2240 Newton.

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]   (1)

Where:

x: is the displacement

[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)                    

[tex]v_{0}[/tex]: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

Answer:

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

Answer:

a) Average speed is 24.04 m/s

b) the rms speed is 25.84 m/s

c) the most probable speed is 16 m/s

Explanation:

Given the data in the question;

a) Determine the average speed.

To determine the average speed, we simply divide total some of speed by number of particles;

Average speed =  [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25    

= 601 / 25

= 24.04 m/s

Therefore, Average speed is 24.04 m/s

b) Determine the rms speed

we know that  (rms speed)² = sum of square speed / total number of particles

so

(rms speed)² =  [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25

(rms speed)² =  16691 / 25

(rms speed)² =  667.64

(rms speed) = √ 667.64

(rms speed) = 25.84 m/s

Therefore, the rms speed is 25.84 m/s

c) Determine the most probable speed.

Most particles (7) have velocity 16 m/s

i.e 7 is the maximum number of particle for a particular speed ,

Therefore, the most probable speed is 16 m/s

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm