Anyone can help me out with this question ? Just number 2,

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

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: take less time to go the same distance

Explanation:

Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.

Answer:

B

Explanation:

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

Car B: Distance = (7 m/s) x (50 s)  =  350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

Answer:

[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]

Explanation:

Distance is equal to the product of speed and time.

[tex]d=s*t[/tex]

1. Car A

Car A has a speed of 8 meters per second and travels for 43 seconds.

[tex]s= 8 \ m/s \\t= 43 \ s[/tex]

Substitute the values into the formula.

[tex]d= 8 \ m/s *43 \ s[/tex]

Multiply and note that the seconds will cancel out.

[tex]d= 8 \ m*43= 344 \ m[/tex]

2. Car B

Car B has a speed of 7 meters per second and travels for 50 seconds.

[tex]s= 7 \ m/s \\t= 50 \ s[/tex]

Substitute the values in and multiply.

[tex]d= 7 \ m/s * 50 \ s[/tex]

[tex]d= 7 \ m * 50 = 350 \ m[/tex]

350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.

Answer:

Explanation:

The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .

1/2 k x² = mgx

.5 x k x .33² = m x 9.8 x .33

k / m = 59.4

frequency of oscillation =  [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]

= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]

= 1.22 per second .

Answer:

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

Explanation:

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:

the speed of electron is 5.021 x 10⁶ m/s

the speed of proton is 2733.91 m/s

Explanation:

Given;

magnitude of electric field, E = 554 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s

The force experienced by each particle is calculated as;

F = EQ

F = (554)(1.6 x 10⁻¹⁹)

F = 8.864 x 10⁻¹⁷ N

The speed of the particles are calculated as;

[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]

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

The time of the day she spent walking is equal to 3.70 hrs.

Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.

Power = work/time

P = W/t

Given, the energy spends part of her day walking, Ew = 280 W

The energy is spent by sitting in the class, Es = 100 W

The total energy spends, Et = 1.1 × 10⁷J

[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]

[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]

280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷

180 t = 0.24 × 10⁷

t =  0.24 × 10⁷/180 × 3600

t = 3.70 hr

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Answer:

I think the bucs are gonna win because Tom Brady is on their team and it's rigged

but maybe I'm just thinking negatively lol

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

Answer:

it will option option A hope it helps

Solution :

1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).

2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).

3. Non-selective scatter takes place when particle size in greater than the wavelength  (λ).

We have the sizes of different particles :

[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]

Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]

Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]

Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]

Wavelength           [tex]$ O_2 $[/tex]         Smoke particles    Cloud droplets     Rain droplets

                            [tex]$10^{-10} \ m$[/tex]        [tex]$ 3 \times 10^{-7} \ m$[/tex]           [tex]$ 2 \times 10^{-5} \ m$[/tex]              [tex]$ 3 \times 10^{-3} \ m$[/tex]

[tex]$5500 \times 10^{-4} \ m$[/tex]      Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$11 \times 10^{-6} \ m $[/tex]         Rayleigh    Rayleigh            Non-selective      Non-selective

[tex]$1600 \times 10^{-10} \ m $[/tex]    Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$10^{-2} \ m $[/tex]                 Rayleigh      Rayleigh               Rayleigh          Mie

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:

isnt heat transfer

Explanation:

sorry if im wrong

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:

[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:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

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:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

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:

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.

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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:

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:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

Answer:

49 Ns

Explanation:

Given data

Force= 14N

time = 3.5seconds

Applying the expression for impulse

P= Ft

substitute

P=14*3.5

P=49 Ns

Hence the impulse is 49 Ns