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3 years ago

Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER

Physics

1 answer:

lukranit [14]3 years ago

3 0

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

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An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$

Therefore, the particle's charge-to-mass ratio is $958.1\times10^5C/kg$

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0

4 years ago

The Earth’s diameter is about 8,000 miles; our Moon’s diameter is about 2,000 miles; how

Vitek1552 [10]

Three moons can fit inside the volume of the sun.

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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6 0

2 years ago

¿Qué características debe tener el movimiento de una persona para que el valor del desplazamiento sea igual al de la distancia r

SIZIF [17.4K]

Explanation:

What characteristics must the movement of a person have so that the value of the displacement is equal to the distance traveled?

Displacement is equal to the shortest path covered by an object. It is given by the difference of final position and the initial position.

Distance is equal to the total path covered by an object during the journey.

When an object moves in a straight line path, in this case, the displacement is equal to the distance traveled.

3 0

3 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

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5 0

2 years ago

g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

4 years ago