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3 years ago

Acceleration is rate of change of A-PositionB-TimeC-VelocityD-Speed

Physics

2 answers:

Drupady [299]3 years ago

8 0

<h3>Answer:</h3>

[C] Velocity.

<h3>Explanation:</h3>

<u>As we know that</u>,

a = v - u/t

<u>where, a = acceleration, v = final velocity, u = initial velocity and t = time taken to travel</u>.

FrozenT [24]3 years ago

6 0

C-Velocity would be the answer

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alukav5142 [94]

Answer:

$8M(r\pi)^2$

Explanation:

First we convert 2 cycles/s to angular velocity knowing that each circle has an angle of 2π

$\omega = 2 * 2\pi = 4\pi rad/s$

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$I = mr^2 = Mr^2$

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3 years ago

Ice is put in a cooler to cool the contents. to speed up the cooling process, the ice can be?

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2 years ago

Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the

Ivan

Answer:

the magnetic field experienced by the electron is 0.0511 T

Explanation:

Given the data in the question;

Wavelength λ = 21 cm = 0.21 m

we know that Bohr magneton μ $_B$ is 9.27 × 10⁻²⁴ J/T

Plank's constant h is 6.626 × 10⁻³⁴ J.s

speed of light c = 3 × 10⁸ m/s

protein spin causes magnetic field in hydrogen atom.

Initial potential energy = -μ $_B$ B × cos0°

= -μ $_B$ B × 1

= -μ $_B$ B

Final potential energy = -μ $_B$ B × cos180°

= -μ $_B$ B × -1

= μ $_B$ B

so change in energy will be;

ΔE = μ $_B$ B - ( -μ $_B$ B )

ΔE = 2μ $_B$ B

now, difference in energy levels will be;

ΔE = hc/λ

2μ $_B$ B = hc/λ

2μ $_B$ Bλ = hc

B = hc / 2μ $_B$ λ

so we substitute

B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]

B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]

B = 510556326.09

B = 0.0511 T

Therefore, the magnetic field experienced by the electron is 0.0511 T

7 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

3 years ago

A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.

stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

$v^{2} = v^{2}_{0} + 2ax$

You plug into the equation to get:

$0 = 15^{2} + 2a(14)$

solve for a to get

-8.04 m/s2

3 0

3 years ago

Acceleration is rate of change of A-PositionB-TimeC-VelocityD-Speed​

Acceleration is rate of change of A-PositionB-TimeC-VelocityD-Speed