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If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s

larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0

3 years ago

Read 2 more answers

An engine flywheel initially rotates counterclockwise at 6.55 rotations/s. Then, during 20.9 s, its rotation rate changes to 2.1

11Alexandr11 [23.1K]

Answer:

The average angular acceleration is -2.628 rad/s²

Explanation:

Counterclockwise = positive

Clockwise = -negative

Given;

initial rotation of the flywheel, θ₁ = 6.55 rotation/s

final rotation of the flywheel, θ₂ = - 2.19 rotation/s

The average angular acceleration is given by;

$\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2$

Therefore, the average angular acceleration is -2.628 rad/s²

7 0

3 years ago

A baseball with a mass of 0.15 kilograms collides with a bat at a speed of 40 meters/second. The duration of the collision is 8.

Vedmedyk [2.9K]

Answer: 1687.5 N

Explanation:

From the second law of motion given by Newton, Force is the rate change of momentum.

$F = \frac{dp}{dt}=\frac {m dv}{dt} = \frac{m (v_f-v_i)}{dt}$

Mass of the baseball, m = 0.15 kg

Initial velocity, $v_i=-40 m/s$ (negative because direction of initial velocity is opposite to the final velocity)

Final velocity, $v_f=50 m/s$

The duration of collision, $dt= 8.0 \times 10^{-3} s$

Force, $F = \frac{0.15 kg (50-(-40) m/s)}{8.0 \times 10^{-3} s}=1687.5 N$

Hence, the value of force is 1687.5 N.

8 0

3 years ago

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid

Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0

3 years ago