Help plsss I need to answer this

Which of the following statements is accurate? A. Sound waves passing through the air will do so as transverse waves, which vibr

egoroff_w [7]

The answer is, C. the wavelength is measured in parallel to the direction of the wave, at any point, under the same repetition for any type of wave.

7 0

3 years ago

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A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of

BigorU [14]

<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>

6 0

3 years ago

A student throws a pumpkin up from 14.2 meters with a speed of 19.4m/s. what is the velocity

RSB [31]

.73 because velocity is determined by distance divide by speed brainliest if this helps

6 0

3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.

kherson [118]

Answer:

a) k = 120 N / m

, b) f = 0.851 Hz

, c) v = 1,069 m / s

, d) x = 0

, e) a = 5.71 m / s²

, f) x = 0.200 m

, g) Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

F = k x

k = F / x

k = 24.0 / 0.200

k = 120 N / m

b) the angular velocity of the simple harmonic movement is

w = √ k / m

w = √ (120 / 4.2)

w = 5,345 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 5.345 / 2π

f = 0.851 Hz

c) the equation that describes the movement is

x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

x = 0.2 cos wt

Speed is

v = dx / dt

v = -A w sin wt

The speed is maximum for sin wt = ±1

v = A w

v = 0.200 5.345

v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

x = A cos wt = 0

x = 0

e) the acceleration is

a = d²x / dt² = dv / dt

a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

a = A w²

a = 0.2 5.345

a = 5.71 m / s²

f) the position for this acceleration is

x = A cos wt

x = A

x = 0.200 m

g) Mechanical energy is

Em = ½ k A²

Em = ½ 120 0.2²

Em = 2.4 J

h) the position is

x = 1/3 A

Let's calculate the time to reach this point

x = A cos wt

1/3 A = A cos 5.345t

t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

a = -1.91 m / s²

5 0

3 years ago

Help plsss I need to answer this​

Help plsss I need to answer this