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ahrayia [7]
2 years ago
12

I need help w the question

Physics
1 answer:
Eddi Din [679]2 years ago
3 0
C- Global Conveyor Belt, hope this helped :)
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Please need help on this
lara [203]

#6 - Lower frequencies correspond to <em>longer</em> wavelengths.

#7 - To change the pitch of a sound, you have to change its <em>frequency</em>.

6 0
4 years ago
Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring
lana [24]

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x  ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m   ⇒ ω = √ 95 N /m  /  1.5 kg  

ω = 7.96  rad /s

A = v / ω   ⇒  A = 6 m /s   /   7.96 rad / s

A = 0.754

d = cos * φ    ⇒  φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ )    ⇒  y (t) = 0.754 * cos ( 7.96 t - 69.52)

4 0
3 years ago
A car is traveling on the highway at 28 m/s The driver sees a tree in the middle of the road and slams on the brakes. The car co
wel

Answer:what are the answer options?

Explanation

3 0
3 years ago
3. What factors can contribute to an infant's ability to develop sensory competencies?
marta [7]

Answer:

Explanation:

Here's how to stimulate your baby's senses in age-appropriate ways:

Decorate the nursery with colorful, bold patterns. ...

Play hand games. ...

Use "baby talk." Studies have shown that the sing-song, fluctuating tones mothers use to talk to their babies are important for language development. ...

Hold your baby as much as possible.

7 0
3 years ago
The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample
Bezzdna [24]

Answer:

4589.05 year

Explanation:

The relation of activity is given by A=A_0e^{\lambda t} we have given \frac{A}{A_0}=0.574

Half life T=5730\ years

We know that half life T=\frac{0.693}{\lambda }

\lambda =\frac{0.693}{5730}=1.2094\times 10^{-4}\ per\ year

So 0.574=e^{-1.2094\times 10^{-4}t}

e^{1.2094\times 10^{-4}t}=\frac{1}{0.574}=1.742

{1.2094\times 10^{-4}t}=ln1.742=0.555

t=\frac{0.555}{1.2094\times 10_{-4}}=4589.05\ year

7 0
3 years ago
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