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Grace [21]
3 years ago
6

A sound wave has a speed of

Physics
1 answer:
ELEN [110]3 years ago
7 0

Explanation:

use the formula

speed = frequency x wavelength

330 = frequency x 0.372

330 / 0.372 = frequency

887 Hz = frequency

hope this helps, please mark it brainliest

You might be interested in
The lowest note on a grand piano has a frequency of 28.1 Hz. It is a fixed string oscillating in its fundamental (longest wavele
tangare [24]

Answer:

2210.91 N

Explanation:

f = v/∧ = 1/2 √ T/ μ

where f= 28.1 Hz , T= tension ,

L= 2m

mass density =  μ = 350÷1000/2.00

= 0.175kg/m

from  f = 1/2L √ T/ μ

make T the subject of the formula

T= (f×2L)² ₓ  μ

T= (28.1×2×2)² ×0.175

T=12633.76×0.175

=2210.91 N

5 0
3 years ago
A car on a straight road starts from rest and accelerates at 1.0 meter per second? for 10. seconds.
Marysya12 [62]

Answer:

11 or 10 meters a second

Explanation:

pretty sure thats my explanation "pretty sure" like my confidence? mark me brainliest bad at spelling lol

7 0
2 years ago
A bucket of water of mass 14.7 is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.280m w
Snowcat [4.5K]

Answer:

Explanation:

Tension T in the rope will create torque in solid cylinder ( axle ). If α be angular acceleration

T R = 1/2 M R²α ( M is mass and R is radius of cylinder )

= 1/2 M R² x a / R ( a is linear acceleration )

T = Ma / 2

For downward motion of the bucket

mg - T = m a ( m is mass and a is linear acceleration of bucket downwards )

mg - Ma / 2  = ma

a = mg / ( M /2 + m )

Substituting the values

a = 14.7 x 9.8 / ( 5.8+ 14.7 )

= 7 m / s²

A )

T = Ma / 2

= 5.8 x 7

= 40.6 N

B ) v² = u² + 2 a h

= 2 x 7 x 10.3

v = 12 m /s

C )

v = u + a t

12 = 0 + 7 t

t = 1.7 s

6 0
3 years ago
The pressure at the bottom of a full barrel of water is Poriginal. Determine what happens to the pressure when the radius or hei
erastovalidia [21]

Answer:

C) Pnew=Poriginal/9

Explanation:

From the given conditions:

Original area, A_o=\pi.r_o\,^2

New area of the barrel bottom panel after increasing by a factor of 3:

r_{new}=3r_o

On increasing the radius the area now becomes:

A_{new}=\pi.(3r_o)^2

A_{new}=9\pi.(r_o)^2=9A_o

As we know that there will be no increase in the down-force since the height remains constant.

Pressure is given as:

P_{original}=\frac{Force}{A_o}

∴New pressure:

P_{new}=\frac{Force}{9A}

\Rightarrow P_{new}=\frac{P_{original}}{9}

4 0
3 years ago
An ideal spring obeys Hooke’s law, F~ = −k~x. A mass of 0.50 kilogram hung vertically from this spring stretches the spring 0.07
masha68 [24]

The spring constant is 66.7 N/m

Explanation:

First of all, we have to find the magnitude of the force acting on the spring. This is equal to the weight of the mass hanging on the spring, which is:

F=mg

where:

m = 0.50 kg is the mass of the object

g=10 m/s^2 is the acceleration of gravity

Substituting,

F=(0.50)(10)=5 N

Now we can use Hookes' law to find the constant of the spring:

F=-kx

where

F is the force applied

k is the spring constant

x is the stretching of the spring

Here we have:

F = 5 N

While the stretching is

x = 0.075 m

Therefore, ignoring the negative sign in the formula (which only tells us the direction), we find the spring constant:

k=\frac{F}{x}=\frac{5}{0.075}=66.7 N/m

#LearnwithBrainly

5 0
3 years ago
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