Ask question

Ask question

All categories

jek_recluse [69]

3 years ago

5

Please help! Will give brainliest.

1 answer:

I am Lyosha [343]3 years ago

9 0

<span>The correct frequency when you tune a guitar is when you hear the right tune in your own hearing and standard. The measure frequency of a guitar string is when you measure the tune of the string correctly. This is not the same because manual tuning is affected by many factors.</span>

You might be interested in

A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res

Luden [163]

The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW

8 0

2 years ago

In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m

ahrayia [7]

Answer:

The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

Explanation:

Given;

density of the liquid, ρ = 1500 kg/m³

frequency of the wave, F = 410 Hz

wavelength of the sound, λ = 7.80 m

The speed of the wave is calculated as;

v = Fλ

v = 410 x 7.8

v = 3,198 m/s

The bulk modulus of the liquid is calculated as;

$V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2$

Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

5 0

2 years ago

The open ocean, sea floor, and coral reefs are all examples of habitats with high rates of primary production.

lisov135 [29]

the answer you are seeking is false

7 0

2 years ago

Read 2 more answers

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance $x$ m by the time of collision, then train B would have traveled $(2881 - x)$ m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

${Distance_{B} = 2881 - x$ m

${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

$x$ ≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0

2 years ago

Does latitude has an effect on weight?

lord [1]

Answer:

I think so but i could be wrong..

Explanation:

5 0

2 years ago

Read 2 more answers