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Anika [276]
3 years ago
8

Heyy,Can you help me to do this physics problem:

Physics
1 answer:
Kobotan [32]3 years ago
5 0

Answer:

Average Speed = (Total distance)/(Total Time) = (1800 + 1800)/(30 + 45) = 3600/75 = 48 km/hr. Hence the correct option here is C) 48 km/hr. So to find the average speed never use the formula of the averages but try to find the total distance covered and the total time taken.

Explanation:

It is equals to distance divided time

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A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. Wha
pychu [463]

A. 409 Hz

The fundamental frequency of a string is given by:

f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

f_n = nf_1

where f_1 is the fundamental frequency.

Here in this case

f_1 = 409 Hz

A person is capable to hear frequencies up to

f = 1.00 \cdot 10^4 Hz

So the highest harmonics that can be heard by a human can be found as follows:

f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24

8 0
3 years ago
If you had 6,000 milligrams of rock to move how many kilograms is that?
True [87]

Answer:

0.006

Explanation:

100mg= 0.0001kgghsjsslslkdn d

5 0
3 years ago
What is the velocity of sound in air and vacuum​
guapka [62]

Answer:

approximately 2.99 × 10⁸ m/s

8 0
3 years ago
Read 2 more answers
HELP + extra pts // Two 10-m high diving platforms are at opposing ends of a 30-m pool. How fast must two clowns run straight of
ZanzabumX [31]

Explanation:

The clowns need to leave the diving boards with enough horizontal velocity such that each travels 15 m (half the width of the pool) in the same time that they fall (vertically) the 10-m from the top of the diving board.

We'll assume no force acts on the clowns horizontally to slow them down while they are in flight. And we'll assume that only gravity acts on the clowns vertically.

We can treat the horizontal and vertical components separately. This will help simplify the problem.

Let's start with the vertical displacement. Let's say the clown is dropped from a height of 10-m. How long would it take them to fall that distance?

Using our equations of motion (with constant, linear acceleration), we can solve for this time. d = vt + \frac{1}{2} at^2

Where d is the distance travelled, v is the initial velocity, a is acceleration, and t is time.

If the clown is dropped, they have an initial velocity of 0 (zero). We assumed only gravity acts on the clown, so acceleration equals the gravitational acceleration on earth. a = 9.8m/s^2

The distance the clown travels is 10-m from the diving board to the surface of the water.

Let's solve. 10 = 0*t + \frac{1}{2} (9.8) t^2 \rightarrow 20/9.8 = t^2 \rightarrow t = \sqrt{20/9.8}

Now that we know time, we can calculate how fast the clown needs to be running when they leap from the diving board to cover a distance of 15-m (Remember, half the width of the pool.)

Using our equations of motion, we know that d = vt + 0.5at^2

We assumed no forces act horizontally on the clown, therefore a = 0. We just need to solve for v. Substituting in the time we just solved for, we get something like this. 15 = v \sqrt{20/9.8}

I'll leave it to you to solve this equation.

6 0
3 years ago
In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sec
Mrac [35]

Answer:

F_s=1075.9493\ N

Explanation:

Given:

  • area of piston on the smaller side of hydraulic lift, a_s=0.075\ m^2
  • area of piston on the larger side of hydraulic lift, a_l=0.237\ m^2
  • Weight of the engine on the larger side, W_l=3400\ N

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

P_s=P_l

\frac{F_s}{a_s}=\frac{W_l}{a_l}

\frac{F_s}{0.075} =\frac{3400}{0.237}

F_s=1075.9493\ N is the required effort force.

5 0
4 years ago
Read 2 more answers
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