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Travka [436]
3 years ago
5

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

nd its approximate location at time t = 1.02.
Physics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

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A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

6 0
2 years ago
Find the force required to do 25 joule work when the force causes a displacement of 0.5 m​
Eduardwww [97]

Answer:

<h2>50 N</h2>

Explanation:

The force required can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{25}{0.5}  \\

We have the final answer as

<h3>50 N</h3>

Hope this helps you

6 0
3 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
3 years ago
Air is composed primarly of which of the following gases?
romanna [79]
Nitrogen and carbon dioxide
6 0
4 years ago
In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s
marshall27 [118]

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

W_N=\Delta K\\\\

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)    (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F.  Thus you can calculate experimentally both sides of the equation.

8 0
3 years ago
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