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Serggg [28]
3 years ago
5

I need help on 1 through 4.

Physics
1 answer:
VashaNatasha [74]3 years ago
8 0
Okay what’s the question, I’ll help
You might be interested in
What is the 4th dimension? I have heard that it's time, it's a from of saying characteristic. I don't know I need help on this,
sashaice [31]

If you want to tell a friend about a fish you caught or a tree you cut down,
you're going to tell him WHERE you were ... its position in space, 3 numbers,
'x', 'y', and 'z' ... and also WHEN you were ... its position in time, one more
number. 

Dimensions are numbers used to describe the location of a point, and the
difference in location between two points.  With four numbers, you can exactly
describe the location of anything, and its distance from any other thing, in
space and time.


3 0
3 years ago
A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) F
Romashka-Z-Leto [24]

Answer:

Explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

=  (-kx + kx³/a²) / m

If U be potential function

- dU / dx =  (-kx + kx³/a²) / m

U(x)  = ∫  (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx²  = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U =  k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x=  ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max)  =  k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.

4 0
4 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.
zloy xaker [14]

Answer:

P=28.085\,hp

Explanation:

Given that:

  • mass of 1 skier, m=80kg
  • inclination of hill, \theta=25^{\circ}
  • length of inclined slope, l=220m
  • time taken to reach the top of hill, t=2.3 min= 138 s
  • coefficient of friction, \mu=0.15

<em>Now, force normal to the inclined plane:</em>

F_N=m.g.cos\theta

F_N=80\times 9.8\times cos25^{\circ}

F_N=710.54\,N

<em>Frictional force:</em>

f=\mu.F_N

f=0.15\times 710.54

f=106.58\,N

<em>The component of weight along the inclined plane:</em>

W_l=m.g.sin\theta

W_l=80\times 9.8\times sin25^{\circ}

W_l=331.33\,N

<em>Now the total force required along the inclination to move at the top of hill:</em>

F=f+W_l

F=106.58+331.33

F=437.91\,N

<em>Hence the work done:</em>

W=F.l

W=437.91\times 220

W=96340.80\,J

<em>Now power:</em>

P=\frac{W}{t}

P=\frac{96340.80}{138}

P=698.12\,W

<u>So, power required for 30 such bodies:</u>

P=30\times 698.12

P=20943.65\,W

P=\frac{20943.65}{745.7}

P=28.085\,hp

8 0
3 years ago
A rain drop of radius 0.3mm has terminal velocity 1m/s in air. The viscosity of air is 18X10-5 poise, then viscous force is​
dimaraw [331]

Answer:

F = 1.0178 × 10^(-2) dyne

Explanation:

From stokes law, the viscous force also known as drag force on rain drop is given by the formula;

F = 6πηrv

Where;

η is viscosity

r is radius

v is velocity

We are given;

η = 18 × 10^(-5) poise

r = 0.3 mm = 0.03 cm

v = 1 m/s = 100 cm/s

Thus;

F = 6π × 18 × 10^(-5) × 0.03 × 100

F = 1.0178 × 10^(-2) dyne

8 0
3 years ago
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