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1 year ago

I think it’s gas but I’m not sure.

Physics

1 answer:

malfutka [58]1 year ago

4 0

Answer:

Liquid

Explanation:

The substance will be liquid because the gas form will be 100° Celsius. So, it is not gas instead it is liquid. Solid is not the answer because it has to be 0° C

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A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0

3 years ago

Signal words most often used in a chronological essay are

Nookie1986 [14]

Chronological essays by the definition of a chronological meaning in order. There is an order in a specific writing. Like a history write up from a certain happening years ago. It is different from procedural essays because these are essays who are giving instructions of certain set up to guide the person accordingly in doing something to make it more accurate. Like recipes, instructions in playing, etc. Example words that are used in chronological essays are first, second, third, fourth, fifth, next, after, then, lastly, finally, consequently, in addition, thus, therefore, however, etc.

4 0

3 years ago

Read 2 more answers

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth d and the other (lets call it l) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section A at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed v of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where q is the water flow (1 cubic foot per minute).

7 0

3 years ago

In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

$\omega=\dfrac{1}{\sqrt{CL}}$

By putting the values

$\omega=\dfrac{1}{\sqrt{CL}}$

$\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}$

ω = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33 A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0

3 years ago