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SVEN [57.7K]
2 years ago
14

A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela

tes u and wint
Physics
1 answer:
larisa86 [58]2 years ago
5 0

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

<span>Now we will write the equation for a Change in Potential energy ΔU,</span>

ΔU = Uf - Ui

= 0-m*g*h

<span>  ΔU = -m*g*h --Equation 2</span>

Now compare the both equation

<span>Wint = -ΔU</span>

we can rewrite the above equation

ΔU = -W.

<span>So our Answer is ΔU = -W. .</span>

<span> </span>

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Suppose an isolated box of volume 2V is divided into two equal compartments. An ideal gas occupies half of the container and the
SpyIntel [72]

Answer:

A. the internal energy stays the same

Explanation:

From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.

Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.

Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.

7 0
2 years ago
Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0
vichka [17]

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

7 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o
mariarad [96]

Answer:

S_i=\frac{9}{10} =0.9

Explanation:

Given:

  • volume of oil in the cylinder, V_o=60\ cm^2
  • volume of the oil level when the ice is immersed, V=90\ cm^3
  • the volume level of oil when the ice melted, V'=87\ cm^3

<u>Now, therefore the volume of ice:</u>

V_i=V-V_o

V_i=90-60

V_i=30\ cm^3

<u>Now the volume of water:</u>

V_w=V'-V_o

V_w=87-60

V_w=27\ cm^3

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

S_i=\frac{\rho_i}{\rho_w} .....................(1)

as we know that density is given as:

\rm \rho=\frac{mass}{volume}

now eq. (1)

S_i=\frac{m}{V_{i}}\div  \frac{m}{V_w}

where, m = mass of the water or the ice which remains constant in any phase

S_i=\frac{V_w}{V_i}

S_i=\frac{27}{30}

S_i=\frac{9}{10} =0.9

7 0
2 years ago
A straight wire carries a current of 3 A which is in the plane of this page, pointed toward the top of the page. A particle of c
Ilia_Sergeevich [38]

Answer:

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x 10^{-8} N

The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward

Explanation:

given information:

current, I = 3 A

q_{0} = +6.5 x 10^{-6} C

r = 0.05 m

v = 280 m/s

and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:

F = qvB sin θ

where

F = magnetic force (N)

q = electric charge (C)

v = velocity (m/s)

θ = the angle between the velocity and magnetic field

to find B we use

B = μ_{0}I/2πr

μ_{0} = 4π x 10^{-7} or 1.26 x 10^{-6} N/A^{2} , thus

B = 4π x 10^{-7} x 3 / 2π(0.05)

  = 1.2  x  10^{-5} T

Now, we can calculate the magnitude force

F = qvB sin θ

θ = 90°, because the speed and magnetic are perpendicular

F = 6.5 x 10^{-6} x 280  x 1.2 x  10^{-5} sin 90°

  = 2.18 x 10^{-8} N

Using the hand law, the magnetic direction is radially inward

8 0
2 years ago
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