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AleksAgata [21]
3 years ago
14

Select the correct answer

Physics
1 answer:
VMariaS [17]3 years ago
4 0

Answer:

Mary's average speed is 0.42 meters/second.

Explanation:

Mary's total travel was 250 meters and the elapsed time was 10 minutes, equivalent to 600 seconds.

d = displacement = 80 meters + 125 meters + 45 meters = 250 meters                  

t = time = 10 minutes . 60 seconds/minute = 600 seconds

We can obtain now the average speed:

s = average speed = displacement / time = 250 meters / 600 seconds ≅ 0.4167 ≅ 0.42

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Which process is represented by the PV diagram shown below?
MrMuchimi

Answer: B. the isovolumetric process

Explanation:

In the graph given, the volume is constant throughout. It represents a constant volume process. Such processes are called the isovolumetric process or isochoric process.

<em>Hence, option B is the correct answer.</em>

Option A is incorrect because in an isobaric process, the pressure is constant.

Option C is incorrect because in an isothermal process, the temperature is constant.

Option D is incorrect because in an adiabatic process there is no heat transfer.

8 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
3 years ago
An Ohmic material is one in which its resistance does not depend on the voltage across it or the current within it. In that case
Alexandra [31]

Voltage

Mark brainliest

6 0
3 years ago
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
A 3858 N piano is to be pushed up a(n) 3.49 m frictionless plank that makes an angle of 31.6 ◦ with the horizontal. Calculate th
uranmaximum [27]
The work done will be equal to the potential energy of the piano at the final position

P.E=m.g.h

.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank

sinx=opposite side / hypotenuse
= h/L

then h=L.sinx=3.49×sin31.6°=0.638m

weight w=m.g
m=w/g=3858/10=385.8kg

Consider Gravity g=10m/s2

then P.E.=m.g.h=385.8kg×10×0.638=2461.404J

then Work W=P.E.=2451.404J
8 0
3 years ago
Read 2 more answers
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