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fredd [130]
3 years ago
9

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

ound. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Physics
1 answer:
Hunter-Best [27]3 years ago
6 0

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

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The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How
Vlada [557]

Answer:

The value is  w =  7.54 \  m        

Explanation:

From the question we are told that

     The length of the crack is  a =  0.3 \  m

     The  frequency is  f =  30.0 \ kHz =  30 *10^{3} \  Hz

      The distance outside the cave that is being consider is  D =  100 \  m

      The speed of sound is v_s =  340 \  m/s

Generally the wavelength of the wave is mathematically represented as

        \lambda =  \frac{v}f}

=>     \lambda =  \frac{340 }{30*10^{3}}

=>     \lambda = 0.0113 \ m/s

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          y =  \frac{ n *  \lambda * D}{a}

=>     y =  \frac{ 1  *  0.0113 * 100}{0.3}

=>     y = 3.77 \  m

Generally the width of the sound beam is mathematically represented as

         w =  2 *  y

=>      w =  2 *  3.77

=>      w =  7.54 \  m        

4 0
3 years ago
The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

5 0
4 years ago
Read 2 more answers
If we push a piece of furniture to move it but we don't manage to move it, are forces acting on
zubka84 [21]

Answer:

Yes

Explanation:

Forces are pushing such as gravity, and forces from your own body and the couch are pushing against each other.

6 0
3 years ago
What is the median of 5 12 7 15 3 6 1
xxMikexx [17]

Answer:

6

Explanation:

The median is found by listing the numbers in order numerically and finding the one that is right in the middle. In this case you have 1, 3, 5, 6, 7, 12, 15 where the middle term is 6.

7 0
3 years ago
A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?
olchik [2.2K]
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
5 0
4 years ago
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