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3 years ago

Please help on this one?

Physics

2 answers:

cricket20 [7]3 years ago

6 0

C. 1.95 because from 10-20 hours the height grows 0.3. from 20-30 it grows 0.3 to get 35 you need to add half of 0.3 which is 0.15 so in 35 hours it would grow 1.95 cm

Maru [420]3 years ago

3 0

The correct answer for this problem is c

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Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

(a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
 $F=qE$
As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
 $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
 $F=qE$
as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

(d) does a magnetic field do so?
Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
 $F=qE$
So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
 $F=qvB \sin \theta$
is different from zero because v is different from zero.

6 0

3 years ago

Why does a green leaf look green in our eyes

mamaluj [8]

They look green because of the “special pair” of chlorophyll molecules.

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3 years ago

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For a substance to change from a gas to a liquid what must happen to its particles

baherus [9]

Answer: The answer is A!

Explanation:

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3 years ago

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A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell

Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0

3 years ago

1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin

omeli [17]

1. The bag's velocity immediately before hitting the ground.

Recall this kinematics equation:

Vf = Vi + aΔt

Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

Δt = 2s

Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

Vf = 19.62m/s

2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (bag is dropped starting from rest)

a = 9.81m/s² (acceleration due to gravity of the earth)

Δt = 2s

Plug in the values and solve for d:

d = 0×2 + 0.5×9.81×2²

d = 19.62m

3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

4.91Δt² + 2Δt - 19.6 = 0

Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

Vf = 19.66m/s

4 0

3 years ago