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inn [45]
3 years ago
13

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

After traveling in a straight line for 2.13 km, they stop and discover that they have been traveling 25.9 ° north of east, because their guide misread his compass. What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?
Physics
2 answers:
Talja [164]3 years ago
5 0

This is a vector subraction problem.

maxonik [38]3 years ago
3 0

Vector  subraction problem


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A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius
vlabodo [156]

Answer:

<u>r < a:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}

<u>r = a:</u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>a < r < b:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>r = b:</u>

E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

<u>r < c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

Explanation:

Gauss' Law will be applied to each region to find the E-field.

\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}

Applying Gauss' Law:

E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u />E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}<u />

<u>At r = b:</u>

<u />E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}<u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>At r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

3 0
3 years ago
Without fluid friction, all objects accelerate at?
lys-0071 [83]

Answer:

Acceleration of gravity=9.8m/s/s

Explanation:

Newton's Second Law-acceleration is proportional to the net force acting on an object.

All objects usually free fall at the same acceleration of 9.8m/s/s-this regardless of their mass. This acceleration is known as acceleration of gravity.

7 0
3 years ago
Snell's Law: Light traveling through water comes to a glass surface at an angle of incidence of
Lerok [7]

Answer:

1. The best definition of refraction is ____.

a. passing through a boundary

b. bouncing off a boundary

c. changing speed at a boundary

d. changing direction when crossing a boundary

 

Answer: D

Bouncing off a boundary (choice b) is reflection. Refraction involves passing through a boundary (choice a) and changing speed (choice c); however, a light ray can exhibit both of these behaviors without undergoing refraction (for instance, if it approaches the boundary along the normal). Refraction of light must involve a change in direction; the path must be altered at the boundary.

6 0
3 years ago
What is the pressure exerted by an elephant who weighs 2500N and whose feet cover an area of 7.5m2?
qaws [65]

Answer:

0.00339905404

Explanation:

kg: 0.00339905404

pascals: 333.333333

pounds: 0.0483459126

8 0
3 years ago
If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.
Schach [20]

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

6 0
3 years ago
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