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KIM [24]
3 years ago
7

Let the four horizontal compass directions north, east, south, and west be represented by units vectors n^, e^, s^, and w^, resp

ectively. Vertically up and down are represented as u^ and d^. Let us also identify unit vectors that are halfway between these directions, such as (ne)^ for northeast. Rank the magnitudes of the following cross products from the largest to the smallest. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)
(a) n^ x n^(b) w^ x (ne)^(c) u^ x (ne)^(d) n^ x (nw)^(e) n^ x e^
Physics
1 answer:
Marrrta [24]3 years ago
5 0

Answer:

c = e > b = d > a

Explanation:

Given vectors are all unit vectors, therefore they have a magnitude of 1

<h3>Let a, b be two vectors and magnitude of cross product of these two vectors is (magnitude of a) × (magnitude of b) × (sine of angle between these two vectors)</h3>

As all are unit vectors their magnitude is 1 and therefore in this case the cross product between any two vectors depends on the sine of angle between those two vectors

In option a as both the vectors are same, the angle between them will be zero and sin0° will also be 0

In option b angle between those two vectors is 135° and sin135° is 1 ÷ √2

In option c angle between those two vectors is 90° and sin90° is 1

In option d angle between those two vectors is 45° and sin45° is 1 ÷ √2

In option e angle between those two vectors is 90° and sin90° is 1

So by comparison of magnitudes of cross products in each option, the order will be  c = e > b = d > a

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Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f
o-na [289]

Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m

μo = 4 π × 10^-7 Tm/A

The formula for the force per unit length between the two wires is given by

F = μo / 4π x (2 i1 x i2) / d

3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082

I = 0.0362 A = 36.22 mA

4 0
3 years ago
A speed boat travels from the dock to the first buoy a distance of 20 meters in 18 seconds it began the trip at a speed of 0 m/s
Lunna [17]

Answer:

1.11 m/s

Explanation:

The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

v=\frac{d}{t}

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

v=\frac{20 m}{18 s}=1.11 m/s

6 0
2 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
HELP ME DO THIS QUESTION PLEASE ​
raketka [301]

Answer:

i dont know

Explanation:

but what you can do is ask you mom or dad to get you a tutor to help you

7 0
3 years ago
The gravitational force between the charged constituents of the atom is negligible compared with the electric force between them
katrin2010 [14]

Answer: The initial force is reduced a factor 1/4 when the separation between charge is doubled

Explanation: As it well known the electric force between two charges is given by:

Finitial=k*q1*q2/d^2 where d is the distance between charges and k is a constant

if the distance is doubled this means 2*dinitial thus the new force is equal to F initial* 1/4

6 0
2 years ago
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