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solmaris [256]
3 years ago
5

Suppose you take two non-zero displacements represented by vectors A & B.The magnitude of A is 5 m and the magnitude of B is

6 m. Under what circumstances can you end up back at your starting point? What is the magnitude of the largest displacement you can end up from the starting point?
Suppose you take two non-zero displacements represented by vectors A & B which are perpendicular to each other. The magnitude of A is 5 m and the magnitude of B is 6 m. What is the component of vector B along the direction of vector A measured in m?
Physics
1 answer:
olga55 [171]3 years ago
8 0

Answer:

a. When the total displacement is -(A + B)

b. A + B = 1 m or -(A + B) = -11 m

c. 0 m

Explanation:

a. Under what circumstances can you end up back at your starting point?

If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B

b. What is the magnitude of the largest displacement you can end up from the starting point?

The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.

c. When A and B are perpendicular, what is the component of B in the direction of A?

Since A is perpendicular to B, the angle between A and B is 90°

So the component of B in A,s direction is Bcos90° = B × 0 = 0 m

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A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

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

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

3 0
3 years ago
Which is an IUPAC name for a covalent compound
Elena L [17]
In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:

1. The first element of the formula will use the normal name of the given element. for example: CO2 ( Carbon Dioxide), Carbon is the element name of the first element of the formula.

2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

3. Prefixes are used to indicate the number of atom of the elements in the compound. for example: mono- 1 atom, di- 2atoms, tri- 3 atoms and etc

4. Prefix "mono"is never used in naming the first element. For example: Carbon dioxide, there should be no monocarbon dioxide.
3 0
3 years ago
Read 2 more answers
You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0
4 years ago
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Usimov [2.4K]

Answer:

Stupid

Explanation:

Because there is never a answer when we are trying to find one

7 0
3 years ago
The magnetic field produced by a long straight current-carrying wire is
alexdok [17]

Answer:

proportional to the current in the wire and inversely proportional to the distance from the wire.

Explanation:

The magnetic field produced by a long, straight current-carrying wire is given by:

B=\frac{\mu_0 I}{2 \pi r}

where

\mu_0 is the vacuum permeability

I is the current intensity in the wire

r is the distance from the wire

From the formula, we notice that:

- The magnitude of the magnetic field is directly proportional to I, the current

- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r

Therefore, correct option is

proportional to the current in the wire and inversely proportional to the distance from the wire.

8 0
3 years ago
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