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mafiozo [28]
3 years ago
8

Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be

positive or negative, or zero, make sure you indicate the correct sign. Draw the vectors and see if your answer makes sense! v1=( -5, 7 , 2) ft v2=(3, 1, 2) ft
Physics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

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An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m
vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

V=E\times d

V=3000\ N/C\times 0.7\ m

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

6 0
3 years ago
A drawback to using nuclear fission is that it...
Westkost [7]

Answer:

D. creates radioactive waste.

Explanation:

Nuclear energy can create nuclear radioactive waste

8 0
1 year ago
1. What is the bond angle between the hydrogen atoms in an ammonia (NH3) molecule? 2. What is the bond angle between the oxygen
Klio2033 [76]

The answer is 107 degrees. The geometric shape for ammonia is Trigonal Pyramidal, even though its electron geometry is “Tetrahedral”. This is because ammonia has a lone pair of electrons that occupy its space like the other 3 hydrogens in the geometric structure.

The answer 180 degrees. This is because of the linear geometric structure of carbon dioxide. The oxygen atom is on either side of the carbon atom, each is bound by a double covalent bond. All the atoms are involved in the bond and there are no one pair electrons.

The answer is tetrahedral geometry. This is because all the 4 valence electrons of the carbon are involved in a bond with a hydrogen atom. The angles in a tetrahedral geometric arrangement, such as in methane, is 109.5 degrees, where the hydrogen atoms are as far apart, from each other, as possible .

7 0
3 years ago
Read 2 more answers
A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc
tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we have

d_i = 2.3 m = 230 cm

d_o = 25 cm

now plug in all values into the formula

-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}

f = 28 cm

now for power of lens

P = \frac{1}{f}

P = \frac{1}{0.28} = 3.5 D

so the power in dioptre is

P = 3.5 D

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3 years ago
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