Please I need help with this :(

Since vector a = 3i - 2j - k and vector b = i + 4j + k, the unit vector, n, normal to the plane is (7i - j + 7k)/3√11 where a, b and n, in this order, form a right-handed system.​

The normal vector to the plane is c = a × b

Since a = 3i - 2j - k and b = i + 4j + k,

c = a × b

c = (3i - 2j - k) × (i + 4j + k)

= 3i × i + (- 2j) × i + (-k) × i + 3i × 4j + (- 2j) × 4j + (-k) × 4j + 3i × k + (- 2j) × k + (-k) × k

= 3i × i - 2j × i - k × i + 12i × j - 8j × j - 4k × 4j + 3i × k - 2j × k - k × k

Given that

c = 3(0) - 2(-k) - (-j) + 12k - 8(0) - 16(-i) + 3(-j) - 2i - 0

c = 0 + 2k + j + 12k - 0 + 16i - 3j - 2i

c = 16i - 2i + j - 3j + 2k + 12k

c = 14i -2j + 14k

The unit vector normal to the plane n = c/|c| where |c| is the magnitude of c = √(14² + (-2)² + 14²) = √(196 + 4 + 196)

= √396

= √(36 × 11)

= 6√11

So, n = c/|c|

= (14i -2j + 14k)/6√11

= 2(7i - j + 7k)/6√11

= (7i - j + 7k)/3√11

So, the unit vector normal to the plane is (7i - j + 7k)/3√11

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Answer:

3

5

4

Explanation:

x = (8=(9+9)

(9+9) = 4, 5, 3

Newtons first law of motion.

The potential difference between points a and b is zero.

The total emf in the circuit is the sum of all the emf in the circuit.

emf(total) = 1.5 + 1.5 = 3.0 V

The potential difference between two points, a and b is calculated as follows;

V(ab) = Va - Vb

V(ab) = 1.5 - 1.5

V(ab) = 0

Thus, the potential difference between points a and b is zero.

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Speed of the rocket under uniform circular motion is 4.97 × [tex]10^7[/tex] m/s.

A fictitious force that moves in a circle and is directed away from the center of the circle is called centrifugal force. When measurements are taken in an inertial frame of reference, the force does not exist. It only becomes relevant when we switch from a ground/inertial reference frame to a rotating reference frame.

As the rocket is in uniform circular motion, the gravitational force due to earth and centrifugal force must be equal in magnitude.

The magnitude of gravitational force due to earth, [tex]F_G = \frac{G M m}{R^2}[/tex]

And, the magnitude of centrifugal force, [tex]F_C = \frac{mv^2}{R}[/tex]

Where,

G = universal gravitational constant =   6.67 * 10-11 N (m/kg)².

M = the mass of the earth = 5.97 x [tex]10^{24[/tex] Kg.

m =  the mass of the rocket = 0.5 x [tex]10^6[/tex]kg.

R = the orbit of the rocket = [tex]{8*10^6}[/tex]m.

For uniform circular motion,

[tex]\frac{G M m}{R^2}[/tex] = [tex]\frac{mv^2}{R}[/tex]

⇒ v² = [tex]\frac{GM}{R}[/tex]

⇒v² =[tex]\frac{6.67 * 10^{-11 } *5.97 x 10^{24 } }{8*10^6}[/tex]

⇒ v = 4.97 × [tex]10^7[/tex] m/s.

Hence, the speed of the rocket is 4.97 × [tex]10^7[/tex] m/s.

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Answer:

a) v = 25.54 ft/s

b) Xmax = 0.4180

c) Xmax = 0.0048 ft

Explanation:

a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;

to calculate the distance travelled

S = vt

given an expression of wavelength

l = vt

l = 2πv / ω

ω = 2πv / l

equation for counter of the road

y = Y sin ωt

= Y sin ( (2πv / l)t)

next we calculate the angular frequency

ω = √ ( k/m)

=  √ ( g / Sst )

= √( 32.2 / ( 1.5/12))

= 16.05 rad/s

Now we calculate the speed v at which amplitude of the boat will be maximum

ω = 2πv / l

16.05 = (2 × π × v ) / 10

160.5 = 2πv

v = 160.5 / 2π

v = 25.54 ft/s

b)

we calculate the maximum amplitude using the following expression;

X/Y =  [ √( 1 + ( 2Sr)²)] / [ √(( 1  - r²)² +  ( 2Sr)²))]

Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1  - 1²)² +  ( 2×0.05×r)²))]

Xmax / 0.0416 = 1.004987 / 0.1

Xmax 0.1 = 0.0418

Xmax = 0.0418 / 0.1

Xmax = 0.4180

c)

we convert speed of the boat from mph velocity m/s

v = 55 mph × 1.46667ft/s  / 1mph

v = 80.66 ft/s

next we calculate angular velocity

ω = 2πv / l

= 2π × 80.66 / 10

= 50.68 rad/s

next is the frequency ratio

r = 50.68 / 16.05

= 3.16

finally we find the amplitude of the boat

X/Y =  [ √( 1 + ( 2Sr)²)] / [ √(( 1  - r²)² +  ( 2Sr)²))]

Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1  - 3.16²)² +  ( 2×0.05×3.16)²))]

Xmax / 0.0416  = √1.0998 / √ ( 80.74 + 0.0998)

Xmax / 0.0416 = 1.0487 / 8.9910

Xmax 8.9910 = 0.0436

Xmax = 0.0436 / 8.9910

Xmax = 0.0048 ft

Answer:

As current increases, the magnetic field strength increases.

So you're correct! :)

As the strength of the magnetic field increases, the current also increases.

From classical electromagnetism, we know that current is induced on conductor owing to relative motion between the conductor and the magnet. This is the principle upon which electromagnetic induction is based.

In the experiment, the student will find that the strength of the magnetic field is directly proportional to the induced current. Hence, as the strength of the magnetic field increases, the current also increases.

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Answer:

its D i just did it on the app

A triangle is the symbol used to indicate a fulcrum in a lever, a triangle has three sides and three vertices.

A fulcrum is a pivot that has a base and a point that serves as a balance for  a lever.

Basically a lever is a simple machine that is a beam with two sides pivoted to fulcrum.

A lever has Load, Effort and the last part is fulcrum

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Answer:

A 100V battery is connected to two oppositely charged plates that are 10cm apart.

i. what is the magnitude of the electric field?

ii. Calculate the electric force exerted on a +200uC point charge.

Explanation:

A example of a scientific way of thinking is making observations, forming questions, making hypotheses, doing an experiment, analyzing the data, and forming a conclusion.

Answer:

a)  The electric field at point P has no component in the x and z directions.

b) dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)Ey = = 2kλd / y₀( d² + 4y₀²))^1/2

d) Ey = 411.84 N/C

Explanation:

a)

from the uploaded image;

the electric field will only be in y-direction.

Therefore the electric field at P have no component in the  x and z directions.

b)

dEy = dEsin θ

dE = kdq / r²

from figure

sinθ = y₀ / √( x² + y₀²)

r = √( x² + y₀²)

dq = λdx

dEy = kdq sinθ / r²

dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)

the net electric field at p is ,

Ey = ∫^d/2_-d/2  kλdxy₀ / (√( x² + y₀²))^3/2

= kλy₀ ∫^d/2_-d/2  dx / (√( x² + y₀²))^3/2

= 2kλd / y₀( d² + 4y₀²))^1/2

d)

let y₀ = 15 × 10⁻²m,  λ = 3.5 nC/m , d = 1.5m

Ey = 2kλd / y₀( d² + 4y₀²))^1/2

Ey = 2(9×10⁹ N.m²/C²)(3.5×10⁻⁹m)(1.5m) / 0.15×(1.5)² + 4(0.15)²))^1/2

Ey = 411.84 N/C

Answer:

Use equation for kinetic energy: Ek=mV²/2

m=700 kg

V=10m/s

Ek=700kg*100m²7s²/2

Ek=35000 J=35kJ

Explanation:

Hope this helps you

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Answer:

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed

b) Student 2 is right that for the car to reach the truck it must have some relationship,  

c)   t = √d/a

Explanation:

In this exercise you are asked to analyze the movement in a mention using kinematics.

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed he can use the equation

         v = d / t

b) Student 2 is right that for the car to reach the truck it must have some relationship, which is given by

        v = v₀ + a t

        x = v₀ t + ½ a t²

c)    let's use the equations

      v = d / t

       v = v₀ + at

      d / t = vo + at

If we can assume that the car starts from rest, so v₀ = 0

       d / t = a t

        t² = d / a

        t = √d / a

if the speed of the car is not zero the result is a little more complicated

      d = vo t + a t²

      at² + vo t - d = 0

let's solve the quadratic equation

      t = [-v₀ ± √ (v₀² + 4a d)] / 2a

Answer:

The amount of potential energy is increasing when they jump on the board. The amount of potential energy is decreasing and the amount of kinetic energy is increasing when they are falling towards the water. The point of maximum kinetic energy is right before they splash into the water. The point of maximum potential energy is when they jump the highest into the air.

Explanation:

The points of maximum potential energy when they jump highest into the air, their potential energy is at its highest.

Potential energy develops in systems where the configuration, or relative position of the pieces, determines the amount of the forces they exert on one another. Potential energy directly proportional to height of object higher the object higher the potential energy with respect to same reference point.

When they jump onto the board, potential energy is building. When they are descending toward the sea, their potential energy is dwindling and their kinetic energy is rising. Just before they splash into the water is when they have the most kinetic energy. When they jump highest into the air, their potential energy is at its highest.

Potential energy is highest at its highest point of jump.

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Answer:

380 N

Explanation:

Forces: gravity mg (downward), Normal force N (upward)

Acceleration: Note velocity is up, but slowing down, so acceleration is opposite to velocity, or downward

Newton's 2nd law:

mg - N = ma

N = m(g - a) = 100(9.8 - 6) = 380 N

Answer:

Sorry, don't understand the question

Explanation:

Answer:

directly proportional to the square of its speed.

Explanation:

;)

Answer:

A

D

A

Is all the information here?

If so than I do not know

The first 3 are correct though

Explanation:

Answer:

42.87

Explanation:

Google

OR

420/9.8 which equals 42.87.

Answer:

P1 = P2      pressure is uniform

F1 / A1 = F2 / A2

F1 = F2 (A1 / A2) = 2,400 N * (2.87 / 314) = 21.9 N

The Kinetic energy of the bowling ball is 4000 J while the momentum of the ball is 200 kg m/s

Kinetic energy, KE is the energy exerted by a body in motion. This is represented mathematically as half the product of mass, m and velocity, v squared.

Kinetic energy

KE =  1 / 2 m v^2

m = 5 kg

v = 40 m/s

KE = 0.5 * 5 *40 ^2

KE = 4000 J

Momentum, M

M = mass * velocity

M = 5 * 40

M = 200 kg m/s

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Answer:m

R

​

ν

R

​

+m

L

​

ν

L

​

=0   ⇒    (0.500kg)ν

R

​

+(1.00kg)(−1.20m/s)=0

which yields ν

R

​

=2.40m/s . Thus , Δx=ν

R

​

t=(2.40m/s)(0.800s)=1.92m .

(b) Now we have m

R

​

ν

R

​

+m

L

​

(ν

R

​

−1.20m/s)=0 which yields

        ν

R

​

=

m

L

​

+m

R

​

(1.2m/s)m

L

​

​

=

1.00kg+0.500kg

(1.20m/s)(1.00kg)

​

=0.800m/s .

Consequently , Δx=ν

R

​

t=0.640m .

Explanation:

Answer:

67.9 Ω

Explanation:

L = 60 mH = 60 x 10⁻³ H = 0.060 H

f = 180 Hz, ω = 2πf = 1131 rad/s

reactance = ωL = 1131 x 0.060 = 67.9 Ω

Answer:

q/m = 2177.4 C/kg

Explanation:

We are given;

Initial speed v_o = 5 × 10³ m/s = 5000 m/s

Now, time of travel in electric field is given by;

t_1 = D_1/v_o

Also, deflection down is given by;

d_1 = ½at²

Now,we know that in electric field;

F = ma = qE

Thus, a = qE/m

So;

d_1 = ½ × (qE/m) × (D_1/v_o)²

Velocity gained is;

V_y = (a × t_1) = (qE/m) × (D_1/v_o)

Now, time of flight out of field is given by;

t_2 = D_2/v_o

The deflection due to this is;

d_2 = V_y × t_2

Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)

d_2 = (qE/m) × (D_1•D_2/(v_o)²)

Total deflection down is;

d = d_1 + d_2

d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]

d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]

Making q/m the subject, we have;

q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]

We have;

E = 800 N/C

d = 1.25 cm = 0.0125 m

D_1 = 26.0 cm = 0.26 m

D_2 = 56 cm = 0.56 m

Thus;

q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]

q/m = 312500/143.52

q/m = 2177.4 C/kg

The ratio of the charge to mass of the object will be equal to:

[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]

what is electric field?

The electric field is defined as the whenever an atom carries a charge whether positive or negative the their influence of force is spread around the charge at a particular distance this effect of force is called as the electric field.

We are given;

Initial speed  [tex]V_o[/tex] = 5 × 10³ m/s = 5000 m/s

Now, time of travel in electric field is given by;

[tex]t_1=\dfrac{D_1}{v_o}[/tex]

Also, deflection down is given by;

[tex]d_1=\dfrac{1}{2}at^2[/tex]

Now,we know that in electric field;

F = ma = qE

Thus,  [tex]a=\dfrac{qE}{m}[/tex]

So;

[tex]d_1=\dfrac{1}{2}\times(\dfrac{qE}{m})(\dfrac{D_1}{v_o})^2[/tex]

Velocity gained is;

[tex]V_y=(a\timest_1)=(\dfrac{qE}{m})\times (\dfrac{D_1}{v_o})[/tex]

Now, time of flight out of field is given by;

[tex]t_2=\dfrac{D_2}{v_o}[/tex]

The deflection due to this is;

[tex]d_2=V_y\times t_2[/tex]

Thus, [tex]d_2=(\dfrac{qE}{m}\times (\dfrac{D_1}{v_o})\times(\dfrac{D_2}{v_o})[/tex]

[tex]d_2=(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{(v_o)^2}[/tex]

Total deflection down is;

[tex]d=d_1+d_2[/tex]

[tex]d=[\dfrac{1}{2}\times(\dfrac {qE}{m}) \times (\dfrac{D_1}{v_o^2})]+[(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{v_o^2}})][/tex]

Making q/m the subject, we have;

[tex]\dfrac{q}{m}=\dfrac{(d\times v_o^2)}{[E(\dfrac{D_1^2}{2})+(D_1\times D_2))]}[/tex]

We have;

E = 800 N/C

d = 1.25 cm = 0.0125 m

D_1 = 26.0 cm = 0.26 m

D_2 = 56 cm = 0.56 m

Thus;

[tex]\dfrac{q}{m}=\dfrac{(0.0125\times 5000^2)}{[800((\dfrac{0.26^2}{2})+(0.26\times 0.56))]}[/tex]

[tex]\dfrac{q}{m}= 2177.4[/tex]

Hence the ratio of the charge to mass of the object will be equal to:

[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]

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