is water wet? If water is not wet does that make it dry?

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:

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:

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:

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:

3

5

4

Explanation:

x = (8=(9+9)

(9+9) = 4, 5, 3

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:

[tex]v=697.2km/h[/tex]

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

[tex]V=\frac{d}{t}[/tex]

The distance is clearly 1743 km and the time is:

[tex]t=2h+30min*\frac{1h}{60min} =2.5h[/tex]

Thus, the velocity turns out:

[tex]v=\frac{1743km}{2.5h}\\ \\v=697.2km/h[/tex]

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

Answer:

directly proportional to the square of its speed.

Explanation:

;)

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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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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Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

a = 4.9 m/s²

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:

its D i just did it on the app

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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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: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:

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:

[tex]\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}[/tex]

Explanation:

[tex](inertia.of.the.disk) Id =\frac{M R^{2}}{2}\\(inertia.of.the.solid) Is = m R^{2}\\\\Itotal = Id + Is = \left(\frac{M}{2}+m\right) R^{2}\\\\\\L=\vec{\gamma} \times \vec{p}={mvr} \quad \& \quad L=I \omega\\\\Mrock \timesv \times R=\left(\frac{M}{2}+m\right) R^{2} \times \omega\\\\\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}\\[/tex]

The required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].

Given data:

The mass of disk is, M.

The radius of disk is, R.

The mass of teenager is, m.

The mass of rock is, [tex]m_{rock}[/tex].

The speed of rock is, v.

Here we  need to apply the concept of angular momentum to obtain the value of angular speed of disk. The expression for the angular momentum is given as,

[tex]L = I_{total} \times \omega[/tex] .......................................(1)

Here, [tex]\omega[/tex] is the angular speed and [tex]I_{total}[/tex] is the total moment of inertia of system.

And its value is,

[tex]I_{total} = I_{disk}+I_{solid}\\\\I_{total} = \dfrac{MR^{2}}{2}+mR^{2}[/tex] ..........................................(2)

And the angular momentum is also expressed as,

[tex]L = p \times R\\L =( m_{rock}v) \times R[/tex] ....................................................(3)

Then, using the equation (1), (2) and (3) we have,

[tex]( m_{rock} \times v) \times R = (\dfrac{MR^{2}}{2}+mR^{2}) \times \omega\\\\m_{rock} \times v = (\dfrac{MR}{2}+mR) \times \omega\\\\\\\omega = \dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex]

Thus, we can conclude that the required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].

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Answer:

A) v = 31.8 m/s

B) v_f = 18.1 m/s

C) F_avg = 4130.7 N

Explanation:

A) From law of conservation of energy, we know that;

mgh = W_f + ½mv²

v is the speed at which she is going at the bottom of the slope.

Thus, making v the subject, we have;

v = √[2gh - (2W_f)/m)]

We are given;

h = 70 m

m = 58 kg

W_f = 1.04 × 10⁴ J = 10400 J

Thus;

v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]

v = 31.8 m/s

B) Total force will be given by the formula;

F_t = F_k + F_r

Where;

F_k is force of kinetic friction = mg•μ_k

μ_k = 0.25

So, F_k = 58 × 9.8 × 0.25

F_k = 142.1 N

We are given force of air resistance(F_r) as 150 N

Thus;

F_t = 142.1 + 150

F_t = 292.1 N

Final velocity is gotten from the formula;

v_i² - v_f² = 2F_t•L/m

Thus;

v_f = √[v_i² - (2F_t•L/m)]

Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m

Thus;

v_f = √[31.8² - (2 × 292.1 × 68/58)]

v_f = 18.1 m/s

C) the average force exerted on her by the snowdrift as it stops her is given by the formula;

F_avg = m•v_f²/2l

F_avg = 58 × 18.1²/(2 × 2.3)

F_avg = 4130.7 N

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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Answer:

B-A-C

Explanation:

The idea is to evaluate the slopes of the three graphs. Notice that the three graphs are a representation of position (x) as a function of time (t).

Then, the slope of those lines are giving information of the change in position over the change in time (rise over the run). For the graph with positive slope the velocity is positive, for those with negative slope (line going down) the velocity is negative. But we are asked to compare speeds, which are the magnitude  of each velocity (slope) so it is important to determine the graph with the largest slope (graph C), and that with the smallest (graph B)

Then the order is: B-A-C (third option answer option)

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:

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:

P1 = P2      pressure is uniform

F1 / A1 = F2 / A2

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

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

Note: N - m g is the net force producing the centripetal force