A 64.7 cm long straight section of wire is located entirely inside a uniform magnetic field of |B| = 0.370 T. The wire is perpendicular to the direction of the magnetic field. When a current runs through the wire a magnetic force of |F| = 0.110 N acts on that section of wire. Calculate the size of the current.

Answer:

A

Explanation:

A)

INCORRECT

No, Gauss's Law can not be derived from Coulomb's Law alone. Since, Coulomb's Law provides electric field produced by a single point charge. Therefore, it is important to assume principle of super position along with Coulomb's law to get the field due to all charges. So, Gauss's Law can be proved by the help of both Super Position Principle and Coulomb's law.

B)

CORRECT

Yes, Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface.

C)

CORRECT

Yes, Coulomb's law can be derived from Gauss' law and symmetry.

D)

CORRECT

Yes, Gauss's Law applies to a closed surface of any shape.

E)

CORRECT

Yes, if a closed surface encloses no charge, the electric field is zero everywhere on the surface.

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex]   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex]         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]

The induced emf is 28.43V

(c) The induced current in the coil is given by:

[tex]I_{in}=\frac{emf}{R}[/tex]      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

Answer:

a) [tex]E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]

b) zero

Explanation:

a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:

[tex]E_T=E_1+E_2[/tex]

E1: electric field of charge 1

E2: electric field of charge 2

It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:

[tex]E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\[/tex]

r13: distance between charges 1 and 3

r12: charge between charges 2 and 3

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Thus, you first calculate the distance r13 and r23, and also the angles:

[tex]r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°[/tex]

Next, you replace the values of all parameters in order to calculate E1 and E2:

[tex]E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}[/tex]

finally, you obtain for ET:

[tex]E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]

b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

[tex]\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s[/tex]

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex]    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

[tex]v=\lambda f\\\\[/tex]

v: speed of sound

λ: wavelength = 0.700 m

[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]

Next, you replace the values of all parameters in the equation (1):

[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

[tex]I=\frac{P}{4\pi r^2}[/tex]

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]

hence, the intensity is 8.12*10^⁻4 W/m^2

Answer:

look at the data for the control group to see how much the fish grew without the new food.

Explanation:

i got it right

Answer:

Nancy Elizabeth Lieberman (nacido el 1 de julio de, 1958), apodado "Señora mágica", ] es un ex profesional jugador de baloncesto y entrenador de la WNBA (WNBA) que es actualmente un organismo de radiodifusión para los New Orleans Pelicans de la Nacional La Asociación de Baloncesto (NBA) y la entrenadora principal de Power , un equipo en el BIG3 que lideró en su Campeonato 0.]bLieerman es considerada como una de las figuras más importantes del baloncesto femenino estadounidense.

Explanation:

your answer is b to be presided

Answer:

259.52N

Explanation:

Force initiated on a spring that causes an extension is related by the expression below;

F= K×e

Where F is the Force

K is the spring constant

e is the extension caused by the spring.

By change of subject formula for K;

K = F/e

Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}

Hence W =7.15×9.8=70.07N

Hence K = 70.07/0.27 =259.5185N/m

=259.52N/m to 2 decimal place.

Answer:

a) [tex]a_{T}=8.50m/s^{2}[/tex], [tex]v_{T}=2.78 m/s[/tex] and [tex]N=279.83N[/tex]

b) [tex]a_{T}=0m/s^{2}[/tex], [tex]v_{T}=5.68 m/s[/tex] and [tex]N=795.2N[/tex]

Explanation:

a)

In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:

[tex]h_{0}=2.5m-h_{1}[/tex]

so we get:

[tex]h_{1}=2.5m(sin (20^{o}))[/tex]

[tex]h_{1}=0.855m[/tex]

so

[tex]h_{0}=2.5m-h_{1}[/tex]

[tex]h_{0}=2.5m-0.855m[/tex]

[tex]h_{0}=1.645m[/tex]

once we got this value, we can find the final height the same way. This time the angle is 30° so we get:

[tex]h_{f}=2.5m-h_{2}[/tex]

[tex]h_{f}=2.5m-2.5m(sin 30^{o}))[/tex]

[tex]h_{f}=1.25m[/tex]

Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:

[tex]U_{0}=U_{f}+K_{f}[/tex]

so now we substitute with the corresponding formulas:

[tex]mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}[/tex]

if we divided both sides of the equation by the mass, then the equation simplifies to:

[tex]gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}[/tex]

and now we can solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2g(h_{0}-h_{f}}[/tex]

and we can now substitute values:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}[/tex]

which solves to:

[tex]v_{f}=2.78m/s[/tex]

which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:

We can build a free body diagram (see attached  picture) and do a balance of forces so we get:

[tex]\sum{F_{x}}=ma_{T}[/tex]

in this case we only have one x-force which is the x-component of the weight, so we get that:

[tex]w_{x}=ma_{T}[/tex]

so we get:

[tex]mgcos\theta=ma_{T}[/tex]

and solve for the tangential acceleration so we get:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}*cos(30^{o})[/tex]

[tex]a_{t}=8.50m/s^{2}[/tex]

Now we can find the centripetal acceleration by using the formula:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(2.78m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=3.09m/s^{2}[/tex]

Next, we can find the normal force which is found by doing a sum of forces on y, so we get:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg sin\theta[/tex]

when factoring we get:

[tex]N=m(a_{c}+g sin\theta)[/tex]

and we substitute:

[tex]N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=279.83N

b)

The procedure for part b is mostly the same with some differences due to the angle. First:

[tex]h_{0}=1.645m[/tex]

[tex]h_{f}=0m[/tex]

so

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:

[tex]U_{0}=K_{f}[/tex]

so we get:

[tex]mgh_{0}=\frac{1}{2}mv_{f}^{2}[/tex]

so we solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2gh_{0}}[/tex]

and we substitute:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}[/tex]

[tex]v_{f}=5.68m/s[/tex]

according to the free body diagram we get that:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}(cos 90^{o})[/tex]

which yields:

[tex]a_{T}=0[/tex]

we can also find the centripetal acceleration, so we get:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(5.68m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=12.91m/s^{2}[/tex]

and we can do a sum of forces on y to find the normal force:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg [/tex]

when factoring we get:

[tex]N=m(a_{c}+g)[/tex]

and we substitute:

[tex]N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=795.2N

Following are the calculation to the angles:

For angle 30°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}\\\\[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N[/tex]

For angle 90°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:  

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}[/tex]

   [tex]=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}\\[/tex]

    [tex]= \frac{270.494}{858.375}\\\\=0.315[/tex]

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Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

ΔS = 49.13 ft

Answer:

A. mass

Explanation:

Mass determines the quantity of inertia for an object. Mass is the quantity that depends upon the inertia of an object. The inertia that an object has is directly proportional to the mass of the object.

An object that has more mass has a greater tendency as compared to the object that has less mass to resist changes in its state of motion.

Answer:

Ok

Explanation:

Please mark brainliest

Answer:

The monkey will get hit by the dart. This is because when the dart is fired the force of gravity makes it fall from its original position. The monkey also has the same fall from the branches which means they will both be falling at the same gravitational rate.

This will eventually make the monkey to get hit by the dart.

Answer:

Useless

Explanation:

Question

Answer:

The magnitude of the acceleration is [tex]a_r = 1.50 \ m/s^2[/tex]

The direction is  [tex]\theta = 32.5 6^o[/tex] north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  [tex]F_{sail} = (330 ) \ N \ north[/tex]

   The force exerted by water is  [tex]F_{keel} = (210 ) \ N \ east[/tex]

      The mass of the boat(+ crew) is  [tex]m_b = 260 \ kg[/tex]

Now Force is mathematically represented as

      [tex]F = ma[/tex]

Now the acceleration towards the north is mathematically represented as

      [tex]a_n = \frac{F_{sail}}{m_b}[/tex]

substituting values

       [tex]a_n = \frac{330 }{260}[/tex]

      [tex]a_n = 1.269 \ m/s^2[/tex]

Now the acceleration towards the east is mathematically represented as

       [tex]a_e = \frac{F_{keel}}{m_b }[/tex]

substituting values

      [tex]a_e = \frac{210}{260}[/tex]

      [tex]a_e =0.808 \ m/s^2[/tex]

The resultant acceleration is  

      [tex]a_r = \sqrt{a_e^2 + a_n^2}[/tex]

substituting values

     [tex]a_r = \sqrt{(0.808)^2 + (1.269)^2}[/tex]

      [tex]a_r = 1.50 \ m/s^2[/tex]

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        [tex]tan \theta = \frac{a_e}{a_n}[/tex]

       [tex]\theta = tan ^{-1} [\frac{a_e}{a_n } ][/tex]

substituting values

     [tex]\theta = tan ^{-1} [\frac{0.808}{1.269 } ][/tex]

    [tex]\theta = tan ^{-1} [0.636 ][/tex]

   [tex]\theta = 32.5 6^o[/tex]

Answer:

Beta Particle

Explanation:

The beta particle decay completes the following reaction,

¹⁴C₆   ⇒ ¹⁴N₇ + Beta particle

Therefore the answer is option D.

Radioactive decay is a type of nuclear reaction in which the unstable nucleus of a radioactive element releases energy in the form of nuclear radiation. Alpha decay, decay, and beta decay are a few examples of radioactive decay.

The radioactive deacy is acompained by the release of the enormous amount of energy,the nuclear power generation with the help of the nuclear reaction is one of the most significant application of the radioactive decay.

¹⁴C₆   ⇒ ¹⁴N₇ + Beta particle

Beta particles are the negatively charged electron emitted from radioactive decay.

For the given reaction the unstable carbon atom losses a beta particle to produce a stable nitrogen atom.

Thus the given radioactive decay can is completed by beta particle emission.

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Answer:

Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.

Answer:

F/L =  8*10^-4 N/m

Explanation:

To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:

[tex]F_N=F_{1,2}+F_{2,3}[/tex]   (1)

F1,2 : force between first and second wire

F2,3 : force between second and third wire

The force per meter between two wires of the same length is given by:

[tex]\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}[/tex]

μo: magnetic permeability of vacuum =  4pi*10^-7 T/A

r: distance between wires

Then, you have in the equation (1):

[tex]\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3][/tex]

But

I1 = I2 = I3 = 10A

r = 10cm = 0.1m

You replace the values of the currents and the distance r and you obtain:

[tex]\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}[/tex]

hence, the net force per meter is 8*10^-4 N/m

Answer:a

Explanation: they are all positive

The equivalent vectors of both the vectors lie in the first quadrant. So, the right choice is a. first quadrant.

The resultant of the vectors can be calculated using the Parallelogram theorem of vector addition. In this theorem, a parallelogram is formed using the vectors, having the vectors as the adjacent side and the diagonal of this parallelogram is the resultant of both the vectors.

The vectors A and B are drawn at the given angle, angles of 20° and 80°  respectively. Using the vectors, A and B, the dotted parallelogram is drawn, and the vector C is the resultant.

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Answer:

The new volume is  [tex]11.6L[/tex]

Explanation:

[tex]pressure \alpha \frac{1}{volume}[/tex]

Given data

initial pressure p1= 100atm

initial volume v1= 3.7L

final pressure p2= 32 atm

final volume v2= ?

Apply Boyle's law we have

[tex]p1v1= p2v2\\[/tex]

[tex]v2= \frac{p1v1}{p2}[/tex]

Substituting our data we have

[tex]v2= \frac{100*3.7}{32} \\v2=\frac{370}{32} \\v2= 11.6 L[/tex]

The new volume is [tex]11.6L[/tex]

Answer:

B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.

Explanation:

Inner planets are smaller and rockier than outer gas planets,outer planets are larger,because of their lower gravity they don't attract extreme amounts of gas in their planets.The four outer planets were so far from the Sun that its winds could not blow away their ice and gases

Answer:

The student is going at the bottom of the slide with a velocity of 8.66 m/s

Explanation:

Given;

mass of the student, m = 74 kg

height of the water slide, h = 11.3 m

work done, W = -5.42 × 10³ J

Apply work energy theorem;

[tex]W = \frac{1}{2}mv_f^2+ mgh_f-(\frac{1}{2}mv_o^2 + mgh_o)\\\\ W + \frac{1}{2}mv_o^2 + mgh_o -mgh_f= \frac{1}{2}mv_f^2\\\\ W + \frac{1}{2}mv_o^2 +mg(h_o-h_f) = \frac{1}{2}mv_f^2\\\\\frac{W}{m} + \frac{1}{2} v_o^2 + g(h_o-h_f) = \frac{1}{2}v_f^2\\\\\frac{2W}{m}+ v_o^2 + 2g(h_o-h_f) = v_f^2\\\\v_f = \sqrt{\frac{2W}{m}+ v_o^2 - 2g(h_f-h_o)} \\\\v_f = \sqrt{\frac{2(-5.42*10^3)}{74}+ (0)^2 - 2*9.8(-11.3)}\\\\v_f=\sqrt{-146.4865+221.48} \\\\v_f = \sqrt{74.9935} \\\\v_f = 8.66 \ m/s[/tex]

Therefore, the student is going at the bottom of the slide with a velocity of 8.66 m/s

Answer:

It is larger

Explanation:

I believe the answer would be "It is larger" since v = 2v (since it's in series), which would increase the magnetic field and thus produced a bigger angle.

Answer:

B

Explanation:

Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.

Answer:

Explanation:

Given the equation modelled by the height of the train given as:

s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9

a) Velocity is the rate of change of displacement.

Velocity = dS(t)/dt

V = dS(t)/dt = 36t - 6t² miles

Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.

V = 36(3) -6(3)²

V= 108 - 72

Velocity = 36mi/hr

b) for Velocity at time = 7hrs

V(7) = 36(7) - 6(7)²

V(7) = 252 - 294

V(7) = -42mi/hr

The velocity at t = 7hrs is -42mi/hr

c) Acceleration is the rate of change of velocity.

a(t) = dV(t)/dt

Given v(t) = 36t - 6t²

a(t) = 36 - 12t

Acceleration at t=1 is given as:

a(1) = 36 -12(1)

a(1) = 24mi/hr²

Answer:

608.4m/s

Explanation:

We are given that

Mass of Sleigh,M=1200 kg

Speed of Sleigh,u=322 m/s

Speed of jet,u'=680 m/s

Mass of jet,m=4800 kg

Total mass=M+m=1200+4800=6000 kg

We have to find the final velocity of the two objects after the collision.

The collision is inelastic .

By using law of conservation of momentum

[tex]Mu+mu'=(m+M)v[/tex]

Using the formula

[tex]1200\times 322+4800\times 680=6000v[/tex]

[tex]6000v=3650400[/tex]

[tex]v=\frac{3650400}{6000}[/tex]

[tex]v=608.4m/s[/tex]

Hence, the final  velocity of two objects after the collision=608.4m/s

Answer:

r = 4.23 m

Explanation:

To find the radius of the circular path of the electron you use the following formula:

[tex]r=\frac{mv}{qB}[/tex]  (1)

This formula can be used because the motion of the electron is perpendicular to the direction of the magnetic field vector.

m: mass of the electron = 9.1*10^-31 kg

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 1.02*10^-5 T

v: velocity of the electron = 7.6*10^6 m/s

You replace the values of m, v, q and B in the equation (1):

[tex]r=\frac{(9.1*10^{-31}kg)(7.6*10^6 m/s)}{(1.6*10^{-19}C)(1.02*10^{-5}T)}\\\\r=4.23m[/tex]

hence, the raiuds of the orbit of the electron is 4.23m

Answer:

Explanation:

2 )

power of an electric device = V² / R where V is volts and R is resistance

putting given data

power = 9²/ 5

= 16.2 J/s

energy produced in 7 minutes

= 16.2 x 7 x 60

= 6804J .

3 ) Power of an electrical device

= V² / R

= V X I where I  is current

= 4.5 x .5

= 2.25 W or J/s

4 )

energy used in 3 minutes with power of 2.25 W

= 2.25 x 3 x 60

= 405 J .

7 )

power of a electrical device

= V x I

IR x I  where R is resistance .

= I²R

putting given data

power = .005² x 50

= 1.25 x 10⁻³ W .

8 )

Energy used up by a 60 W bulb in 2.5 hours

= 60 x 2.5 x 60 x 60

= 5.4 x 10⁵ J .