As part of calculations to solve an oblique plane triangle (ABC), the following data was available: b=50.071 horizontal distance, C=90.286° (decimal degrees), B=62.253° (decimal degrees). Calculate the distance of c to 3 decimal places (no alpha).

Answers

Answer 1

Answer:

The distance of c is 56.57

Explanation:

Given that,

Horizontal distance b = 50.071

Angle C = 90.286°

Angle B = 62.253°

We need to calculate the distance of c

Using sine rule

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

Put the value into the formula

[tex]\dfrac{50.071}{\sin62.253 }=\dfrac{c}{\sin90.286}[/tex]

[tex]c= \dfrac{50.071\times\sin90.286}{\sin62.253}[/tex]

[tex]c=56.575[/tex]

Hence, The distance of c is 56.575.


Related Questions

You drop a stone into a deep well and hear it hit the bottom 6.65 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about 343 m/s in air. How deep is the well?

Answers

Answer: ok, so total time is Tt = 6.65 seconds = time falling + time of sound traveling.

the time of the rock falling is given by:  [tex](g/2)*t₀^{2} -h = 0[/tex] (1)

where h is the heigth of the hole, and t₀ is the time it took to reach the bottom.

the time of the sound traveling is given by 343*t₁ - h = 0 (2)

so t₁ + t₀ = 6.65s = total time

then you know that t₁ = 6.65s -  t₀

so now you have two variables, t₀ and h, and we want to know the value of h, so we want to write t₀ as a function of h.

in the second equation we have now: 343*(6.65 -  t₀) = h

so  t₀ = (-h/343 + 6.65)s

replacing this on the first equation you have:

 [tex](g/2)*(-h/343 + 6.65)^{2} -h = 0[/tex]

now you want to take h to the right side so

w

so if we replace g by 9.8 meters over seconds square we get

h = [tex]\frac{-1.19 -+ \sqrt[2]{1.56} }{2*0.00041}[/tex]

where you will take the positive root

h = 61 meters

Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which one of the naturally occurring electrons is removed. The radius of the orbit is 2.99 × 10-11 m. Determine the magnitude of the electron's centripetal acceleration.

Answers

Answer:

[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]

Explanation:

In this case we will use the Bohr Atomic model.

We have that: [tex]F=m*a[/tex]

We can calculate the centripetal force using the coulomb formula that states:

[tex]F=k*\frac{q*q'}{r^2}[/tex]

Where K=[tex]9*10^9 \frac{Nm^2}{C}[/tex]

and r is the distance.

Now we can say:

[tex]m*a=k*\frac{q*q'}{r^2}[/tex]

The mass of the electron is = [tex]9.1*10^{-31}[/tex] Kg

The charge magnitud of an electron and proton are= [tex]1.6*10^{-19}C[/tex]

Substituting what we have:

[tex][tex]a=\frac{9*10^{9}*(1.6*10^{-19} )*(2(1.6*10^{-19} ))}{9.1*10^{-31}*(2.99*10^{-11})^2 }[/tex][/tex]

so:

[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]

A 0.07-kg lead bullet traveling 257 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted to heat that it alone absorbs, what is its temperature change?

Answers

Answer:

ΔT = 258°C

Explanation:

mass of bullet, m = 0.07 kg

velocity of bullet, v = 257 m/s

According to the energy conservation law, the kinetic energy of bullet is totally converted into form of heat energy.

let ΔT be the rise in temperature of the bullet, c be the specific heat of lead.

c = 0.128 J / g°C = 128 J/kg°C

[tex]\frac{1}{2}mv^{2}=mc\Delta T[/tex]

By substituting the values

0.5 x 0.07 x 257 x 257 = 0.07 x 128 x ΔT

ΔT = 258°C

Calculat the acceleration of a person at latitude
40degreesowing to the rotation of the earth.

Answers

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R    ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × [tex]10^{-5}[/tex] rad/s and R is earth radius that is 6.37 × [tex]10^{6}[/tex] m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × [tex]10^{-5}[/tex])² × 6.37 × [tex]10^{6}[/tex]

acceleration of person = 9.77 m/s²

Final answer:

To calculate the centripetal acceleration at a latitude of 40°, use the formula ω2 r, considering Earth's angular velocity and the adjusted radius at that latitude. The resulting acceleration is less than that at the equator and significantly weaker than gravity, ensuring no risk of 'flying off' due to Earth's rotation.

Explanation:

To calculate the acceleration due to Earth's rotation for a person at a latitude of 40 degrees, we can use the formula for centripetal acceleration, ac = ω2 r, where ω is the angular velocity of Earth and r is the distance from the axis of rotation. The angular velocity of Earth is approximately 7.29 × 10-5 radians per second (based on a sidereal day). The distance from the axis of rotation at latitude λ can be expressed as r = REcos λ, where RE is the Earth's radius.

The formula for a person at latitude 40° becomes ac = (ω2)(RE)(cos 40°). By using the Earth's radius RE = 6380 km and the given values, we can calculate the centripetal acceleration and compare it with the acceleration due to gravity, g.

At latitude 40°, the velocity due to Earth's rotation is reduced as a function of the cosine of the latitude, hence, the centripetal acceleration will also be less than what it is at the equator. To find out if there is any danger of 'flying off', we must understand that the gravitational force at Earth's surface is significantly stronger than the centripetal force due to Earth's rotation, ensuring that we remain safely on the ground.

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Answers

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so [tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

[tex]\frac{P_1}{P_2}=\frac{T_1}{T_2}[/tex]

As [tex]T_1[/tex] is decreasing therefore Pressure must also decrease so that ratio remains constant.

An ice skater rotating about a vertical axis increases her angular velocity at constant rate from 7.85 rad/s to 10.65 rad/s in 2.3 s, thus her angular acceleration is 1.23 rad/s^2. During her rotation, her right leg outstretched horizontally. Calculate the radial ______ and tangential ______ accelerations of her skate at the end of 2.3 s. Her leg length is 90 cm.

Answers

Answer:

(a). The tangential acceleration is 1.107 m/s².

(b). The radial acceleration is 102.08 m/s².

Explanation:

Given that,

Initial angular velocity = 7.85 rad/s

Final angular velocity = 10.65 rad/s

Time = 2.3 s

Acceleration = 1.23 rad/s²

Radius = 90 cm

(a). We need to calculate the tangential acceleration

Using formula of radial acceleration

[tex]a_{t}=\alpha R[/tex]

Put the value into the formula

[tex]a_{t}=1.23\times90\times10^{-2}[/tex]

[tex]a_{t}=1.107\ m/s^2[/tex]

The tangential acceleration is 1.107 m/s².

(b). We need to calculate the radial acceleration

Using formula of radial acceleration

[tex]a_{c}=\omega^2 r[/tex]

Put the value into the formula

[tex]a_{c}=(10.65)^2\times90\times10^{-2}[/tex]

[tex]a_{c}=102.08\ m/s^2[/tex]

Hence, (a). The tangential acceleration is 1.107 m/s².

(b). The radial acceleration is 102.08 m/s².

Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about 150 N/C at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Express your answer using two significant figures.

Answers

Answer:

Magnitude of electric field on Earth = Q = 6.8 × 10⁵ C

Explanation:

Electric field = E = 150 N/C

Distance from the center of the earth to the surface = Radius of the earth

Radius of the earth = R = 6.38× 10⁶ m

E = k Q / R²  is the basic formula for the electric field. k = 9 × 10⁹ N m²/C²

150 = (9 × 10⁹)(Q) / (6.38× 10⁶ )²

⇒ Charge = Q = (150)(6.38× 10⁶ )²/(9 × 10⁹)

                        = 6.8 × 10⁵ C(2 significant figures).

Answer:

The charge of earth is[tex]-6.8\times 10^{5}Columbs[/tex]

Explanation:

Assuming earth as a spherical body we have

For a sphere of radius 'r' and charge 'q' the electric field generated at a distance 'r' form the center of sphere is given by the equation

[tex]E=\frac{1}{4\pi \epsilon _o }\cdot \frac{Q}{r^{2}}[/tex]

where

'Q' is the total charge on sphere

Now at a distance 'r' equal to radius of earth(6371 km)  we have the electric field strength is 150N/C

Using the given values we obtain

[tex]150=\frac{1}{4\pi \epsilon _o}\frac{Q}{(6371\times 10^{3})^2}\\\\\therefore Q=150\times (6371\times 10^{3})^{2}\times 4\pi \epsilon _o\\\\\therefore Q=6.8\times 10^{5}Columbs[/tex]

Now since the electric field is inwards thus we conclude that this charge is negative in magnitude.

Suppose you have three identical metal spheres, A, B, and C Initially sphere A carries a charge q and the others are uncharged. Sphere A is brou contact and separated. Finally spheres A and C are brought in contact and then separated.

Answers

Answer:

The charge on sphere A and C is [tex]\frac{3q}{8} C[/tex]

Solution:

As per the question:

Charge on sphere A = q C

Now,

When sphere A with 'q' charge is brought in contact with B then conduction takes place and equal charges are distributed on both the spheres as:

[tex]\frac{q + q}{2} = \frac{q}{2}[/tex]

Charge on each A and B is [tex]\frac{q}{2}[/tex]

Now, after separating both the spheres, when sphere B with charge [tex]\frac{q}{2}[/tex] touches uncharged sphere C, again conduction occurs and the charge is distributed as:

[tex]\frac{\frac{q}{4} + \frac{q}{4}}{2} = \frac{q}{4}[/tex]

Now, after the charges are separated with charge [tex]\frac{q}{4}[/tex] on each sphere B and C

Now,

When sphere A and Sphere C with charge [tex]\frac{q}{2}[/tex] and [tex]\frac{q}{4}[/tex] comes in contact with each other, conduction takes place and charge distribution is given as:

[tex]\frac{\frac{q}{2} + \frac{q}{4}}{2} = \frac{3q}{8}[/tex]

A plastic rod is charged to -12 nC by rubbing. a) Have electrons been added or protons removed? Explain! b) How many electrons have been added or protons removed?

Answers

Answer:

(a) electrons being added.

(b) 7.5 × 10¹⁰

Explanation:

(a)

The plastic rod acquires a charge of -12 nC which means it contains more negative charge.

Thu, the electrons have been added to the rod. Removing protons are very difficult as they are in the nucleus and are bind by nuclear forces. Thus, by rubbing, electrons are transferred and thus the rod become negatively charged which means that the electrons being added.

(b)

The total charge = -12 nC

Also,

1 nC = 10⁻⁹ C

So,

Charge = -12 × 10⁻⁹ C

Charge on electron = -1.6 × 10⁻¹⁹ C

So, number of electrons transferred = -12 × 10⁻⁹ C / -1.6 × 10⁻¹⁹ C = 7.5 × 10¹⁰

a) Electrons have been added to the plastic rod.

b) 7.5 x 10¹⁰ electrons have been added to the plastic rod.

a) When a plastic rod is rubbed with a cloth, electrons are transferred from the cloth to the rod. This gives the rod a negative charge, because it now has an excess of electrons.

b) The charge on an electron is -1.6 x 10⁻¹⁹ C. The rod has a charge of -12 nC, which is -12 x 10⁻⁹ C. So, 7.5 x 10¹⁰ electrons have been added to the rod.

Charge: Charge is a property of matter that can be positive, negative, or neutral. Objects with the same charge repel each other, while objects with opposite charges attract each other.

Electrons: Electrons are negatively charged particles that are found in all atoms. They are much smaller than protons and neutrons, the other two types of particles found in atoms.

Protons: Protons are positively charged particles that are found in the nucleus of atoms. They are about the same size as neutrons.

Rubbing: When two objects are rubbed together, electrons can be transferred from one object to the other. This is because the electrons in the two objects are not always evenly distributed.

When a plastic rod is rubbed with a cloth, electrons are transferred from the cloth to the rod. This is because the cloth has a stronger attraction for electrons than the rod does. The rod now has an excess of electrons, which gives it a negative charge.

The number of electrons that are transferred can be calculated by dividing the rod's charge by the charge on an electron. In this case, the rod has a charge of -12 nC, and the charge on an electron is -1.6 x 10⁻¹⁹ C. So, 7.5 x 10¹⁰ electrons have been added to the rod.

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A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 32.3 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 3.45 seconds after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E2 to the magnitude of E1? You may neglect the effects of gravity on the proton.

Answers

Final answer:

To find the ratio of the magnitude of E2 to the magnitude of E1, we analyze the motion of the proton in the electric fields. By equating the work done in the first and second electric fields and using the equations of motion, we can determine the ratio of E2 to E1. The ratio is -1.

Explanation:

To find the ratio of the magnitude of E2 to the magnitude of E1, we need to analyze the motion of the proton in the electric fields. Since the proton returns to its starting point, its net displacement is zero. This means that the net work done on the proton in the combined electric fields is also zero. Using the work-energy theorem, we can equate the work done in the first electric field (E1) to the work done in the second electric field (E2).

The work done by a constant electric field is given by the formula W = μEd, where W is the work done, μ is the magnitude of the electric charge of the proton, E is the magnitude of the electric field, and d is the displacement. Since the displacement is zero, the work done in both fields is zero. Therefore, we have the equation: μE1d₁ + μE2d₂ = 0.

Since the proton is released from rest, its initial velocity is zero. Using the equation of motion, v = u + at, where v is the final velocity (which is also zero), u is the initial velocity, a is the acceleration, and t is the time, we can solve for the times taken in each field: t₁ = 32.3 s and t₂ = 3.45 s.

We can rewrite the equation as: E₁d₁ = -E₂d₂.

Using the equation of motion, d = ut + 0.5at², we can substitute the values of t₁ and t₂ to obtain: d₁ = 0.5a₁t₁² and d₂ = 0.5a₂t₂².

Substituting these values into the equation, we get: E₁(0.5a₁t₁²) = -E₂(0.5a₂t₂²).

The accelerations a₁ and a₂ can be determined using the formula a = μE/m, where μ is the magnitude of the charge of the proton, E is the magnitude of the electric field, and m is the mass of the proton. Since the proton is released from rest, its initial velocity is zero, so the acceleration is the same in both fields. Using this formula, we can express the accelerations in terms of the electric fields: a₁ = (μE₁)/m and a₂ = (μE₂)/m.

Substituting these values into the equation, we get: E₁(0.5((μE₁)/m)t₁²) = -E₂(0.5((μE₂)/m)t₂²). Simplifying the equation, we have: E₁²t₁² = -E₂²t₂².

Dividing both sides of the equation by E₁²t₁², we obtain: 1 = -(E₂/E₁)²(t₂/t₁)².

Taking the square root of both sides of the equation, we get: 1 = -(E₂/E₁)(t₂/t₁).

Dividing both sides of the equation by (t₂/t₁), we finally obtain the ratio of the magnitude of E₂ to the magnitude of E₁: E₂/E₁ = -1.

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Final answer:

Since the proton returned to its starting point, it must have decelerated and then accelerated in the opposite direction due to the electric field E2. Upon calculation, we found that the electric field E2 must be zero. Therefore, the ratio of the magnitude of E2 to E1 is 0.

Explanation:

The problem given can be resolved using the principles of Electromagnetism and Classical mechanics. The proton under the influence of the electric field E1 pointing due North will gain a certain amount of velocity during the first 32.3 seconds. When the direction of the electric field changes to E2 pointing due South, the proton will experience a force in the opposite direction, causing it to decelerate, stop, and then accelerate back to its starting point.

We know that the proton's velocity upon reaching the starting point is zero, and the total time it takes to return to the starting point is 3.45 seconds. Using the equations of motion, we can calculate the deceleration rate caused by the second electric field E2. The acceleration or deceleration experienced by a charged particle in an electric field is given by equation F=qE, where F is force, q is the charge, and E is the electric field.

Considering the proton is experiencing uniform deceleration, we have a = 2s/t². We know s=0 (as it returns to the same point) and the time t = 3.45s. From this, we can find a = 2*0/(3.45)² = 0 m/s². The deceleration caused by the second electric field E2 is therefore 0 m/s². The force acting on the proton is given by F = m*a, where m is the mass of the proton, and a is the deceleration. Since a = 0, F = 0. We know from the equation F = qE, the electric field E2 must be zero. Therefore, the ratio of the magnitude of E2 to E1 is 0.

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A 3.00 kg steel ball strikes a massive wall at 10.0m/s at
anangle of 60.0 degree with the plane of the wall. It bouncesoff
the wall with the same speed and angle. If the ball is incontact
with the wall for 0.200s, what is the average force exertedby the
wall on the ball?

Answers

Answer:259.80 N

Explanation:

Given

mass of ball=3 kg

ball velocity =10 m/s

angle made by ball with plane of wall [tex]\theta [/tex]=60

Momentum change in Y direction remains same and there is change only in x direction

therefore

initial momentum[tex]=mvsin\theta [/tex]

=30sin60

Final momentum=-30sin60

Change in momentum is =30sin60+30sin60

=60sin60

and Impulse = change in momentum

Fdt=dP

where F=force applied

dP=change in momentum

[tex]F\times 0.2=60sin60[/tex]

[tex]F\times 0.2=51.96[/tex]

F=259.80 N

Final answer:

The average force exerted by the wall on the 3.00 kg steel ball after it bounces off is 150 N. We calculated this using the principle of conservation of momentum and Newton's second law.

Explanation:

The subject of this question is Physics, specifically, the topic of motion and force. To answer this question, we should apply the law of conservation of momentum and Newton's second law (Force = change in Momentum / change in Time).

Firstly, the ball is projected against the wall at an angle of 60 degrees, but we are concerned with the component of velocity perpendicular to the wall. This means we will consider the velocity component of the ball towards the wall, which is 10 cos(60), giving us 5.0 m/s. The incoming momentum of the ball can then be calculated as the mass times the velocity (3.0 kg * 5.0 m/s = 15 kg*m/s).

Since the ball bounces off with the same speed at the same angle, its outgoing momentum is -15 kg*m/s. The change in momentum is therefore Outgoing momentum - Incoming momentum = -15 kg*m/s - 15 kg*m/s = -30 kg*m/s. The force exerted by the wall on the ball equals the Change in momentum divided by the time it takes for the change to occur (-30 kg*m/s / 0.200 s = -150 N). Given that force is a vector and we are asked for the magnitude of the force, the answer is 150 N.

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A force F=(cx-3.00x2)i acts on a partical as theparticle
moves along an x axis, with F in newtons, x in meters andc a
constant..At x=0, the paricle's kinetic energy is 20.0 J;atx=3.00m
it is 11.0 J......FIND c...

Answers

Answer:

The value of 'c' is 4.

Explanation:

We know from the basic relation between work and force is

[tex]W_{2}-W_{1}=\int_{x_{1}}^{x_{2}}F\cdot dx[/tex]

Now since the energy of the particle changes from 20 Joules to 11 Joules  as position changes from  x=0 to x=3.00 thus applying the given vales in above relation we get

[tex]11-20=\int_{0}^{3}(cx-3x^{2})\cdot dx\\\\-9=[\frac{cx^2}{2}]_0^3-[\frac{3x^3}{3}]_0^3\\\\-9=\frac{9c}{2}-27\\\\4.5c=18\\\\\therefore c=4[/tex]

Suppose that a car performs a uniform acceleration of 0.42 m/s from rest to 30.0 km/h in the first stage of its motion (From point A to point B). Then it moves at constant speed for half a minute (From point B to point C). After that the car applies the breaks, stopping the vehicle in a uniform manner while the vehicle travels an additional 7.00m distance (From point C to point D). (a) How far did the car travel from the starting point? (b) How long was the car in motion? (c) What is the average speed of the car during the entire motion?

Answers

Answer:

a)Total distance = 399. 5 m

b)Total time =51.51 sec

c)Average speed = 7.75 m/s

Explanation:

For A to B:

[tex]S=ut+\dfrac{1}{2}at^2[/tex]

[tex]v^2=u^2+2as[/tex]

v= u + at

8.33 = 0.42 x t

t=19.83 sec

1 Km/h=0.27 m/s

30 Km/h=8.33 m/s

[tex]8.33^2=2\times 0.42\times s[/tex]

s=82.6 m

For B to C

V= 8.33 m/s

s= V x t

s=8.33 x 30

s=249.9 m

For C to D

[tex]S=ut-\dfrac{1}{2}at^2[/tex]

v= u - at

Final speed v=0

So

s=v x t/2

7= 8.33 x t/2

t=1.68 sec

Total distance = 82.6 + 249 .9 +7

Total distance = 399. 5 m

Total time = 19.83 + 30 + 1.68

Total time =51.51 sec

Average speed =Total distance/Total time

Average speed = 399.5/51.5

Average speed = 7.75 m/s

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver’s motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. How long is the diver in the air, from the moment he leaves the board until

Answers

Answer:

The diver is in the air for [tex]1.65s[/tex].

Explanation:

Hi

Known data [tex]v_{i}=6.3m/s, y_{i}=3.0m[/tex] and [tex]g=9.8m/s^{2}[/tex].We need to find the time when the diver reaches the highest point, so we use the following equation [tex]v_{f} =v_{i}-gt[/tex] with [tex]v_{f}=0ms[/tex] so [tex]t_{1}=\frac{v_{i}-v_{f} }{g}=\frac{6.3m/s}{9.8m/s^{2} }=0.64 s[/tex]. Then we need to find the highest point, so we use [tex]y=v_{i}t-\frac{1}{2} gt^{2}=(6.3m/s)(0.64s)-\frac{1}{2} (9.8m/s^{2})(0.64s)^{2}=2.03m[/tex], this is above the springboard so the highest point is [tex]y_{max}=5.03m[/tex].

Finally, we need to find the time in free fall, so we use [tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}[/tex], at this stage [tex]v_{i}=0m/s, y_{i}=5.03m[/tex] and [tex]y_{f}=0m[/tex], therefore we have [tex]0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex], and solving for [tex]t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s[/tex].

Last steep is to sum [tex]t_{1}[/tex] and [tex]t_{2}[/tex], so [tex]t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s[/tex].

The total time spent by the diver in the air from the moment he leaves the board until he gets to the water is 1.65 s

We'll begin by calculating the time taken to get to the maximum height from the board.

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Time to reach maximum height (t₁) =?

v = u – gt (since the diver is going against gravity)

0 = 6.3 – 9.8t₁

Collect like terms

0 – 6.3 = –9.8t₁

–6.3 = –9.8t₁

Divide both side by –9.8

t₁ = –6.3 / –9.8

t₁ = 0.64 s

Next, we shall determine the maximum height reached by the diver from the board

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum Height from the board (h₁) =?

v² = u² – 2gh (since the diver is going against gravity)

0² = 6.3² – (2 × 9.8 × h₁)

0 = 39.69 – 19.6h₁

Collect like terms

0 – 39.69 = –19.6h₁

–39.69 = –19.6h₁

Divide both side by –19.6

h₁ = –39.69 / –19.6

h₁ = 2.03 m

Next, we shall determine the height from the maximum height reached by the diver to the water.

Maximum height from the board (h₁) = 2.03 m

Height of board from water (h₂) = 3 m

Height of diver from maximum height to water (H) =?

H = h₁ + h₂

H = 2.03 + 3

H = 5.03 m

Next, we shall determine the time taken by the diver to fall from the maximum height to the water.

Height (H) = 5.03 m

Acceleration due to gravity (g) = 9.8 m/s²

Time to fall from maximum height to water (t₂) =?

H = ½gt²

5.03 = ½ × 9.8 × t₂²

5.03 = 4.9 × t₂²

Divide both side by 4.9

t₂² = 5.03 / 4.9

Take the square root of both side

t₂ = √(5.03 / 4.9)

t₂ = 1.01 s

Finally, we shall determine the total time spent by the diver in the air.

Time to reach maximum height (t₁) = 0.64 s

Time to fall from maximum height to water (t₂) = 1.01 s

Total time in air (T) =?

T = t₁ + t₂

T = 0.64 + 1.01

T = 1.65 s

Therefore, the total time spent by the diver in the air is 1.65 s

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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving in a circular path perpendicular to a magnetic field of magnitude B. (Use any variable or symbol stated above along with the following as necessary: q and c.)

Answers

Explanation:

Let [tex]m_p[/tex] is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

[tex]\dfrac{mv^2}{r}=qvB[/tex]

r is the radius of path,

[tex]r=\dfrac{mv}{qB}[/tex]

Time period is given by :

[tex]T=\dfrac{2\pi r}{v}[/tex]

[tex]T=\dfrac{2\pi m_p}{qB}[/tex]

Frequency of proton is given by :

[tex]f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}[/tex]

The wavelength of radiation is given by :

[tex]\lambda=\dfrac{c}{f}[/tex]

[tex]\lambda=\dfrac{2\pi m_pc}{qB}[/tex]

So, the wavelength of radiation produced by a proton is [tex]\dfrac{2\pi m_pc}{qB}[/tex]. Hence, this is the required solution.

To charge an electroscope negatively by induction you need: a) a positively charged rod and a ground b) a negatively charged rod c) a negatively charged rod and a ground d) two objects of the same charge

Answers

Answer:

Option a)

Explanation:

In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.

The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.

So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.

Final answer:

To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere, it polarizes the sphere and attracts electrons from the earth's ample supply. By breaking the ground connection and removing the positive rod, the sphere is left with an induced negative charge.

Explanation:

To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere (the electroscope), it polarizes the sphere. By connecting the sphere to a ground, electrons are attracted from the earth's ample supply, resulting in an induced negative charge on the sphere. The ground connection is then broken, and the positive rod is removed, leaving the sphere with the induced negative charge.

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he catches his friend. How much time does it take until he catches his friend (after his friend passes him)?

Answers

Final answer:

It will take approximately 1.61 seconds for the bicyclist to catch up to his friend after his friend passes him.

Explanation:

To determine the time it takes for the bicyclist to catch up to his friend, we can use the equation:

distance = initial velocity * time + 0.5 * acceleration * time^2

Since the friend is traveling at a constant speed of 3.1 m/s, the distance traveled by the bicyclist during the 2-second delay is 6.2 m. Using the equation above:

6.2 m = 0 m/s * t + 0.5 * 2.4 m/s^2 * t^2

Simplifying the equation:

2.4 m/s^2 * t^2 = 6.2 m

t^2 = 6.2 m / 2.4 m/s^2

t^2 = 2.5833 s^2

t ∼ 1.61 s

Therefore, it will take approximately 1.61 seconds for the bicyclist to catch up to his friend.

A certain copper wire has a resistance of 10.5 Ω. It is cut into two pieces such that one piece has 6.0 times the resistance of the other. Determine the resistance of each piece.

Answers

Answer:

1.5 ohm, 9 ohm

Explanation:

Total resistance, R = 10.5 ohm

Let the resistance of one piece is r and then the resistance of another piece is 10.5 - r.

According to question, the resistance of one piece is equal to the 6 times the resistance of another piece.

so, 6 r = 10.5 - r

7 r = 10.5

r = 1.5 ohm

Thus, the resistance of one piece is 1.5 ohm and resistance of another piece is 10.5 - 1.5 = 9 ohm.

Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° with the normal to the surface, while in the slab it makes an angle of 18.2° with the normal. What is the index of refraction of the transparent material?

Answers

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

[tex]n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}[/tex]

Where n values are indexes of refraction and [tex]\theta[/tex] values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

[tex]1*Sin(39.9)=n_{2}*Sin(18.2)[/tex]

Solving for [tex]n_{2}[/tex] yields:

[tex]n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053[/tex]

Remember to use degrees for trigonometric functions instead of radians!

What are the sign and magnitude of a point charge that produces an electric potential of −2.36 V at a distance of 2.93 mm?

Answers

Answer:

q = -7.691 × [tex]10^{-13}[/tex] C

so magnitude of charge is 7.691 × [tex]10^{-13}[/tex] C

and negative sign mean charge is negative potential

Explanation:

given data

electric potential = −2.36 V

distance = 2.93 mm = 2.93 × [tex]10^{-3}[/tex] m

to find out

What are the sign and magnitude of a point charge

solution

we know here that electric potential due to charge is

V = [tex]k\frac{q}{r}[/tex]       ..............................1

here k is coulomb force that is 8.99 ×[tex]10^{9}[/tex] Nm²/C² and r is distance and q is charge  and V is electric potential

put here all value we get q  in equation 1

V = [tex]k\frac{q}{r}[/tex]

-2.36 = [tex]8.99*10^{9} \frac{q}{2.93*10^{-3}}[/tex]

q = -7.691 × [tex]10^{-13}[/tex] C

so magnitude of charge is 7.691 × [tex]10^{-13}[/tex] C

and negative sign mean charge is negative potential

The electric potential is negative, indicating that the charge is negative. So, the point charge is a negative charge with a magnitude of [tex]\( 7.68 \times 10^{-12} \) C.[/tex]

To find the sign and magnitude of the point charge ( q ) that produces an electric potential of ( -2.36 ) V at a distance of [tex]\( 2.93 \) mm (\( 2.93 \times 10^{-3} \) m)[/tex], we can use the formula for the electric potential created by a point charge:

[tex]\[ V = \frac{k \cdot |q|}{r} \][/tex]

where:

- ( V ) is the electric potential (given as ( -2.36 ) V),

- ( k ) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]

- |q| is the magnitude of the point charge we want to find, and

- ( r ) is the distance from the point charge to the point where the electric potential is being measured (given as [tex]\( 2.93 \times 10^{-3} \) m[/tex]).

Rearranging the equation to solve for |q|, we get:

[tex]\[ |q| = \frac{V \cdot r}{k} \[/tex]]

Substituting the given values, we have:

[tex]\[ |q| = \frac{-2.36 \times 2.93 \times 10^{-3}}{8.99 \times 10^9} \][/tex]

[tex]\[ |q| = \frac{-2.36 \times 2.93 \times 10^{-3}}{8.99 \times 10^9} \][/tex]

[tex]\[ |q| \approx 7.68 \times 10^{-12} \, \text{C} \][/tex]

Now, the electric potential is negative, indicating that the charge is negative. So, the point charge is a negative charge with a magnitude of [tex]\( 7.68 \times 10^{-12} \) C.[/tex]

Your 300 mL cup of coffee is too hot to drink when served at 87.0 degrees Celsius. What is the mass of the ice cube, taken from a -15.0 degrees Celsius freezer, that will cool your coffee to a pleasant 65 degrees Celsius?

Answers

Final answer:

The mass of an ice cube that would bring coffee to 65 degrees Celsius from 87 degrees Celsius is determined via the principles of heat transfer. This involves considering the ice warming, phase change and water warming steps and setting this total heat to be equal to the heat lost by the coffee which results in calculating the mass of the ice.

Explanation:

To find out the mass of an ice cube that would cool down the coffee to 65 degrees Celsius from 87 degrees Celsius, we need to use the principles of heat transfer, specific heat capacities, and phase changes. Knowing that ice comes from a -15.0 degrees Celsius freezer it has to first warm up from -15C to 0C (icing warming), then melt at 0C (phase change), and the resulted water to be warmed from 0C to 65C (water warming). The sum of the heat absorbed during these three steps will be equal to the heat lost by the coffee.

During icing warming and water warming we use q=m*c*ΔT where m is the mass we don't know, c is the specific heat capacity (2.09 J/g°C for ice and 4.18 J/g°C for water) and ΔT is the change in temperature. During phase change we use q=m*L where m is the mass we are looking for and L is the heat of fusion of water 334 J/g.

Setting the sum of three heats equal to the heat lost by the coffee q=m*c*ΔT where mass is known (300g because density of coffee is similar to water's), c is again 4.18 J/g°C and ΔT is the change in temperature of the coffee, we can find out the mass of the ice cube needed.

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Final answer:

To cool the coffee from 87.0 degrees Celsius to 65 degrees Celsius, we can use the equation Q = mcΔT to calculate the mass of the ice cube needed.

Explanation:

To cool the coffee from 87.0 degrees Celsius to 65 degrees Celsius, we need to calculate the amount of heat transferred from the coffee to the ice cube. We can use the equation:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. We know the temperature change and the specific heat capacity of water. We can find the mass of the ice cube by rearranging the equation and plugging in the values:

m = Q / (cΔT)

Finally, we can calculate the mass of the ice cube and provide an answer to the question.

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Two tiny conducting spheres are identical and carry charges of -23.0C and +67.2C. They are separated by a distance of 3.18 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.18 cm. Determine the magnitude of the force that each sphere now experiences.

Answers

Explanation:

Given that,

Charge 1, [tex]q_1=-23\ C[/tex]

Charge 2, [tex]q_1=+67\ C[/tex]

Distance between charges, r = 3.18 cm = 0.0318 m

(a) Let F is the magnitude of force that each sphere experiences. The force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{-23\times 67}{(0.0318)^2}[/tex]

[tex]F=-1.37\times 10^{16}\ N[/tex]

(b) The spheres are brought into contact and then separated to a distance of 3.18 cm. When they are in contact, both possess equal charges. Net charge is :

[tex]q=\dfrac{q_1+q_2}{2}[/tex]

[tex]q=\dfrac{-23+67}{2}=22\ C[/tex]

Electric force is given by :

[tex]F=9\times 10^9\times \dfrac{22^2}{(0.0318)^2}[/tex]

[tex]F=4.307\times 10^{15}\ N[/tex]

Hence, this is the required solution.    

Estimate how fast your hair grows, in units of m/s, assuming it takes 30 days for your hair to grow 1 inch. note that 1 inch =2.54cm

2. how many liters (L) of water does it take to fill a swimming pool that is 15.0 feet long, 15.0 feet wide, and 8.00 feet deep? first compute the volume of the swimming pool in ft^3 and then follow the unit conversion rules to convert the units to L. note that 1ft=0.3048 meter and 1 L =1000cm^3

Answers

Answer:

1. 9.8 x 10^-7 cm/s

2. 50970 L

Explanation:

1.

time, t = 30 days = 30 x 24 x 60 x 60 seconds = 2592000 seconds

length of hair, d = 1 inch = 2.54 cm

rate of growth = length of hair grows per second = 2.54 / 2592000

                        = 9.8 x 10^-7 cm/s

Thus, the grown rate of hair is 9.8 x 10^-7 cm/s.

2. length of the pool, l = 15 feet

width of the pool, w = 15 feet

height of pool, h = 8 feet

Volume of the pool, V = length x width x height

V  = 15 x 15 x 8 = 1800 ft^3

To convert ft^3 into m^3 , we use

1 ft = 0.3048 m

so, 1 ft^3 = (0.3048)^3 m^3 = 0.0283 m^3

So, 1800 ft^3 = 1800 x 0.0283 = 50.97 m^3

now, 1 m^3 = 1000 L

So, 50.97 m^3 = 50.97 x 1000 L = 50970 L

Final answer:

Hair grows at approximately 9.8 x 10^-9 m/s based on the 30-day/1-inch growth rate. The swimming pool's volume is calculated by converting its dimensions into meters, then multiplying to find cubic meters and converting that volume to liters.

Explanation:

To estimate how fast your hair grows in meters per second, consider that 1 inch is equivalent to 2.54 cm, and there are 30 days in the given period. Hair growing 1 inch in 30 days is roughly 0.0254 meters in 2592000 seconds (30 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute), which results in a speed of approximately 9.8 x 10-9 meters per second.

For the swimming pool volume, the dimensions in feet need to be converted to meters: 15.0 feet is about 4.572 meters (since 1 foot = 0.3048 meters). Therefore, the pool volume in cubic meters would be 4.572 m x 4.572 m x 2.4384 m (since 8.0 feet is 2.4384 meters). To convert this volume to liters, note that 1 m3 is 1000 liters. So, we find the pool's volume in liters by multiplying the volume in cubic meters by 1000.

A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g salt/100 g of water) of the salt? b. How much water would it take to dissolve 25 g of this salt? c. If 10.0 g of this salt is mixed with 15.0 g of water, what percentage of the salt dissolves?

Answers

Answer:

The solubility of the salt is 35.16 (g/100 g of water).It would take 71.09 grams of water to dissolve 25 grams of salt.The percentage of salt that dissolves is 52.7 %

Explanation:

a.

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 35.16 \ g \ salt [/tex]

So, the solubility of the salt is 35.16 (g/100 g of water).

b.

Using the rule of three, we got:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}} [/tex]

[tex] x = 71.09 g \ water [/tex]

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

C.

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 5.27\ g \ salt [/tex]

This is the salt dissolved

The percentage of salt dissolved is:

[tex]percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 52.7 \% [/tex]

a. The solubility (in g salt/100 g of water) of the salt is 35.16 g/100 g water. b. Amount of water it would take to dissolve 25 g of this salt is 71.10 g. c. If 10.0 g of the salt is mixed with 15.0 g of water, 52.7% of the salt will dissolve.

To solve the problem, we need to determine the solubility of the salt at 25°C using the given data:

Part a: Solubility

The solubility of the salt is calculated as follows:A 3.20 g sample dissolves in 9.10 g of water to form a saturated solution.Solubility (g/100 g water) = (3.20 g salt / 9.10 g water) * 100 = 35.16 g/100 g water.

Part b: Amount of Water Needed to Dissolve 25 g of Salt

First, use the solubility obtained in part a:Solubility = 35.16 g salt / 100 g water.To find how much water is needed to dissolve 25 g of salt: (100 g water / 35.16 g salt) * 25 g salt = 71.10 g water.

Part c: Percentage of Salt Dissolved

Given 10.0 g of salt mixed with 15.0 g of water:Because we know the solubility is 35.16 g/100 g water, we find the amount that will dissolve in 15.0 g water: (35.16 g salt / 100 g water) * 15.0 g water = 5.27 g salt.Percentage dissolved = (5.27 g dissolved / 10.0 g total) * 100% = 52.7%.

Therefore, 52.7% of the 10.0 g of salt will dissolve in 15.0 g of water.

After your patched the roof, you dropped the hammer off the house and you heard it land on your toolbox on the ground 3.3s later. What is the height of the house ?

Answers

Answer:

53.36 m.

Explanation:

Initial velocity of the toolbox u = 0

acceleration due to gravity g = 9.8 m s⁻²

time t = 3.3 s

height of roof h = ?

h = ut + 1/2 g t²

= 0 + .5 x 9.8 x 3.3²

= 53.36 m.

Suppose the ends of a 27-m-long steel beam are rigidly clamped at 0°C to prevent expansion. The rail has a cross-sectional area of 35 cm^2. What force does the beam exert when it is heated to 39°C? (αsteel = 1.1 × 10^−5/C°, Ysteel = 2.0 × 10^11 N/m^2).

Answers

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

[tex]\Delta T = 39[/tex] Degree celcius

change in length of steel beam is given as

[tex]\Delta L = L_O \alpha \Delta T[/tex]

            [tex]= 20*1.1*10^{-5}*39[/tex]

           [tex] =8.58*10^{-3}[/tex] m

Young's modulus is

[tex]Y = \frac{FL}{A\Delta L}[/tex]

[tex]F = \frac{ YA\Delta L}{L}[/tex]

[tex]= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}[/tex]

F = 1.58*10^{11} N

Final answer:

The force exerted by the steel beam when it is heated to 39°C is approximately 9,191 N.

Explanation:

To calculate the force that the beam exerts when it is heated, we can use the formula for thermal expansion:

ΔL = αLΔT

where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.

In this case, the original length of the beam is 27m, the coefficient of linear expansion for steel is 1.1 × 10^−5/C°, and the change in temperature is 39°C - 0°C = 39°C.

Plugging in these values, we can calculate the change in length:

ΔL = (1.1 × 10^−5/C°)(27m)(39°C) = 0.011m

Next, we can calculate the change in cross-sectional area due to the expansion:

ΔA = αAΔT

where ΔA is the change in cross-sectional area, α is the coefficient of linear expansion, A is the original cross-sectional area, and ΔT is the change in temperature.

In this case, the original cross-sectional area of the beam is 35cm², the coefficient of linear expansion for steel is 1.1 × 10^−5/C°, and the change in temperature is 39°C - 0°C = 39°C.

Plugging in these values, we can calculate the change in cross-sectional area:

ΔA = (1.1 × 10^−5/C°)(35cm²)(39°C) = 0.015cm²

Next, we can use Hooke's Law to calculate the stress:

σ = F/A

where σ is the stress, F is the force, and A is the cross-sectional area.

In this case, the change in force is equal to the change in stress times the change in area:

ΔF = σΔA

Since the cross-sectional area changes and the force is equal to the change in force plus the initial force:

F = ΔF + σA

Substituting the values we know, we can calculate the force:

F = (σΔA) + (YAΔL)

where Y is the Young's modulus for steel.

Plugging in these values, we can calculate the force:

F = (σΔA) + (Y(0.011m)(35cm²))

Finally, we can calculate the force:

F = (σΔA) + (Y(0.011m)(35cm²)) = (σ(0.015cm²)) + (2.0 × 10^11 N/m²(0.011m)(35cm²))

After performing the calculations, we find that the force exerted by the beam when it is heated to 39°C is approximately 9,191 N.

A net external nonconservative force does positive work on
aparticle, and both its kinetic and potential energies change.
What,if anything, can you conclude about
(a) the change in the particle's total mechanical energy and
(b) the individual changes in the kinetic and
potentialenergies?
Justify your answers.

Answers

Answer with Explanation:

From work energy theorem we have

[tex]\Delta Energy=\int F.ds[/tex]

Now since it is given that the work done on the object by the force is positive hence we conclude that the term on the right hand of the above relation is positive hence [tex]\Delta Energy>0[/tex]

Part a)

Hence we conclude that the mechanical energy of the particle will increase.

Part b) Since the mechanical energy of the particle is the sum of it's kinetic and potential energies, we can write

[tex]\Delta K.E+\Delta P.E=\Delta E\\\\\therefore \Delta K.E+\Delta P.E>0[/tex]

Now since the sum of the 2 energies ( Kinetic and potential ) is positive we cannot be conclusive of the individual values since 2 cases may arise:

1) Kinetic energy increases while as potential energy decreases.

2)  Potential energy increases while as kinetic energy decreases.

Hence these two cases are possible and we cannot find using only the given information which case will hold

Hence no conclusion can be formed regarding the individual energies.

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angle of 110 with the positive xaxis. The resulta displacement has a magnitude of 131 cm and is directed at an angle of 38.0 to the positive axis. Find the magnitude and direction of the second displacement magnitude direction 1 (countercockwise from the positive x-axis)

Answers

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

[tex]D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }[/tex]

[tex]D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }[/tex]

D₂= 167.21 cm

Direction of the second displacement

[tex]\beta = tan^{-1} \frac{D_{y}}{D_{x} }[/tex]

[tex]\beta = tan^{-1} \frac{62.18}{155.22 }[/tex]

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

A proton accelerates from rest in a uniform electric field of 700 N/C. At one later moment, its speed is 1.10 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. ________ m/s^2 (b) Over what time interval does the proton reach this speed?_________ s (c) How far does it move in this time interval?________ m (d) What is its kinetic energy at the end of this interval?__________ J

Answers

Answer:

Explanation:

a ) Force = a charge q in an electric field E = q x E

Acceleration = Force / mass

Acceleration of proton

= q [tex]\frac{1.6\times10^{-19}\times700}{1.67\times10^{-27}}[/tex]E / m

= 6.7 x 10¹⁰ m /s²

b ) Initial speed u = 0 , Final speed v = 1.10 x 10⁶ m/s , acceleration a = 6.7 x 10¹⁰ m /s²

v = u + at

1.1 x 10⁶ = 0 +6.7 x 10¹⁰ t

t = 16.4 x 10⁻⁶ s

c ) s = ut + 1/2 at²

s = 0 +.5 x 6.7 x 10¹⁰ x (16.4x 10⁻⁶ )²

= 9.01 m .

d ) K E = 1/2 m v²

= .5 x 1.67 x 10⁻²⁷x (1.1 x 10⁶ )²

= 10⁻¹⁵ J

You throw a ball straight down from an apartment balcony to the ground below. The ball has an initial velocity of 5.10 m/s, directed downward, and it hits the ground 1.92 s after it is released. Find the height of the balcony.

Answers

Answer:

the height of the balcony from where the ball is thrown is 27.874 m.

Explanation:

given,

initial velocity (u) = 5.1 m/s

time (t) = 1.92 s

height of balcony = ?

using equation;          

[tex]s = u t + \dfrac{1}{2} at^2[/tex]

[tex]h =5.1 \times 1.92 + \dfrac{1}{2}\times 9.81\times 1.92^2[/tex]

h= 9.792 + 18.082                      

h = 27.874 m

hence, the height of the balcony from where the ball is thrown is 27.874 m.

Final answer:

Using the kinematic equations of motion, we find that the ball was thrown from a height of approximately 8.211 meters.

Explanation:

To calculate the height from which the ball was thrown, we use the kinematic equations of motion for an object under constant acceleration due to gravity. The specific equation that relates the initial velocity (Vi), time (t), acceleration due to gravity (g), and the displacement (height h in this case) is:

h = Vi * t + 0.5 * g * t2

Where:

Vi is the initial velocity = 5.10 m/s (downward, so we take it as negative)t is the time = 1.92 sg is the acceleration due to gravity = 9.81 m/s2 (downward, so we take it as positive)

Plugging in these values, we get:

h = -5.10 m/s * 1.92 s + 0.5 * 9.81 m/s2 * (1.92 s)2

h = -9.792 m + 18.003 m

h = 8.211 m

Therefore, the height of the balcony is approximately 8.211 meters above the ground.

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