I need help with physics question.

Answer:

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Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

Explanation:

The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

[tex]y:\:\:\:\:N - mg = 0[/tex]

[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]

or

[tex]m \dfrac{v^2}{r} = \mu mg[/tex]

Solving for [tex]\mu[/tex],

[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]

I am guessing wind is caused by climate change in the atmosphere

Explanation:

wind is cause by climate change in the atmosphere that depends weather is is breezy really cold or rain and cold

Answer:

caused by the uneven heating of the Earth by the sun and the  own rotation.

Answer:

The appropriate answer is "3761.69 N/C".

Explanation:

Given that:

Mass,

m = 4.99 g

or,

   = [tex]4.99\times 10^{-3} \ kg[/tex]

Charge,

q = 13 µC

or,

  = [tex]13\times 10^{-6} \ C[/tex]

As we know,

⇒ [tex]F=mg=Eq[/tex]

then,

⇒ [tex]E=\frac{mg}{q}[/tex]

By putting the values, we get

        [tex]=\frac{4.99\times 10^{-3}\times 9.8}{13\times 10^{-6}}[/tex]

        [tex]=3761.69 \ N/C[/tex]

Answer: hello your question is incomplete below is the missing part

A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.

answer:

To 2 significant Figures = 1.6 m/s^2

Explanation:

Calculate the magnitude of minimum acceleration at which the thief can descend

we apply the relation below

Mg - T = Ma  --- ( 1 )

M = 69kg

g = 9.81

T = 58 * 9.81

a = ? ( magnitude of minimum acceleration)

From equation 1

a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69

  = 1.5639 m/s^2

To 2 significant Figures = 1.6 m/s^2

Answer:

[tex]a=2.8*10^{-9}m/s[/tex]

Explanation:

From the question we are told that:

First Mass [tex]m=8.50kg[/tex]

2nd Mass [tex]m=14.5kg[/tex]

Distance

[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]

Generally the Newtons equation for Gravitational force is mathematically given by

[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]

Therefore

Initial force on m

[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]

Final force on m

[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]

Acceleration of m

[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]

[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]

[tex]a=2.8*10^{-9}m/s[/tex]

Answer:

[tex]t=1.9 sec[/tex]

Explanation:

From the question we are told that:

Height [tex]h=28m[/tex]

Time [tex]t=3s[/tex]

Generally the Newton's equation for Initial velocity upward is mathematically given by

 [tex]s=ut+\frtac{1}{2}at^2[/tex]

 [tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]

 [tex]u=24.03m/s[/tex]

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 [tex]v=\sqrt{24.03-2*9.8*28}[/tex]

 [tex]v=5.35m/s and -5.35m/s[/tex]

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 [tex]v=u+at[/tex]

 [tex]5.35=24.03-9.8t[/tex]

 [tex]t=\frac{28.03-5.35}{9.8}[/tex]

 [tex]t=1.9 sec[/tex]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

[tex]Mv+mu=Mv_1+mv_2[/tex]

(900 x 47) + (200 x -30)  = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

36300 =  (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)

[tex]9v_1 = 363 - 2v_2[/tex]

[tex]v_1=\frac{363 - 2v_2}{9}[/tex]

The mathematical expression for the conservation of kinetic energy is

[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]

[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex]    ................(ii)

[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]  

[tex]21681 = 9v_1^2+2v_2^2[/tex]

Substituting (i) in (ii)

[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]

[tex](363-2v_2)^2+18v_2^2=195129[/tex]

[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]

[tex]22v_2^2-145v_2-63360=0[/tex]

Solving the equation, we get

[tex]v_2=96 \ m/s, -30 \ m/s[/tex]

The negative velocity is neglected.

Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get

[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]

     = 19

Thus, only impulse of importance is used to find final velocity.

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]

Height of helicopter [tex]h=105\ m[/tex]

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]

So, package will take 5.21 s to reach the ground

Answer:

The energy of a photon is 2.94x10⁻¹⁹ J.

Explanation:

The energy of the photon is given by:

[tex] E = \frac{hc}{\lambda} [/tex]  

Where:

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

λ: is the wavelength = 675 nm

Hence, the energy is:

[tex] E = \frac{hc}{\lambda} = \frac{6.62 \ccdot 10^{-34} J.s*3.00 \cdot 10^{8} m/s}{675 \cdot 10^{-9} m} = 2.94 \cdot 10^{-19} J [/tex]

Therefore, the energy of a photon is 2.94x10⁻¹⁹ J.

I hope it helps you!

Answer:2

Explanation:

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s²          { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.

Convex lens is the answer.

See the attached diagram.

Answer:

0.913

Explanation:

k.e=1/2mv square

k.e=1/2×0.078g×23.4256m/s square

k.e=0.913J

The kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).

To calculate the kinetic energy (KE) of the lemming when it is 2.00 m above the ground, we need to consider the change in its potential energy (PE) as it falls.

The potential energy at a height h is given by:

PE = m g h

Where:

m is the mass of the lemming (0.0780 kg)

g is the acceleration due to gravity (9.8 m/s²)

h is the height above the ground

Given:

Height of the cliff (h) = 5.36 m

Velocity of the lemming (v) = 4.84 m/s

Height above the ground (h') = 2.00 m

The lemming will lose potential energy as it falls from the cliff, which is converted into kinetic energy. Therefore, the kinetic energy when it is 2.00 m above the ground is equal to the difference between its total initial kinetic energy and the potential energy at that height.

Initial potential energy at the top of the cliff:

PE_initial = m g h

Potential energy when it is 2.00 m above the ground:

PE_final = m * g * h'

The change in potential energy is given by:

ΔPE = PE_final - PE_initial

The kinetic energy (KE) when it is 2.00 m above the ground:

KE = ΔPE = -ΔPE (due to energy conservation)

Let's calculate the potential energy at the top of the cliff and when it is 2.00 m above the ground:

PE_initial = m ×g × h

= 0.0780 kg × 9.8 m/s² × 5.36 m

PE_initial ≈ 4.09 J

PE_final = m ×g × h'

= 0.0780 kg ×9.8 m/s² ×2.00 m

PE_final ≈ 1.53 J

The change in potential energy (ΔPE) is:

ΔPE = PE_final - PE_initial = 1.53 J - 4.09 J

ΔPE ≈ -2.56 J

Since the change in potential energy is equal to the kinetic energy, the kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).

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Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

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Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.

1) Can change the state of an object : For example, pushing a heavy stone in order to move it.

2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.

3) May change the direction of motion of an object.

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Answer:

The mass decreases

Explanation:

Just smart

Answer:

the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

         X = X_c - X_L

         X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

         X_c = X_L

         1 / wC = w L

         w² = 1 / CL

         w = [tex]\sqrt{\frac{1}{CL} }[/tex]

let's look for the extreme values

C = 1  10⁻¹² F

         w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]

         w = [tex]\sqrt{1 \ 10^{18}}[/tex]

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6

         w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018

         w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

           w = 2π f

           f = w / 2π

the light velocity  is

           c = λ f

           λ = c / f

we substitute

          λ = c 2π/w

we calculate the two values

 C = 1 pF

          λ₁ = 3 10⁸ 2π / 10⁹

          λ₁= 18.849 10⁻¹ m

          λ₁ = 1.88 m

C = 10 pF

           λ₂ = 3 10⁸ 2π / 0.316 10⁹

           λ₂ = 59.65 10⁻¹ m

           λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

Answer:

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Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]

The work done by gravity is:

[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]