A sled having a certain initial speed on a horizontal surface comes to rest after traveling 20 m. If the coefficient of kinetic friction between the object and the surface is 0.20, what was the initial speed of the object

Answer: 6,277,647m

Explanation:

Radius of Earth = 6400km

To calculate the gravitational acceleration of a planet, we use the following formula:

g = mG/r^2

Gravitational acceleration is equal to the mass of the planet multiplied by the gravitational constant all divided by the radius of the planet squared.

We already know what the gravitational acceleration will be, 2.45m/s^2.

So, 2.45m/s^2 = mG/r^2

the mass of the earth is equal to 5.9*10^24.

And the gravitational constant is equal to 6.67408 * 10^-11.

We don't know the radius though.

2.45m/s^2 = 5.9*10^24 * 6.67408 * 10^-11 divided by r^2

2.45m/s^2 = 3.93 * 10^14 divide by r^2

Now, we can cross-multiply.

2.45m/s^2 * r^2 = 3.93 * 10^14

divide r^2 from both sides.

r^2 = 3.93 * 10^14 divided by 2.45m/s^2

r^2 = 1.6*10^14.

Now, take the square root of both sides.

r = 12,677,647 meters from the center of the Earth.

To calculate the height from the surface of the Earth, we need to subtract r by the Earth's radius.

That will be 12,677,647-6.400.000m = 6,277,647m from the surface of Earth.

The mass M of the star is mathematically given as

M=24.269kg

Question Parameters

Orbital period P= 14.3*24*60*60 s

P= 1235520s

Speed V=160000

Generally, the equation for the Radius is  mathematically given as

R =[tex]\frac{ V*t}{ 2*pi}[/tex]

Therefore

R =[tex]\frac{ ( 160000 )*(1235520 s)}{ 2*3.142}[/tex]

R=[tex]3.14581795*10^{10}[/tex]

Where

M*v^2/R = GM^2/(2R)^2

Therefore

M=[tex]=4(v^2)R/G\\\\ =4(240^2)R/G\\\\ = \frac{4*(160000)^2)*3.14581795*10^{10}}{6.67*10^{-11}}[/tex]

M=4.82956159*10^{31}kg

Considering, that the mass of our Sun is 1.99*10^{30} kg.

The mass M of the star is

M[tex]=\frac{4.82956159*10^{31}}{1.99*10^{30}}[/tex]

M=24.269

In conclusion, the mass M of the star is

M=24.269

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The width of the central diffraction peak is 0.0754 m.

What is central diffraction:

It is distance between the 1st order minima from the center of the screen on both sides of the center.

The first diffraction minimum from the center is given as:

where

y is the location of first minimum diffraction.

Distance of screen from slit (L) = 2.40 m

Width of the slit (a) = 0.0328 mm = 0.0328 × 10⁻³ m

Wavelength of light (λ) = 588 nm = 588 × 10⁻⁹ m

Width of central diffraction peak is,

Now, plug in the values given and solve for width 'W'.

This gives,

W = 2 * 2.40 * 588 × 10⁻⁹ / ( 0.0328 × 10⁻³ )

W = 0.0860 m

Therefore,

The width of the central diffraction peak is 0.0754 m.

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Explanation:

there are 4 phases in which matter could be classified.

1. Solid: an example of this would be the metal copper.

2. Liquid: an example of this would be H2O or water.

3. Gas: an example of this would be oxygen gas or O2

4. Plasma: an example of this would be lightning.

hope this helps :)

Answer:

 W = 1960 N

T = 1960 N     in the upward direction

Explanation:

Let's apply Newton's second law to this exercise with acceleration equal to zero, translational equilibrium

Weight is

            W = m g

            W = 200 9.8

            W = 1960 N

In the attachment we can see a diagram

             T -W = 0

             T = W

             T = 1960 N

in the upward direction

credit - somebody else

The ratio of the mass of copper to water will be 10.8

The quantity of heat energy needed to raise a body's temperature per unit of mass is known as specific heat. Other names for specific heat include mass specific heat and specific heat capacity.

The quantity of heat in joules needed to elevate one gram of a material by one Kelvin is known as specific heat in SI units (symbol: c). The standard way to report specific heat is in joules (J).

We know that

The product of mass of copper and mass of water is equal to the product of the specific heat of copper and Water

Therefore ,

[tex]m_wc_w = m_cc_c[/tex]

We know that

Specific heat of water is 4.18 J

Specific heat of copper is 0.39 J

Therefore,

[tex]m_c*(0.39)=m_w*(4.18)\\\frac{m_c}{m_w} = \frac{4.18}{0.39}\\ = 10.8[/tex]

Hence  The ratio of the mass of copper to water will be 10.8

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After 5 seconds the velocity of the projectile is 71m/s

Given that Initial velocity is 120m/s

Time t = 5s

We need to find the velocity after 5 sec

Projectile motion is the movement of an item hurled or projected into the air, which is only subject to the acceleration of gravity. The trajectory of the object is referred to as its trajectory.

The horizontal velocity of a projectile is constant,

There is a vertical acceleration caused by gravity and  its value is 9.8 m/s

Now the velocity for 5 seconds

[tex]v(t)=v_0-gt\\[/tex]

[tex]v(t)=120-(9.8)(5)\\\\v(t) =71m/s[/tex]

Hence the velocity after 5 seconds is 71m/s

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20 m/s is the speed of the heavier fragment just after the explosion.

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

F = ma                   where, F = Force (Newton)

                                          m= mass

                                          a = acceleration

Given:

mass of rocket = M = 1500 kg

acceleration of rocket = a = 10 m/s²

elapsed time = t = 2.00 s

mass of lighter fragment = m₁ = m = 500 kg

mass of heavier fragment = m₂ = 2m = 1000 kg

maximum height of lighter fragment = h = 530 m

Let's calculate the final speed of the rocket just before the explosion:

v = u + at

v = 0 + 10(2)

v = 20 m/s

Then, we will calculate the height of the rocket just before the explosion:

[tex]h' = ut + \frac{1}{2}at^{2}[/tex]

[tex]h' = 0 + \frac{1}{2} (10)(2.00)^{2}[/tex]

[tex]h' =20m[/tex]

The initial speed of lighter fragment just after the explosion:

[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex]Δ[tex]h[/tex]

[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex] [tex](h - h')[/tex]

[tex]0^{2} = u^{2}[/tex]₁[tex]- 2(9.8) (530-20)[/tex]

[tex]u^{2}[/tex]₁[tex]=9996[/tex]

[tex]u[/tex]₁ [tex]=\sqrt{9996} m/s[/tex]

Using Conservation of Momentum Law :

[tex]M v=m[/tex]₁ [tex]u[/tex]₁ [tex]+ m[/tex]₂[tex]u[/tex]₂

[tex]1500 (20)= 500(\sqrt{9996} ) + 1000u[/tex]₂

[tex]u[/tex]₂ ≅ [tex]- 20 m/s[/tex]

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Answer:

Absorbed solar radiation undergoes irreversible degradations, transferring from reservoir to another converting from light energy to radiation energy.

Explanation:

The solar energy cycle functions in the following form:

Therefore, absorbed solar radiation undergoes irreversible degradations, transferring from reservoir to another converting from light energy to radiation energy.

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The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

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The kinetic energy gained by the air molecules is 0.0437 J                

Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

[tex]P =mgh[/tex]

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

[tex]E= \frac{1}{2} mv^{2}[/tex]

[tex]E= \frac{1}{2}[/tex]×[tex]1.4[/tex] × 10⁻³ × (0.7)²

E = 0.000343 J

The  kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 -  0.0441

  = 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

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The friction force between the box and the incline if the box does not slide down the incline will be 0.577

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

We have to find the friction force between the box and the incline if the box does not slide down the incline

Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:

F₁ = F₂

μmgcosΘ = mgsinΘ

μ = (mgsinΘ)/(mgcosΘ)

μ = tanΘ

μ = 0.577

Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577

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:The resistance created by the box's friction with the incline The slope will be 0.577 if the box doesn't descend.

Friction is the force that prevents solid surfaces, fluid layers, and material constituents from sliding past one another. Different types of friction exist. Dry friction opposes the relative lateral motion of two touching solid surfaces.

Given that the 20-kg box lying on the inclined plane above and the incline's 30 degree inclination

If the box does not slide down the incline, we must determine the friction force between them.

Given that the gravitational force F2 down the gradient must be equal to or greater than the frictional force F1,

F₁ = F₂

If mgcosx = mgsinx

equals (mgsinx)/(mgcosx)

μ = tanx

μ = 0.577

Therefore, if the box does not slide down the incline, the friction force between the box and the incline will be 0.577.

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The average velocity of the jet during the diversion is 518.75 m/s.

The displacement of the pilot from the starting position is calculated as follows;

d² = 20² + 60² - 2(20)(60) x cos(θ)

where;

θ = (90 - 38) + (90 - 55) = 87⁰

d² = 20² + 60² - 2(20)(60) x cos(87)

d² = 3874.39

d = 62.25 km

V = total displacement/total time

where;

V = (62.25 km)/(0.033 hr)

V = 1,867.5 km/hr = 518.75 m/s

Thus, the average velocity of the jet during the diversion is 518.75 m/s.

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A current magnetic relay is often used to assist in the starting of the compressor, but most new appliances use positive temperature coefficient (PTC) device.

A compressor can be defined as a mechanical device that is designed and developed to provide power to refrigerators, especially by increasing the pressure on air or other applicable gases.

According to heating, ventilation, and air conditioning (HVAC), a current magnetic relay is mostly used to assist in the starting of the compressor, but many new electrical appliances use positive temperature coefficient (PTC) device.

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The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz

Given Conditions:

mass of string; m = 0.0133 kg

Force on the string; F = 8.89 N

Length of string; L = 1.97 m

1. To find the wavelength at the fourth normal node.

At the fourth harmonic, there will be 2 nodes.

Thus, the wavelength will be;

λ = L/2

λ = 1.97/2

λ = 0.985 m

2. To find the velocity of the wave from the formula;

v = √(F/(m/L)

Plugging in the relevant values gives;

v = √(8.89/(0.0133/1.97)

v = 36.2876 m/s

Now, formula for frequency here is;

f = v/λ

f = 36.2876/0.985

f = 36.84 Hz

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The resultant wave at point (5, 2) is Ψ = 5.99 cos [ 7.15 - (20/s) t]

What are the plane waves:

Here,

Two plane waves are given:

c (5, t) = 4 cos [(8π/3) - (20/s) t]

c (2, t) = 2 cos [(3π/2) - (20/s) t]

now, the waves as imaginary exponentials,

separating the spatial parts, and then adding them together

we get The resultant:

Ψ = [  4 sin (8π/3) +  2 sin (3π/2) ]^2 +  [ 4 cos(8/3 π) +  2 cos(3/2π) ]^2

Ψ = 5.99 tan(a) = 0.747/ 5.95

a = 7.15

Ψ = 5.99 cos [ 7.15 - (20/s) t]

hence,

The resultant wave is Ψ = 5.99 cos [ 7.15 - (20/s) t]

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Your question is incomplete, but most probably the full question was:

Two plane waves with the same frequency and with vibrations (measured by Psi) in the z-direction are given by c (x, t) = (4cm.) cos [pi/3cm. x - 20/s t + pi] c (y, t) = (2cm.) cos[pi/4cm. y - 20/s t + pi]

Express the waves as imaginary exponentials, separate the spatial parts, and add them together using a phasor diagram to find the resultant at the point x = 5cm. y = 2cm

The deflection angle is 37.29º if the prism was made of polystyrene plastic.

What is a laser beam?

Laser beam is a light beam propagating dominantly in one direction.

It is a beam of radiation produced from a laser.

Here given that,

refraction index of glass, n1= 1.52

refraction index of polystyrene plastic, n2 = 1.59

deflection angle, B = 22.6º

For the second surface,

B  = 45º - 22.6º

B  = 22.40º

Now from the formula of Snell's law:

n1 sinФ = n2 sin B

sin B / sinФ = n1 / n2

sinФ = (sin 22.40º) * ( 1/ 1.59)

Ф = 37.29º

Hence,

The deflection angle is 37.29º if the prism was made of polystyrene plastic.

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In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

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Answer:

The answer is B

Explanation:

The section C of the heating curve represents the liquid phase, which is where liquid water is heated. Hence option D is correct.

The body or particle is considered to be heating up as the temperature rises.

100°C is the boiling point of water at 1 atm of pressure.

The term "enthalpy of vaporisation" refers to the amount of heat energy needed to change liquid water into steam.

Enthalpy of vaporisation has a greater value than enthalpy of fusion.

The amount of power needed to melt ice into liquid water is known as the enthalpy of fusion.

As a result, section C illustrates the heating of liquid water.

Hence option D is correct.

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Answer:

Most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.

What are the characteristics of d-block elements?

Therefore, most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.

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The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

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Answer:

Explanation:

On return to Earth, gravity once again “pulls” the blood and fluids into the abdomen and legs. The loss of blood volume, combined with atrophy of the heart and blood vessels that can occur in space, reduces the ability to regulate a drop in blood pressure that happens when we stand on Earth.

May have to swing wide to complete the right turn

In any vehicle which makes any turn, the rear wheels follow a shorter path as compared to the front wheels. While the truck being larger than any usual vehicle, the distance between the wheels are even more, which makes it harder to complete the turn by going in its own lane or the other two right lanes.

So, the truck will try to go wide in left and then complete the right turn as by doing this the distance between the rear and front will get compensated by the width of the lanes making it easier for the truck to complete its turn.

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[Your question is incomplete, but most probably your full question was

May not turn without engaging its hazard lights

May complete its turn in either of the two lanes

Must stay in the right lane at all times while turning

May have to swing wide to complete the right turn]

a)The total distance she walks will be 1100 m.

b)The magnitude and direction of the budgie's path will be 854.4 m at 69.44°.

c)Sally's average speed for the journey will be 2.82 m/sec

d)The average velocity of Sally's budgie willl be 2.19 m/sec North of East.

Displacement is defined as the shortest distance between the two points.

Distance travelled in north direction,d₁ = 8.0 x 10² m

Time elapsed  travelled in north direction,t₁ = 5.0 minutes

Distance travelled in east direction,d₂ = 0.30 km = 3.0 × 10² m

Time elapsed  travelled in east direction,t₂= 1.5 minutes

The total distance she walks,d =? m

The magnitude and direction of the budgies path will be, D

Sally's average speed for the journey, [tex]\rm S_{avg}[/tex]

The average velocity of Sally's budgie,[tex]\rm V_{avg}[/tex]

a)

The total distance she walks is found as;

d=d₁+d₂

d=800 m +300 m

d = 1100 m

b)

The magnitude and direction of the budgies path will be;

[tex]\rm D = \sqrt{d_1 ^2 + d_2^2 } \\\\ D = \sqrt{800^2 +300^2 } \\\\ D = 854.4 \ m[/tex]

The direction of the displacement is;

[tex]\rm \theta = tan^{-1} \frac{800}{300} \\\\ \theta = tan^{-1}(2.6) \\\\ \theta = 69.44 ^ 0[/tex]

c)journey is found as;

Sally's average speed for the

[tex]\rm S_{avg} = \frac{d}{t} \\\\ S_{avg} =\frac{100}{300+900} \\\\ S_{avg} =2.82 \ m/sec[/tex]

d)

The average velocity of Sally's budgie is found as;

[tex]\rm V_{avg} = \frac{854.4}{300+ 900 } \\\\ V_{avg} = 2.19 \ m/sec[/tex] North of east

Hence, the total distance she walks, the magnitude and direction of the budgie's path, Sally's average speed, and the average velocity of Sally's budgie will be 1100 m,854.4 m at 69.44°,2.82 m/sec, and 2.19 m/sec North of East.

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Light bulbs 1, 2, and 4 will continue to shine when Linh switches off bulb 3.

The battery-powered electrical circuit where a voltage differential is generated across a capacitor, a resistor, or an inductor.

As soon as Linh cuts the middle branch, the bulbs in positions 1, 2, and 4 will function as a resistor in a series circuit. Except for resistor number 3, the current through the other resistors will be constant.

In conclusion, if Linh switches off Bulb 3, only Bulbs 1, 2, and 4 will still be on.

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Answer:

1. no

2. yes

Explanation:

on edge 2022

Answer:

Cosmic ray is not considered electromagnetic radiation.

Explanation:

Electromagnetic radiation refers to

X rays are the rays produced when a negatively charged electrode is heated by electricity and electrons are released, thereby producing energy. It is a type of radiation called EM waves.

Radio wave are wave from the portion of the electromagnetic spectrum at lower frequencies than microwaves. It is an EM wave.

Ultraviolet wave Invisible rays that are part of the energy that comes from the sun. It is an example of EM wave.

Cosmic ray is a high-speed particle either an atomic nucleus or an electron that travels through space.

Cosmic ray is not an electromagnetic wave.

Hence

Cosmic ray is not considered electromagnetic radiation

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The skater's final angular speed is equal to 12 rad/s.

When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.

The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

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The magnitude of other charge will be 1 × 10⁻² coulomb

The formula of electrostatic force is

Electrostatic force = K q1 q1 / r²

where k is the coulomb's constant whose value is 9 × 10⁹

q1 and a2 are the magnitude of charges

and r is the distance between them

magnitude of the force given to us is 9.0 × 10⁻⁵ newtons

magnitude of one charge = 1.0 × 10⁻⁶ coulomb

Force = K q1 q2 / r²

9.0 × 10⁻⁵ = ( ( 9 × 10⁹ ) × ( 1.0 × 10⁻⁶ ) × q2 ) / 1

9.0 × 10⁻⁵ =  9 × 10³ × q2  

10⁻² = q2  

Charge on q2 is 1 × 10⁻² coulomb

So the magnitude of the second charge is came out to be 1 × 10⁻² coulomb after applying the formula of electrostatic force.

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Answer:

Approximately [tex]2.05\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

The net force on the girl would be:

[tex]\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}[/tex].

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be [tex]123.0\; {\rm N}[/tex]. The girl would exert a reaction force on the rope at the same magnitude ([tex]123.0\; {\rm N}\![/tex]) in the opposite direction. This force would translate to a [tex]123.0\; {\rm N}\!\![/tex] force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be [tex]123.0\; {\rm N}[/tex]. Since the mass of the boy is [tex]m(\text{boy}) = 60\; {\rm kg}[/tex], the acceleration of the boy would be:

[tex]\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

Answer:

graph of cosine function

Explanation:

f(x) = sin(x)

The curve of the sine function crosses the y-axis at the origin, heads up to y = 1 at x = π/2 and down to y = -1 at x = 3π/2.

Domain: (-∞, ∞)

Range: [-1, 1]

f(x) = cos(x)

The curve of the cosine function crosses the y-axis at (0, 1) which is its maximum, heads down to y = -1 at x = π then heads up again.

Domain: (-∞, ∞)

Range: [-1, 1]

Given information:

If the wheel starts when time t = 0 seconds, and they begin their journey at the top of the wheel (the maximum point), the most appropriate graph to model their ride is the graph of the cosine function since it is at its highest point when x = 0.