A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times greater is the sound intensity of the train whistle than that of the library?

The energy  stored in the spring when it is compressed 20.0 cm is 15.54J, the maximum height of the ball is 1.056 meters. while the muzzle velocity of the ball is  4.55 m/s.

a) Potential Energy in the Spring

Potential energy = 1/2 * spring constant * (compression distance)^2

Potential energy = 1/2 * 777 N/m * (0.2 m)^2 = 15.54 J

b)  Maximum Height of the Ball

Maximum height = potential energy / weight of the ball

Maximum height = 15.54 J / (1.5 kg * 9.81 m/s^2) = 1.056 m

Therefore, the maximum height reached by the ball is 1.056 meters.

(c) Muzzle Velocity of the Ball

muzzle velocity = sqrt(2 * potential energy / mass)

muzzle velocity = sqrt(2 * 15.54 J / 1.5 kg) = 4.55 m/s

Hence, the muzzle velocity of the ball is 4.55 m/s.

A) The energy stored in the spring when compressed is 15.54 J. B) The maximum height reached by the ball is 1.05 m. C) The muzzle velocity of the ball is 4.56 m/s.

A) The energy stored in a spring, also known as elastic potential energy, can be calculated using the formula:

Espring = ½ * k * x2

where k is the spring constant, and x is the compression distance. By plugging in the given values (k = 777 N/m and x = 0.20 m), we get:

Espring = ½ * 777 N/m * (0.20 m)2 ≈ 15.54 J

B) At the maximum height, all the elastic potential energy is converted into gravitational potential energy (Egravitational), given by:

Egravitational = m * g * h

Where m is the mass of the ball, g is the acceleration due to gravity (9.81 m/s2), and h is the height. Since Espring = Egravitational, we can solve for h:

h = Espring / (m * g) = 15.54 J / (1.5 kg * 9.81 m/s2) ≈ 1.05 m

C) The muzzle velocity of the ball can be found using the conservation of energy principle. The elastic potential energy stored in the spring is converted entirely into the kinetic energy of the ball at the spring's equilibrium position:

Kinetic energy, Ekinetic = ½ * m * v2

Since Espring = Ekinetic, solving for v gives:

v = √(2 * Espring / m) = √(2 * 15.54 J / 1.5 kg) ≈ 4.56 m/s

Answer:

Centripetal force, F = 9.81 N

Explanation:

It is given that,

Mass of the object (yo-yo), m = 0.12 kg

It is whirled around by its string in a circle

Radius of circle, r = 0.44 m

Speed of object, v = 6 m/s

We need to find the required centripetal force in Newtons to keep this yo -yo whirling in a circle. The centripetal force is given by :

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

[tex]F=\dfrac{0.12\ kg\times (6\ m/s)^2}{0.44\ m}[/tex]

F = 9.81 Newton

So, the centripetal force required to keep this yo -yo whirling in a circle is 9.81 N. Hence, this is the required solution.

Answer:

a) The velocity is zero when the object is at the highest vertical point (peak). b) Yes, at the peak, the velocity for an instant gets to zero and then the direction of motion and the velocity will be downwards.

C) No, the direction of ‘g’ will be opposite in direction when the object is on the way down when compared to what it was during the upward motion.

Explanation:

a) Since gravity is acting on the object and trying to pull it down to earth, the ball will slow down to a velocity of zero when the ball starts its free fall downward.

b) When the direction of the object changes, its velocity will also change direction.

c) When the ball has a positive velocity, gravity is pulling it in the opposite direction, so it has a negative acceleration trying to slow it down. When the ball's velocity is negative (downward free fall), the acceleration will be positive at the instant when the ball hits the ground. The only way to stop the ball from falling is a positive force upward, which results in a positive acceleration when the ball hits the ground.

Hope this helps :)

(a) The velocity of the object will be zero at the maximum height

(b) The velocity of the object changes direction.

(c) The acceleration due to gravity will not have the same sign on the way up as on the way down.

(A) The velocity of the object decreases as the object ascends upwards and eventually becomes zero at the maximum height. As the object descends downwards, the velocity increases and becomes maximum before the object hits the ground.

(B) The velocity of the object is always directed in the direction of the motion of the object. As the object moves upward, the direction of the object is upward and as the object moves downwards the direction of the object is downwards. Thus, the velocity of the object changes direction.

(c) The upward motion of the object is opposite direction to acceleration due to gravity and the sign of acceleration due to gravity becomes negative.

As the object moves downwards, it will be moving in the same direction as the acceleration due to gravity. The sign of acceleration due to gravity is positive.

Thus, the acceleration due to gravity will not have the same sign on the way up as on the way down.

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Answer:

37.91594 keV

Explanation:

[tex]E_i[/tex] = Incident energy = 400 keV

θ = 30°

h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s

Incident photon wavelength

[tex]\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m[/tex]

Difference in wavelength

[tex]\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m[/tex]

[tex]\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}[/tex]

Final photon wavelength

[tex]\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV[/tex]

Energy of the recoiling electron

[tex]\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV[/tex]

Energy of the recoiling electron is 37.91594 keV

Answer:

The energy of recoiling electron=192.44 keV

Explanation:

Energy of x-ray[tex]E_o=400 keV[/tex]

Web know that compton shift is

[tex]\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}[/tex]

[tex]m_e[/tex] is the mass of electron and C is the velocity of sound.

Given that θ=30°

Now by putting the values

[tex]\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}[/tex]

[tex]\Delta \lambda =2.8\times 10^{-3}nm[/tex]

[tex]\Delta \lambda_o=\dfrac{hc}{E_o}[/tex]

By putting the values

[tex]\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}[/tex]

[tex]\Delta \lambda_o=3.31times 10^{-3}nm[/tex]

[tex]\lambda =\Delta \lambda +\lambda _o[/tex]

[tex]\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm[/tex]

[tex]\lambda=6.11`\times 10^{-3}nm[/tex]

Energy [tex]E=\dfrac{hC}{\lambda }[/tex]

So [tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}[/tex]

E=207.55 keV

The energy of recoiling electron=400-207.55 keV

The energy of recoiling electron=192.44 keV

To find the speed at the bottom of the ramp, we can use the principles of physics. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp. Using the given values and trigonometry, we can calculate the height of the ramp and then determine the velocity.

To find the speed of the skateboarder at the bottom of the ramp, we can use the principles of physics. Since we are neglecting friction, we can assume that the mechanical energy of the skateboarder is conserved. The mechanical energy consists of both kinetic energy and potential energy. At the top of the ramp, the skateboarder has only potential energy, which is given by the formula: potential energy = mass * gravity * height. At the bottom of the ramp, the skateboarder has only kinetic energy, which is given by the formula: kinetic energy = 1/2 * mass * velocity^2. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp.

In this case, the mass of the skateboarder is not provided, but it cancels out when equating the potential and kinetic energies. So we can assume any mass value as long as it is consistent throughout the calculation. Let's assume the mass of the skateboarder is 1 kg for simplicity. The height of the ramp is not provided, but we can calculate it using trigonometry. The height is the vertical component of the ramp's length, which is given by: height = length * sin(angle). Plug in the values and calculate the height.

Next, we calculate the potential energy at the top of the ramp using the calculated height. The potential energy is then equated to the kinetic energy at the bottom of the ramp and solve for the velocity. Plug in the values and calculate the velocity at the bottom of the ramp.

Answer:

443.3 N

567.4 N

Explanation:

Consider the triangle ABC

AC = hypotenuse = magnitude of force vector = F = 720 N

AB = adjacent = Component of force along east-west line = [tex]F_{x}[/tex]

BC = Opposite = Component of force along north-south line = [tex]F_{y}[/tex]

θ = Angle = 38 deg

In triangle ABC

[tex]Sin38 = \frac{BC}{AC}[/tex]

[tex]Sin38 = \frac{F_{y}}{F}[/tex]

[tex]0.616 = \frac{F_{y}}{720}[/tex]

[tex]F_{y}[/tex] = 443.3 N

Also, In triangle ABC

[tex]Cos38 = \frac{AB}{AC}[/tex]

[tex]Cos38 = \frac{F_{x}}{F}[/tex]

[tex]0.788 = \frac{F_{x}}{720}[/tex]

[tex]F_{x}[/tex] = 567.4 N

Answer:

You will have 4.5 million dollar

Explanation:

The thickness of a $1 bill is 0.11 mm

So we have

              1 $ = 0.1 mm

              0.1 mm = 1 $

              0.0001 m =  1 $

              1 m = 10000 $

           450 m = 450 x 10000 = 4500000 $

So you will have 4.5 million dollar

Answer:

4090909

Explanation:

Thickness of one bill = 0.11 mm

Total thickness = 450 m

No of $1 bills = total thickness / thickness of one bill

No of $1 bills = 450 / 0.11 × 10^-3

= 4090909

Answer:

option (a)

Explanation:

The direction of current is given by the Maxwell's right hand thumb rule.

If we hold a current carrying conductor in our right hand such that the thumb indicates the direction of magnetic field then the curling of fingers give the direction of current.

Here the direction of current is in counter clockwise direction.

Answer:

Velocity at the point of maximum x cordinate is 9.578m/s

Position vector of the particle when it reaches point of maximum x ordinate is [tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]

Explanation:

We shall resolve the motion of the particle along x and y direction separately

The particle will reach it's maximum x coordinate when it's velocity along x axis shall become 0

We have acceleration along x-axis = [tex]-2.6m/s^{2}[/tex]

acceleration along y-axis = [tex]4.7m/s^{2}[/tex]

Thus using the first equation of motion along x axis we get

[tex]v_{x}=u_{x}+a_{x}t\\\\[/tex]

Applying values we get

[tex]0=5.3-2.6t\\\\\therefore t=\frac{5.3}{2.6}sec\\\\t=2.038sec[/tex]

Now to obtain it's position we shall use third equation of motion

[tex]v_{x}^{2}=u_{x}^{2}+2as_{x}\\\\0=(5.3)^{2}+2(-2.6)s_{x}\\\\\therefore s_{x}=\frac{-28.09}{-5.2}m\\\\s_{x}=5.402m[/tex]

Now it's location along y- axis can be obtained using 2nd equation of motion along the y axis

[tex]s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}[/tex]

Applying values as follows we get

[tex]u_{y}=0\\a_{y}=4.7m/s^{2}\\t=2.038s[/tex]

[tex]s_{y}=0\times 2.038+\frac{1}{2}\times 4.7m/s^{2}\times2.038^{2}\\\\s_{y}=9.76m[/tex]

thus the position vector of the particle when it reaches it's maximum x co-ordinate is

[tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]

Now velocity of the particle at the position of maximum x co-ordinate shall be zero along x-axis and along the y-axis it can be found along the first equation of motion along y axis

[tex]v_{y}=u_{y}+a_{y}t\\\\v_{y}=0+4.7\times 2.038\\\\v_{y}=9.578m/s[/tex]

Answer:

4557.88 m/s

Explanation:

M = mass of earth = 5.98 x 10²⁴ kg

R = radius of earth = 6.4 x 10⁶ m

r = orbital radius of satellite = 3R = 3 x 6.4 x 10⁶ = 19.2 x 10⁶ m

[tex]v[/tex] = speed of the satellite

speed of the satellite is given as

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

[tex]v=\sqrt{\frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{19.2\times 10^{6}}}[/tex]

[tex]v[/tex] = 4557.88 m/s

Answer:

The initial speed is 7.174 m/s.

Explanation:

Given that,

Mass = 92.0 kg

Coefficient of friction = 0.61

Time = 1.2 s

We need to calculate the acceleration

Using of friction force

[tex]F = \mu mg[/tex]...(I)

Where, [tex]\mu[/tex] =Coefficient of friction

g = acceleration due to gravity

Using newton's law

[tex]F = ma[/tex]....(II)

m = mass of the baseball

a = acceleration of the baseball

From equation (I) and (II)

[tex]ma=\mu mg[/tex]...(I)

Put the value in the equation

[tex]a=0.61\times9.8[/tex]

[tex]a=5.978\ m/s^2[/tex]

We need to calculate the initial velocity

Using equation of motion

[tex]v = u-at[/tex]

Where, v = final velocity

u = initial velocity

a = acceleration

Here, a is negative because the player comes to rest

t = time

Put the value in the equation

[tex]0=u-5.978\times1.2[/tex]

[tex]u=7.174\ m/s[/tex]

Hence, The initial speed is 7.174 m/s.

The initial speed of the baseball player was approximately 7.176 m/s, calculated using the principles of Physics (Kinematics) and understanding of friction.

The subject of the problem involves a concept in Physics known as Kinematics, specifically dealing with friction and motion. The initial speed of the baseball player can be calculated using the equation of motion: final velocity (v_f) = initial velocity (v_i) + acceleration (a) x time (t), where the final velocity in this case is 0 (as the player comes to rest), time is 1.2 seconds, and the acceleration is the frictional force divided by the mass of the player. Frictional force is obtained by multiplying the mass of the player, the gravitational acceleration, and the coefficient of friction. In this case, substituting acceleration = -0.61 * 9.8 m/s² (because it's a decelerating force), the equation becomes 0 = v_i - 0.61 * 9.8 * 1.2 from which initial velocity (v_i) comes out to be approximately 7.176 m/s.

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Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

Final answer:

Soda is being pulled out of a cylindrical glass at the rate of -3.14 cubic inches per second.

Explanation:

The question asks at what rate soda is being absorbed from a cylindrical glass. Given the cylindrical glass's dimensions and the rate at which the depth of soda decreases, we need to find the volume rate of change.

He formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height of the cylinder. To find the rate at which volume changes, we differentiate with respect to time to obtain dV/dt = πr2(dh/dt). Plugging in the values, we have r = 2 inches and dh/dt = -0.25 inches/second (negative because the depth is decreasing).

Using the given values, the rate at which soda is being pulled out of the glass is

dV/dt = π(2 in)2(-0.25 in/sec) = π(4 in2)(-0.25 in/sec) = -3.14 in3/sec.

Therefore, the rate at which the soda is being pulled out of the glass is -3.14 in3/sec.

Answer:

period of rotation is 9.9843 years

Explanation:

Given data

sides  = 1.0×10^12 m

to find out

period of rotation

solution

first we use the gravititional formula i.e

F(g) = 2 F cos 30

here F = G  × mass(sun)² / radius²

F = 6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])²

so

F(g) = 2  ×  6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])²  ×  (1.73/2)

F(g) = 4.57 ×  [tex]10^{26}[/tex] N

and

by the law of sines

r/sin30 = s/sin120

so r will be

r = 1.0×10^12  (0.5/.87)

r = 5.75 ×  [tex]10^{11}[/tex] m

we know that gravititional force = centripetal force

so  here

centripetal force = mv² /r

centripetal force = 1.99× [tex]10^{30}[/tex] v² / 5.75 ×  [tex]10^{11}[/tex]

so

4.57 ×  [tex]10^{26}[/tex]  = mv² /r

4.57 ×  [tex]10^{26}[/tex]  = 1.99× [tex]10^{30}[/tex] v² / 5.75 ×  [tex]10^{11}[/tex]

v² = 4.57 ×  [tex]10^{26}[/tex] × 5.75 ×  [tex]10^{11} / 1.99× [tex]10^{30}[/tex]

and

v = 11480 m/sec

and we know period of rotation formula

t = d /v

t = 2 [tex]\pi[/tex] r /v

t = 2 × [tex]\pi[/tex] ×  5.75 ×  [tex]10^{11} / 11480

t = 314 × [tex]10^{5} / 31536000 year

t = 9.9843 year

so period of rotation is 9.9843 years

The period of rotation for the three stars can be calculated using Kepler's Third Law and recognizing that the stars form a circular orbital pattern. The relevant distance is from the center of the mass distribution (center of the equilateral triangle) to any of the stars and all three stars have identical masses.

This question relates to astrophysics and can be solved using Kepler's Third Law that provides a relationship between the period of revolution (P), semimajor axis (D), and the total mass (M) of a binary system: D³ = (M₁ + M₂)P². However, since we are dealing with three stars of equal mass (each equivalent to the mass of our sun), they form a circular orbital pattern. We use the formula P = √((4π²/G)(r³/3M)), where r is the distance from the center of the mass distribution to any of the stars.

Firstly, we must understand that in an equilateral triangle, the distance from the center to a vertex is r = √3/2 * side. So, the distance from the center to any star is 0.866 * 1.0×10¹²m. The gravitational constant, G, is approximately 6.67 * 10^-11 N(m/kg)².

Substituting these values into our formula, we can calculate the period of rotation of the stars. One should note that all three stars have identical masses which simplifies the calculation.

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Answer:

Correct option is 'd' Surroundings

Explanation:

The space surrounding a thermodynamic system is known as Surroundings

Answer:

The the specific volume and weight are 79.16m³/k mol and 49 N.

Explanation:

Given that,

Diameter = 3 m

Mass of N₂ = 5 kg

We need to calculate the volume of balloon

Using formula of volume

[tex]V= \dfrac{4}{3}\pi r^3[/tex]

[tex]V= \dfrac{4}{3}\pi \times{\dfrac{d}{2}}^3[/tex]

[tex]V= \dfrac{4}{3}\pi \times(\dfrac{3}{2})^3[/tex]

[tex]V=14.137\ m^3[/tex]

We need to calculate the specific volume in the balloon

[tex] Specific\ volume=\dfrac{Volume}{mass}[/tex]

[tex]Specific\ volume=\dfrac{14.137}{5}[/tex]

[tex]Specific\ volume=2.8274\ m^3/kg[/tex]

Now, Molar mass of N₂ gas is 0.028 kg/mol

Now, Specific volume of N₂ gas in the balloon is

[tex]Specific\ volume =0.028\times2.8274=0.0791672\ m^3/mol[/tex]

[tex]Specific\ volume =79.16m^3/kmol[/tex]

(b). We need to calculate the weight of the gas in the balloon

Weight of the balloon is dependent on the mass of the gas.

The weight of the gas is given by

[tex]W = mg[/tex]

[tex]W = 5\times9.8[/tex]

[tex]W = 49\ N[/tex]

Hence, The the specific volume and weight are 79.16m³/k mol and 49 N.

According to Newton's Law of Gravitation, the Gravity Force is:

[tex]F=\frac{GMm}{{r}^{2}}[/tex]     (1)

This expression can also be written as:

[tex]F=GMm{r}^{-2}[/tex]     (2)

If we derive this force [tex]F[/tex] respect to the distance [tex]r[/tex] between the two masses:

[tex]\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr[/tex]     (3)

Taking into account [tex]GMm[/tex] are constants:

[tex]\frac{dF}{dr}dFdr=-2GMm{r}^{-3}[/tex]     (4)

Or

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit [tex]N/km[/tex]:

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

We have a force that decreases with a rate 1 [tex]\frac{dF_{1}}{dr}dFdr=4N/km[/tex] when [tex]r=20000km[/tex]:

[tex]4N/km=-2\frac{GMm}{{(20000km)}^{3}}[/tex]     (6)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=(4N/km)({(20000km)}^{3})[/tex]     (7)

In addition, we have another force that decreases with a rate 2 [tex]\frac{dF_{2}}{dr}dFdr=X[/tex] when [tex]r=10000km[/tex]:

[tex]XN/km=-2\frac{GMm}{{(10000km)}^{3}}[/tex]     (8)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=X({(10000km)}^{3})[/tex]     (9)

Making (7)=(9):

[tex](4N/km)({(20000km)}^{3})=X({(10000km)}^{3}[/tex]       (10)

Then isolating [tex]X[/tex]:

[tex]X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}[/tex]  

Solving and taking into account the units, we finally have:

[tex]X=-32N/km[/tex]>>>>This is how fast this force changes when [tex]r=10000 km[/tex]

Answer:

(a) 5 hours 20 min

(b)I would have a great inconvenience with the delay of the signal

Explanation:

Hello

To find the time it takes for light to reach Earth from Pluto, we use a rule of three

Step 1

define

distance from sun to earth=x

distance from pluto to earth=40x

Step 2

Rule of three

distance    time

x       ⇒   8 minutes

40x   ⇒   A minutes    

[tex]A=\frac{40x*8*min}{x}\\ A=320 min\\\\to\ convert\ to\ hours\\\\A=320\ min*\frac{1 h}{60 mi}\\ A=5.33 hours\\\\A=5\ hours\ and\ 20min\\\\[/tex]

(a) 5 hours 20 min

Step 3

I would have a great inconvenience with the delay of the signal, since the message I send will arrive 5 hours and 20 minutes later, if the message is answered immediately I will listen to it 320 min after that, more than ten and a half hours to give and receive a hello

Answer:

work done by applied force = 1120J

work done by frictional force = 943.25J

work done by the gravitational force = work done by the normal force = 0J

Explanation:

Given:

Mass of the box = 55kg

Displacement of the box = 7.0m across the floor

Applied force = 160N

Coefficient of kinetic friction ([tex]\mu_k[/tex]) = 0.25

Now the work done (W) is given as:

W = F×d

Where,

F = Applied force on the body

d = Displacement of the body in the direction of the applied force

Forces acting on the box are:

1) Applied force

2) Frictional Force

3) Gravitational force

4) Normal force due to the weight

Now the work done by the respective forces:

1) By the applied force

W = 160 kg × 7 m = 1120 J

2) By the frictional force

[tex]W_f =-\mu mg\times displacement[/tex]

Where

g = acceleration due to the gravity

and the negative sign here depicts that the frictional force is acting the against the applied force and the direction of displacement.

thus,

[tex]W_f =-0.25\times 55\times 9.8\times 7[/tex]

or

[tex]W_f =-943.25J[/tex]

The work done by the normal force and the gravitational force will be zero as there is no displacement in the direction of the application of both the forces

The work done by both the gravitational force and normal force is equal to zero (0).

Given the following data:

In this scenario, there are four forces acting on the box and these include:

For the applied force:

Mathematically, the work done due to an applied force is given by this formula:

[tex]Work\;done = force \times distance\\\\Work\;done = 160 \times 7[/tex]

Work done = 1,120 Nm.

For the frictional force:

Mathematically, the work done due to a frictional force is given by this formula:

[tex]Work\;done = -\mu mg \times distance\\\\Work\;done = -0.25 \times 55 \times 9.8 \times 7[/tex]

Work done = -943.25.

However, the work done by both the gravitational force and normal force is equal to zero (0) because there isn't any displacement or distance covered in the direction of the application of these forces.

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Answer:

A. Both spheres land at the same time.

Explanation:

The horizontal motion doesn't affect the vertical motion.  Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.

Answer:

A. Both spheres land at the same time.

Explanation:

If air resistance and friction are ignored, then both spheres land at the same time. Falling objects fall toward the center of the Earth with the same constant acceleration, independent of their mass, when they are defined to be in free-fall. For an object to be in free-fall, it has to be in a situation in which both air resistance and friction are considered negligible. This is true regardless of the direction of the fall.

The index of refraction can be calculated using the speed of light in vacuum divided by the speed of light in a specific medium. Given the speed of light in a medium is 2.26 x 10^8 m/s, the index of refraction would be calculated as roughly 1.33.

To calculate the index of refraction for a certain medium, you would use the speed of light in vacuum (c) divided by the speed of light in that medium (v). This equation is denoted as n = c/v.

The speed of light in vacuum is approximately 3.00×10^8 m/s. Given that the speed of light in a certain substance is 2.26 x 10^8 m/s, we can substitute these values into the equation to find the material's index of refraction.

So, n = (3.00×10^8 m/s) / (2.26 x 10^8 m/s). Therefore, the index of refraction of that substance is approximately 1.33. This property of a substance tells us how much the path of light is bent, or refracted, when entering the material.

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Answer:

0.545 m

Explanation:

K. E = 26 MeV = 26 x 1.6 x 10^-13 J = 4.16 x 10^-12 J

B = 1.35 T

Let r be the radius of curvature

The formula for the kinetic energy of a cyclotron is given by

[tex]K.E. = \frac{B^{2}q^{2}r^{2}}{2m}[/tex]

m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 c

[tex]4.16\times 10^{-12} = \frac{1.35^{2}\times (1.6\times 10^{-19})^{2}r^{2}}{2\times 1.67\times 10^{-27}}[/tex]

r = 0.545 m

Answer:

a) [tex]V=0.314 m/s[/tex]

b) [tex]\omega=2.09rad/s[/tex]

c) [tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]

Explanation:

Given:

Radius of the circle, r = 15cm = 0.15m

Time period, T  =3.0s

a) The velocity (V) of a particle moving in the circular motion is given as:

[tex]V=\frac{2 \pi r}{T}[/tex]

substituting the given values in the above equation we get

[tex]V=\frac{2\times \pi \times 0.15m}{3.0s}[/tex]

or

[tex]V=0.314 m/s[/tex]

b) Angular speed ([tex]\omega[/tex]) is given as:

[tex]\omega=\frac{2\pi}{T}[/tex]

or

[tex]\omega=\frac{2\times \pi}{3.0s}[/tex]

or

[tex]\omega=2.09rad/s[/tex]

c) The position of the particle on the x-position is given as:

[tex]x(\theta)=rcos(\theta)[/tex]         (reffer the attached figure)

now the relation between the Θ and the time T is given as:

[tex]\omega = \frac{2\pi}{T}=\frac{\theta}{t}[/tex]

or

[tex]\theta= \frac{2\pi}{T}\times t[/tex]

or

[tex]\theta= \frac{2\pi}{3}\times t[/tex]

substituting the values of r and Θ, we get

[tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]

Final answer:

The particle's speed is 0.1π m/s, its angular speed is 2/3π rad/s, and the x-component of the position function is x(t) = 0.15 * cos((2/3π) * t + π).

Explanation:

(a) Speed of the particle: The speed is the distance traveled by a particle along a circular path per unit time. Given a particle completes 1 revolution (which is 2π times the radius r) every 3 seconds, the speed v is given by: v = (2π * 15 cm) / 3 s. Convert 15 cm to meters by dividing by 100 to get 0.15 meters. So, v = (2π * 0.15 m) / 3 s = 0.1π m/s.

(b) Angular speed of the particle: The angular speed ω is the rate of change of the angular displacement and is measured in radians per second. Since the particle completes one revolution every 3 seconds and one revolution is 2π radians, ω = 2π radians / 3 s = 2/3π rad/s.

(c) Position function for the x-component: Assume the particle starts on the negative x-axis, so the angle θ at t=0 is π. The particle moves in a circle with radius r = 0.15 m. The x-component of the position as a function of time t, using the cosine function is x(t) = r * cos(ω*t + π), where ω = 2/3π rad/s. Plugging values, we get x(t) = 0.15 * cos((2/3π) * t + π).

Answer:4800

Explanation:

Given data

k=0.4

surface area(A)=100[tex]ft&^{2}[/tex]

Temprature on near side of material[tex](T_1)[/tex]=150°F

temprature on far side of material[tex](T_2)[/tex]=165°F

thickness(t)=1.5in=0.125[tex]ft^2[/tex]

Conductive heat transfer rate(Q) =[tex]\frac{k\times A\times\left ( T_2-T_1\right )}{y}[/tex]

Q=[tex]\frac{0.4\times 100\left ( 165-150\right )}{1.5}[/tex]

Q=4800°F per feet

Answer:

4.3 N/m

Explanation:

m = 230 g = 0.230 kg, x = 8.2 cm

in 13 oscillations, time taken = 19 s

In 1 oscillation, time taken = 19 / 13 = 1.46 s

By the use of formula of time period , Let k be the spring constant.

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

[tex]1.46 = 2\pi \sqrt{\frac{0.230}{k}}[/tex]

0.054 = 0.230 / k

k = 4.26 N/m

k = 4.3 N/m

Answer:

156.3 J

Explanation:

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

Answer:

Frictional force, F = -10.96 N

Explanation:

It is given that,

Mass of the sled, m = 19 kg

Initial velocity of the sled, u = 4.5 m/s

After 7.80 s has elapsed, the sled stops. We need to find the magnitude of the average friction force acting on the sled while it was moving. The change in momentum is equal to the impulse imparted i.e.

[tex]F.\Delta t=m(v-u)[/tex]

[tex]F=\dfrac{m(v-u)}{\Delta t}[/tex]

[tex]F=\dfrac{19\ kg(0-4.5\ m/s)}{7.8\ s}[/tex]

F = -10.96 N

So, the average frictional force acting on the sled is 10.96 N. Hence, this is the required solution.

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

so

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e²  = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

Answer:

It must be moved by 2.59 m to the right

Explanation:

To keep the plank in uniform state, the moment at sawhorse = 0

Let y be the distance of saw horse from left of plank.

Weight per unit length of plank [tex]=\frac{99}{8}=12.375N/m[/tex]

Taking moment about saw horse

Moment due to left portion,

         182 x y + 12.375 x y x 0.5 y = 182 y + 6.1875 y²

Moment due to right portion,

         12.375 x (8-y) x 0.5 (8-y) = 396 + 6.1875 y² - 99y

So to keep balance these moments should be same

         182 y + 6.1875 y² = 396 + 6.1875 y² - 99y

         281 y = 396

                y = 1.41 m

Distance moved to right = 0.5 x 8 - 1.41 = 2.59 m

So it must be moved by 2.59 m to the right

To answer this question, let us break down the initial velocity into its horizontal and vertical components.

v_{ix} = vcos(θ)

v_{iy} = vsin(θ)

v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time

v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.

v is the magnitude of the initial velocity.

θ is the angle the velocity vector is oriented at with respect to the horizontal.

Given values:

v = 19.6m/s

θ = 40.8°

Plug in these values and solve for v_{ix} and v_{iy}:

v_{ix} = 19.6cos(40.8°) = 14.8m/s

v_{iy} = 19.6sin(40.8°) = 12.8m/s

To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:

D = X + Vt + 0.5At²

When solving for the horizontal displacement, the following values are:

t = elapsed time

D = horizontal displacement

X = initial horizontal displacement

V = initial horizontal velocity

A = horizontal acceleration

There is no initial horizontal displacement, so X = 0m

The initial horizontal velocity V = v_{ix} = 14.8m/s

Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²

Therefore the equation for the stone's horizontal displacement is given by:

D = 14.8t

When solving for the vertical displacement, the following values are:

t = elapsed time

D = vertical displacement

X = initial vertical displacement

V = initial vertical velocity

A = vertical acceleration

There is no initial vertical displacement, so X = 0m

The initial vertical velocity V = v_{iy} = 12.8m/s

Gravity acts downward on the stone, therefore A = -9.81m/s²

Therefore the equation for the stone's vertical displacement is given by:

D = 12.8t - 4.905t²

Now we just plug in various values of t...

a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m

b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m

c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m

d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m

Before you write down the following results, read the following explanation.

e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m

f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m

Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m

The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m

Answer:

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 [tex]P_1,T_1 [/tex] Pressure and temperature at inlet and

 [tex]P_2,T_2 [/tex] Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

[tex]\mu _j=\left(\frac{\partial T}{\partial p}\right)_h[/tex]

Now from T-ds equation

dh=Tds=vdp

So

[tex]Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp[/tex]

⇒[tex]dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

So Joule -Thompson coefficient

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

[tex]\dfrac{\partial v}{\partial T}=\dfrac{v}{T}[/tex]

So by putting the values in

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

[tex]\mu _j=0[/tex] For ideal gas.