What is the gravitational force fg between the two objects described in part b if the distance between them is only 50 km ?

Answer:

Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.

Explanation:

Formula

Cube :

The volume of a cube is = [tex]side^3[/tex]

The side of length is x in.

Then volume of the cube is (V) = [tex]x^3[/tex]

∴ V = [tex]x^3[/tex]

Differentiate with respect to t

[tex]\frac{d}{dt}(V)=\frac{d}{dt} (x^3)[/tex]

[tex]\Rightarrow \frac{dV}{dt} =3x^2\frac{dx}{dt}[/tex]....(1)

Given that the side of the cube is increasing at the rate of 0.2 in/s.

i.e [tex]\frac{dx}{dt} = 0.2[/tex]  in/s.

And the sides of the cube are 7 in i.e x= 7 in

Putting [tex]\frac{dx}{dt} = 0.2[/tex]  and  x= 7 in equation (1)

[tex]\therefore \frac{dV}{dt} =3 \times 7^2 \times 0.2[/tex]  cube in./s

       =29.4 cube in./s

Therefore the volume of cube is change at the 29.4 cube in./s at that instant time.

To find the rate at which the volume of the cube is changing, differentiate the volume formula with respect to time. For a cube with sides of 7 in length growing at 0.2 in/s, the rate of volume change is 6.3 in³/s.

The volume of a cube is determined by the formula V = x³, where x is the length of the side of the cube.

Given that the sides of the cube are 7 in long and increasing at 0.2 in/s, we can calculate how fast the volume is changing by differentiating the volume formula with respect to time. By taking the derivative of V = x³, we find that the rate of change of the volume at that instant of time is 6.3 in³/s.

Answer:

Explanation:

kinetic energy of alpha particle

= Q X V ( Q is charge on the particle and V is potential difference )

= 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³

= 11.04 x 10⁻²² J

1/2 m v² = 11.04 x 10⁻²²

1/2 x 6.68 x 10⁻²⁷ x v² = 11.04 x 10⁻²²

v² = 3.305 x 10⁵

v = 5.75 x 10² m /s

Final answer:

To find the speed of the alpha particle after it has moved through a potential difference, we can use the conservation of energy. Using the given values for charge and potential difference, we can calculate the speed using the formula for change in kinetic energy. The speed of the alpha particle is approximately 1.91x10^5 m/s.

Explanation:

To find the speed of the alpha particle after it has moved through a potential difference of -3.45x10^-3 V, we can use the conservation of energy.

The potential difference is given by ΔV = qΔV, where q is the charge of the alpha particle and ΔV is the potential difference. Plugging in the values, we get ΔV = (3.20x10^-19 C)(-3.45x10^-3 V).

The change in kinetic energy is given by ΔKE = (1/2)mv^2, where m is the mass of the alpha particle and v is its velocity. Setting ΔV equal to ΔKE, we can solve for v. Plugging in the values, we get (1/2)(6.68x10^-27 kg)v^2 = (3.20x10^-19 C)(-3.45x10^-3 V).

Solving for v, we find that the speed of the alpha particle is approximately 1.91x10^5 m/s.

Answer:

R1 = 4.8Ω

Explanation:

The loop circuit has an initial voltage of  V = IR

I = 2 A , R1 = R

V = 2R1

with the current reduced to 1.5A with an additional 1.6Ω resistor

the total resistance of the circuit is 1.6 + R1

the voltage of the two scenarios has to be equal , since the same voltage flows through the circuit

therefore V = 2R1

from Ohms law V = IR

2R1= 1.5 (1.6 + R1)

2R1 = 2.4 + 1.5R1

collecting like terms

2R1 - 1.5R1 = 2.4

0.5R1 = 2.4

R1 = [tex]\frac{2.4}{0.5}[/tex]

R1 = 4.8Ω

Answer:

4.8 Ω

Explanation:

From Ohm's Law,

Using,

I = E/(R+r)................. Equation 1

E = I(R+r)................. Equation 2

Where I = current, E = emf, R = external resistance, r = internal resistance

Given: I = 2 A, R = R1, r = 0 Ω

Substitute into equation 2

E = 2(R1)

E = 2R1.

When an additional 1.6 Ω  resistor is added in series,

E = 1.5(R1+1.6)

2R1 = 1.5R1+2.4

2R1-1.5R1 = 2.4

0.5R1 = 2.4

R1 = 2.4/0.5

R1 = 4.8 Ω

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

Answer:

Explanation:

Answer:

Impedance = 19.44ohms

Current = 5.14A

Power factor = 0.62

Explanation:

Impedance in an RLC AC circuit is defined as the total opposition to the flow of current in the resistor, inductor and capacitor.

Impedance Z = √R²+(Xl-Xc)²

Where R is the resistance = 12Ω

Inductance L = 0.15H

Capacitance C = 100uF = 100×10^-6F

Since Xl = 2πfL and Xc = 1/2πfC where f is the frequency.

Xl = 2π×50×0.15

Xl = 15πΩ

Xl = 47.12Ω

Xc = 1/2π×50×100×10^-6

Xc = 100/π Ω

Xc = 31.83Ω

Z =√12²+(47.12-31.83)²

Z = √144+233.78

Z = 19.44Ω

Impedance = 19.44ohms

To calculate the circuit current, we will use the expression V=IZ where V is the supply voltage = 100V

I = V/Z = 100/19.44

I = 5.14Amperes

To calculate the power factor,

Power factor = cos(theta) where;

theta = arctan(Xl-Xc)/R

theta = arctan(47.12-31.83)/12

theta = arctan(15.29/12)

theta = arctan1.27

theta = 51.78°

Power factor = cos51.78°

Power factor = 0.62

Answer:

The circuit impedance [tex]=19.4 \Omega[/tex]

The circuit's current [tex]=5.14 A[/tex]

Circuit Power Factor [tex]=0.62[/tex]

Explanation:

Given:

Resistance [tex]R=12 \Omega[/tex]

Inductance [tex]=0.15H[/tex]

Capacitance [tex]=100uF[/tex]

Voltage [tex]=100V[/tex]

Step 1:

To calculate the inductive reactance, [tex]$X_{L}$[/tex].

[tex]X_{L}=2 \pi f L=2 \pi \times 50 \times 0.15=47.13 \Omega[/tex]

To calculate the Capacitive reactance,

[tex]X_{C}=\frac{1}{2 \pi f C}[/tex]

[tex]=\frac{1}{2 \pi \times 50 \times 100 \times 10^{-6}}[/tex]

[tex]=31.83 \Omega[/tex]

Step 2:

Circuit impedance,

[tex]$$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$$[/tex]

where R is the resistance,

[tex]$$&Z=\sqrt{12^{2}+(47.13-31.83)^{2}}[/tex]

[tex]&Z=\sqrt{144+234}=19.4 \Omega\end{aligned}$$[/tex]

Step 3:

Circuits Current, I

[tex]$I=\frac{V_{S}}{Z}[/tex]

[tex]=\frac{100}{19.4}[/tex]

[tex]=5.14 \ A[/tex]

Step 4:

Voltages across the Circuit, [tex]$\mathrm{V}_{\mathrm{R}}, \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$[/tex]

[tex]V_{R}=I \times R=5.14 \times 12=61.7$ volts[/tex]

[tex]V_{L}=I \times X_{L}=5.14 \times 47.13=242.2$ volts[/tex]

[tex]V_{C}=\ I \times X_{C}=5.14 \times 31.8=163.5$ volts[/tex]

Step 5:

Circuits Power factor

[tex]$=\frac{R}{Z}=\frac{12}{19.4}=0.619$[/tex]

Therefore,

The circuit impedance [tex]=19.4 \Omega[/tex]

The circuit's current [tex]=5.14\ A[/tex]

Power Factor [tex]=0.62[/tex]

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Answer: Force on the truck is equal to force on the car.

Explanation: According to the Newton's third law of motion which states that; For every action, there is an equal and opposite reaction. These pair of forces are regarded as action - reaction forces. These size or magnitude of the forces on the colliding objects are equal or the same, while the direction of the colliding objects are opposite.

In the scenario above, both truck and carry have the same mass, however, the damage suffered by the car is based on its smaller mass which makes it unable to withstand the acceleration resulting from the collision.

ma = m(-a)

m= mass, a= acceleration

Explanation:

The given data is as follows.

        Dielectric constant, K = 3.0

   Area of the plates (A) = 0.021 [tex]m^{2}[/tex]

   Distance between plates (d) = [tex]2.75 \times 10^{-3} m[/tex]

  Maximum electric field (E) = [tex]3.25 \times 10^{5} V/m[/tex]

Now, we will calculate the capacitance as follows.

             C = [tex]\frac{k \epsilon_{o} \times A}{d}[/tex]

                 = [tex]\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}[/tex]

                 = [tex]\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}[/tex]

                 = [tex]0.203 \times 10^{-9}[/tex] F

Formula to calculate electric charge is as follows.

               E = [tex]\frac{\sigma}{k \epsilon_{o}}[/tex]

or,           Q = [tex]E \times k \times \epsilon_{o}A[/tex]      (as [tex]\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}[/tex])

                   = [tex]3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021[/tex]

                   = [tex]181.2 \times 10^{-9} C[/tex]

Formula to calculate the energy is as follows.

                  U = [tex]\frac{1 \times Q^{2}}{2 \times C}[/tex]

                     = [tex]\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}[/tex]

                     = [tex]\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}[/tex]

                     = [tex]9835.67 \times 10^{-9}[/tex]

or,                  = [tex]98.35 \times 10^{-7} J[/tex]

Thus, we can conclude that the maximum energy that can be stored in the capacitor is [tex]98.35 \times 10^{-7} J[/tex].

Explanation:

Below is an attachment containing the solution.

Answer:

The answers to the question are

a) The volume flow rate of the carbon dioxide at the compressor inlet is 0.2834 m³/s ≈ 0.3 m³/s

b) The power input to the compressor is 73.35 kW ≈ 70 kW

Explanation:

We note the following

Mass flow rate = 0.5 kg/s

Inlet pressure = 100 pKa

Outlet pressure = 600 kPa

Inlet temperature = 300 K

Outlet temperature  =  450 K

Molar mass of CO₂ = 44.01 g/mol

R Universal Gas Constant = 8.314 4621. J K−1 mol−1

a) Number of moles = [tex]\frac{Mass}{Molar.Mass}[/tex] = [tex]\frac{500g}{44.01g}[/tex] = 11.361 moles

P·V= n·R·T ∴ V = [tex]\frac{n*R*T}{P}[/tex] = [tex]\frac{11.361*8.3145*300}{ 100 }[/tex] = 0.2834 m³

Therefore the volume flow rate = 0.2834 m³/s ≈ 0.3 m³/s

b) Cp at 300 K = 0.846 kJ/(kg K)

Cp at 600 K = 0.978 kJ/(kg K)

Cv = 0.657

K = 1.289

While the power input to the compressor can be calculated by

m'×Cp×(T₂-T₁)

Where m' = mass flow rate = 0.5 kg/s

Therefore power = 0.5 kg/s×0.978 kJ/(kg K)×(450 K - 300 K)

= 73.35 kJ/s = 73.35 kW ≈ 70 kW

Answer:

D) F

Explanation:

Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :

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

Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.

Since, New mass of ball A = 2M

           New mass of ball b = 2m

Distance between the two balls = 2r

Substitute these values in equation (1).

[tex]F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }[/tex]

[tex]F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }[/tex]

Using equation (1) in the above equation.

F₁ = F

Answer:

T = 153.77 [N]

Explanation:

To solve this type of problems, we must make a free body diagram, with the forces acting on the box. Then performing a sum of forces on the Y axis equal to zero we can find the value of the normal force. After finding the friction force, we performed a sum of forces equal to the product of mass by acceleration (newton's second law). We can find the T-Force value.

To calculate the tension in the rope, we can use Newton's second law. Using the given values of mass, angle, coefficient of kinetic friction, and acceleration, we can set up an equation and solve for FT. The tension in the rope is approximately 173.5 N.

To calculate the tension in the rope, we can use Newton's second law. The only external forces acting on the mass are its weight and the tension in the rope. Using the given information, we can set up the equation FT - mgsinθ - μkmgcosθ = ma, where FT is the tension, m is the mass, θ is the angle, μk is the coefficient of kinetic friction, a is the acceleration, and g is the acceleration due to gravity.

Substituting the given values, we have FT - (17.5 kg)(9.81 m/s2)sin(23.0°) - (0.295)(17.5 kg)(9.81 m/s2)cos(23.0°) = (17.5 kg)(2.29 m/s2).

Solving for FT, we find that the tension in the rope is approximately 173.5 N.

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Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

[tex]d=30,000 pc =9.26\cdot 10^{20} m[/tex]

The speed of the cosmic ray is

[tex]v=0.98 c[/tex]

where

[tex]c=3.0 \cdot 10^8 m/s[/tex] is the speed of light. Substituting,

[tex]v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s[/tex]

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

[tex]T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s[/tex]

Converting into years,

[tex]T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years[/tex]

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

[tex]T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}[/tex]

And substituting v = 0.98c, we find:

[tex]T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years[/tex]

Answer:

The incidence angle is 27°

Explanation:

As the complete question is not given, the complete question is given here

The angle between the Sun and a rescue aircraft is 54 degrees. What should be the angle of incidence for sunlight on a plane mirror so that the rescue pilot sees the reflected light?

From the question, the total angle between the Sun and the aircraft is 54 degrees which is the sum of the incidence and the reflected angle so

[tex]\theta_{total}=\theta_{incidence}+\theta_{reflection}=54[/tex]

Also from the law of reflection

[tex]\theta_{incidence}=\theta_{reflection}[/tex]

So now the equation becomes

[tex]\theta_{total}=\theta_{incidence}+\theta_{reflection}=54\\\theta_{total}=\theta_{incidence}+\theta_{incidence}=54\\2\theta_{incidence}=54\\\theta_{incidence}=27\\[/tex]

So the incidence angle is 27°

The angle of incidence for sunlight on a plane mirror should be set so that the angle of reflection directs the light towards the rescue pilot. The mirror's orientation is crucial, and light reflects at the same angle relative to the normal. Practical adjustments have to be made based on the positions of the sun and pilot.

For sunlight to be reflected from a plane mirror to a rescue pilot, the angle of incidence should be such that the angle of reflection directs the light towards the pilot. Since the angle of reflection is equal to the angle of incidence, the mirror must be tilted to reflect the light into the pilot's eyes.

According to the law of reflection, light incident on a mirror will reflect off at the same angle relative to the normal (an imaginary line perpendicular to the surface of the mirror). Hence, for the rescue pilot to see the reflected light, the angle of incidence must be adjusted accordingly based on the position of the sun and the pilot's location in the sky.

It is important to note that if we wish to maximize the reflection towards the pilot, utilizing the mirror's orientation is key. At very high angles of incidence, approaching 90 degrees, almost all the light is reflected, according to physical principles. However, such angles may not be practical when trying to target a specific viewer such as a pilot.

Answer:

The term is geotropism (also known as gravitropism)

Explanation:

Answer:

Explanation:

11 oscillations in 18.2 s

Time period is defined as the time taken to complete one oscillation.

T = 18.2 / 11 = 1.655 s

mass, m = 290 g = 0.29 kg

Δx = 21.1 cm = 0.211 m

ω = 2π / T = (2 x 3.14) / 1.655 = 3.796 rad/s

[tex]\omega =\sqrt{\frac{K}{m}}[/tex]

Where, K is the spring constant

K = ω² m = 3.796 x 3.796 x 0.29 = 4.18 N/m

Now, mg = K Δx

0.29 x g = 4.18 x 0.211

g = 3.04 m/s²

Answer:

The answer for this is photosphere.

Explanation:

Most of the visible light we see coming from the sun originates from photosphere.The Photosphere is 300km dense and the temperature  at the bottom of the Photosphere is 6400K and the top of the Photosphere is 4600K respectively.

Following are the feature of photosphere that is given below.

Answer:

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

Momentum: p = mv

Kinetic energy in terms of momentum: [tex]E=\frac{1}{2}\frac{(mv)^{2}}{m}=\frac{p^{2}}{2m}[/tex]

Explanation:

The kinetic energy is given by this equation:

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

Now, we know that the momentum of a particle is p = m*v. This equation is true only with a classical particle, it meas particles with a speed less than the speed of light. If we had a particle traveling at speeds near that of light, the momentum would be p = γm₀v, where γ is the Lorentz factor.

So, if we see, we can rewrite the equation (1) to get this expression in terms of p.

Let's multiply and divide by mass (m) in the equation (1).

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

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

Using the p = mv here:

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

[tex]E=\frac{p^{2}}{2m}[/tex]

Therefore the kinetic energy can express in terms of momentum.

Let's see that it could not be possible using the the relativistic momentum, because it has a relativistic factor.

I hope it helps you!

Answer: 3.26 light years

Explanation:

Each star has a parallax of one arcsecond at a distance of one parsec, which is equivalent to 3.26 light years.

so the parallax of 1 arcsecond will be at a distance of 1/1 × 3.26 light years

A star with a Parallax Angle of 1 arcsecond is 1 parsec or about 3.26 light-years away. This method of determining star distance is fundamental in astronomy.

The distance of a star from us can be determined using its parallax angle.

The unit of measurement is the parsec, which stands for 'parallax-second'.

If a star has a parallax angle of one arcsecond (which is essentially a measurement of the angular shift in the star's position due to the Earth's orbit around the Sun), it's defined to be 1 parsec away from us.

A parallax angle of 1 arcsecond, therefore, means that the star is 1 parsec away.

Because 1 parsec equals 3.26 light-years, such a star would actually be about 3.26 light-years distant from Earth.

This method of calculating the distance of stars using their parallax angles was revolutionized by the Hipparcos spacecraft and is critical in the field of astronomy.

Learn more about Parallax Angle here:

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Answer:

Muscular Arteries

Explanation:

Muscular arteries continue from elastic arteries and control the distribution of blood throughout the body.

Muscular arteries are the vessels that have more smooth muscle than elastic tissue in their tunica media, along with an elastic membrane on each face of the tunica media. They are found farther away from the heart and have a significant role in vasoconstriction due to their increased amount of smooth muscle. They have an internal and external elastic membrane.

The vessels that have a tunica media with more smooth muscle than elastic tissue and an elastic membrane on each face of the tunica media are known as muscular arteries. These arteries exist farther from the heart and due to their increased amount of smooth muscle, they play a significant role in vasoconstriction.

In these arteries, the percentage of elastic fibers decreases, while the presence of smooth muscle increases. This results to the artery having a thick tunica media. It's important to note that the diameter of muscular arteries can range from 0.1mm to 10mm.

Additionally, muscular arteries possess an internal elastic membrane (also known as the internal elastic lamina) at the boundary with the tunica media, as well as an external elastic membrane in larger vessels. This gives these arteries increased structure while allowing them the ability to stretch. Due to the decreased blood pressure, muscles arteries can accommodate, elasticity is less crucial in these types of vessels.

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Answer:

Explanation:

Let the first object have a mass of M

And a period of T1=8sec

The second object has a mass 10kg and a period of T2=12 sec

It is know that,

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

T=2π√(m/k)

Then the constant in this equation is the spring constant (k) and 2π, which does not change for the same material.

Then, make k subject of formulas

T²=4π²(m/k)

T²k=4π²m

Then, k/4π²=m/T²

So the k is directly proportional to m and inversely proportional to T²

M1/T1²=M2/T2²

Since, M1 is unknown, M2=10kg, T1=8sec and T2=12

Then,

M1/T1²=M2/T2²

M1/8²=10/12²

M1/64=0.06944

M1=0.06944×64

M1=4.444kg

The mass of the first object is 4.44kg

Answer:

Therefore the magnitude of tangential velocity is 20 m/s.

Explanation:

Tangential velocity:The tangential velocity is the straight line velocity of at any point of rotating object.

It is denoted by [tex]v_t[/tex]

[tex]v_t= \omega r[/tex]

ω= angular velocity

r = radius of rotating object.

Angular velocity: Angular velocity is ratio of angle to time.

Here ω= 50 rad/s and r = 0.4 m

Tangential velocity=(50 ×0.4)m/s

                                =20 m/s

Therefore the magnitude of tangential velocity is 20 m/s.

Answer:

The magnitude and direction of the conventional current is then 1.186 Amps moving to the left direction

Explanation:

To answer the question, it should be noted that the direction of conventional current is in the opposite direction of the flow of electrons. Therefore, the direction of flow of conventional current will be to the right

The magnitude of the electric current is equal to the rate of flow of the electrons or the time it takes for the electrons to flow past a section of the wire. Therefore the magnitude is that of the 7.4 × 1018 electrons/s

However the unit of the electricity = ampere which is = coulombs/seconds

The 7.4 × 1018 electrons carry

7.4 × 10¹⁸×1.60217662 × 10⁻¹⁹ coulombs = 1.1856106988 coulombs

Therefore the magnitude of electric current = 1.186 coulombs/Seconds

= 1.186 Amps

Explanation:

Below is an attachment containing the solution.

a) [tex]15.4^{\circ}[/tex]

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

[tex]mg sin \theta[/tex]

where

m =16 kg is the mass of the box

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

[tex]\theta[/tex] is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

[tex]F=ma\\mgsin \theta = ma[/tex]

where a is the acceleration.

From the equation above we get

[tex]a=g sin \theta[/tex]

And we are told that the acceleration must not exceed

[tex]a=2.6 m/s^2[/tex]

Substituting this value and solving for [tex]\theta[/tex], we find the maximum angle of the ramp:

[tex]\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}[/tex]

b)

Here we are told that the vertical distance of the ramp is

[tex]h=1.4 m[/tex]

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

[tex]GPE=KE\\mgh=\frac{1}{2}mv^2[/tex]

where:

m = 16 kg is the mass of the box

[tex]g=9.8 m/s^2[/tex]

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s[/tex]

c)

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

[tex]mg cos \theta[/tex]

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

[tex]N-mg cos \theta[/tex]

and by substituting:

m = 16 kg

[tex]g=9.8 m/s^2[/tex]

[tex]\theta=15.4^{\circ}[/tex]

We can find the normal force:

[tex]N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N[/tex]

The maximum angle of inclination of the ramp and the final velocity of the boxes are calculated based on the given acceleration and distance. The normal force on a box moving down the ramp remains unaffected by the angle of inclination in this case as the ramp is devoid of friction. It's an application of physics concepts in real-world situations.

This problem is a practical application of the concepts of physics, specifically mechanics. Let's break it down.

a. The maximum angle of inclination of the ramp can be found by utilizing the relationship between the acceleration, the gravitational constant, and the angle of inclination. The formula is as follow:
sin(θ) = acceleration / g
Substitute the given acceleration (2.6 m/s²) and the gravitational constant g (9.8 m/s²) and solve for θ.
b. To find the final velocity of the boxes, we can apply the equations of motion. Using the formula v² = u² + 2gs (where u is the initial velocity, g is the gravitational constant, s is the distance) and substituting the given values (u=0, g=2.6 m/s², s=1.4m), we find the final velocity.
c. The normal force on a box moving down the ramp can be found from the formula: Normal force = mg cos (θ) (where m is the mass, g is gravitational constant and θ is the angle of inclination). Here, θ does not affect the normal force because the box is moving down a frictionless ramp.

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Answer:

Farther,

Because the stars are far from one another

Explanation:

The star look dim because a star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.

The sun appears very bright to us because it is closer to us, the sun distance from the earth is one light year which is around 92,955,807 miles. Now the closest star to the earth is 4.22 light-years, which is four times that of the sun and so it slowly spread out over time.

Therefore, if two objects have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!

The same applies to star.

Answer:

[tex]\theta = 86.491^{\textdegree}[/tex]

Explanation:

The equations for the horizontal and vertical position of the ball are, respectivelly:

[tex]4.570\,m = [(7.157\,\frac{m}{s})\cdot\cos \theta]\cdot t\\3.050\,m = 2.440\,m +[(7.157\,\frac{m}{s})\cdot \sin \theta]\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]

By isolating each trigonometric component and summing each equation:

[tex]20.885\,m^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \cos^{2} \theta]\cdot t^{2}[/tex]

[tex][0.61\,m + \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}]^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \sin^{2} \theta]\cdot t^{2}[/tex]

[tex]21.257\,m^{2} + (5.982\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = (2623.796\,\frac{m^{2}}{s^{2}})\cdot t^{2}[/tex]

[tex]21.257\,m^{2} - (2617.814\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = 0[/tex]

The positive real roots are:

[tex]t_{1} = 10.434\,s,t_{2} = 0.09\,s[/tex]

The needed angle is:

[tex]\theta = \cos^{-1} [\frac{4.570\,m}{(7.157\,\frac{m}{s} )\cdot t} ]\\\theta_{1} = 86.491^{\textdegree}\\\theta_{2} = NaN[/tex]

Answer:

57.39 rev

Explanation:

From circular motion,

s = rθ................... Equation 1

Where s = distance, r = radius, θ = angular distance.

make θ the subject of the equation

θ = s/r............... Equation 2

Where can look for s using any of the equation of motion

s = (v+u)t/2............ Equation 3

Where v and u = Final and initial velocity respectively, t= time.

Given: v = 21.5 m/s, u = 0 m/s (at rest), t = 11.4 s

Substitute into equation 3

s = (21.5+0)11.4/2

s = 122.55 m.

given: r = 66.5/2 = 33.25 cm = 0.3325 m

Substitute into equation 2

θ = 122.55/0.3325

θ = 368.57 rad

θ = (360.57×0.159155) rev

θ = 57.39 rev

Answer:

58.6886 revolutions

Explanation:

First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:

V2= Vo2 + 2aDS

V = 21.5

Vo = 0

a = 21.5/11.4 = 1.886

(21.5)^2 = 2*1.886*DS

DS = 462.25/3.772 = 122.5477 m

For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 =  208.81 cm =  2.0881 m

So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:

122.5477/2.0881 =  58.6886

Explanation:

Below is an attachment containing the solution.

Answer:

200 J

Explanation:

In this problem, I assume there is no air resistance, so the  system is isolated (=no external forces).

For an isolated system, the total mechanical energy is constant, and it is given by:

[tex]E=KE+PE[/tex]

where

KE is the kinetic energy

PE is the potential energy

The kinetic energy is the energy due to the motion of the object,  while the potential energy is the energy due to the position of the object relative to the ground.

At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its  speed is zero; it means that all its mechanical energy is just potential energy, and it is:

[tex]E=PE_{max}=200 J[/tex]

As the boulder falls  down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.

Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have

[tex]E=KE_{max}=200 J[/tex]

Therefore, the kinetic energy just before hitting the ground is 200 J.

Answer:

Explanation:

Given that

Initial velocity  u=13m/s

Length of slope

L=460m

Height of slope =30m

Mass of cyclist and bike =70kg

Drag force, fictional force=12 N

Final velocity?

Because the system is not isolated, there is some workdone by the drag force.

Therefore,

∆E=W

K.E(f) - K.E(i) + P.E(f) - P.E(i)=W

½mVf² - ½mVi² + mgy(f) - mgy(i)=W

Note, y(f) = 0, the cyclists is already on the floor

½mVf² -½mVi² - mgy(i) = -Fd × d

½×70×Vf² - ½×70×13²-70×9.81×30=-12×450

35Vf²- 5915 - 20601=-5400

35Vf²=-5400+5915+20601

35Vf²=21116

Vf²=21116/35

Vf²=603.314

Vf=√603.314

Vf=24.56m/s

The final velocity is 24.46m/s at the bottom of the track.