Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

[tex]\sigma=E*\epsilon[/tex] (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

[tex]\delta L=L*\epsilon[/tex] (2)

Here L is the member extension and δL the change total longitudinal elongation.  

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

[tex]\sigma=P/A[/tex]  

[tex]\sigma=22.44N / 1290 mm^2[/tex]  

[tex]\sigma=0.0174 N/mm^2[/tex]  

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

[tex]=\sigma=E*\epsilon[/tex]

[tex]\epsilon=\sigma/E[/tex]

[tex]\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2} [/tex]

[tex]\epsilon=8.53*10^-{5}[/tex]

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

[tex]\delta L=3048 mm * 8.53*10^{-5} [/tex]  

[tex]\delta L= 0.25 mm [/tex]

Answer:

A) weight=mass X gravitational force

10Kg X 10 M/S 2

100 KgM/S2

100Newton

B) Force =Mass X gravitational force

10Kg X 10 M/S2

100 Newton

The weight of a 10kg stone sitting at rest on the ground is 98 Newtons, and the normal force acting on it is also 98 Newtons.

The weight of an object is calculated by multiplying its mass by the acceleration due to gravity. In this case, the mass of the stone is 10kg and  The standard acceleration due to gravity on Earth is approximately 9.8 m/s2.  Therefore, the weight of the stone would be 10 kg x 9.8 m/s^2 = 98 Newtons.

b) The normal force acting on the stone is the force exerted by the surface of the ground on the stone in the upward direction. When the stone is sitting on the ground, the normal force is equal in magnitude and opposite in direction to the weight of the stone. Therefore, the normal force is also 98 N.

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Answer:

(a) 6.42 x 10^-18 N

(b) 9.73 x 10^8 m/s^2

Explanation:

v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree

(a) F = q v B Sin theta

F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51

F = 6.42 x 10^-18 N

(b) Acceleration, a = Force / mass

a = (6.42 x 10^-18) / (6.6 x 10^-27)

a = 0.973 x 10^9

a = 9.73 x 10^8 m/s^2

Answer:

[tex]\omega_O = 0.16 rad /sec[/tex]

Q = 0.24

Explanation:

given data:

resonant  angular frequency is given as  \omega_O = \frac{1}{\sqrt{LC}}

where L is inductor = 150 mH

C is capacitor = 0.25\mu F

[tex]\omega = \frac{1}{\sqrt{150*10^{6}*0.25*10^{-6}}}}[/tex]

[tex]\omega_O = 0.16 rad /sec[/tex]

QUALITY FACTOR is given as

[tex]Q = \frac{1}{R}{\sqrt\frac{L}{C}}[/tex]

Putting all value to get quality factor value

Q =[tex] \frac{1}{1000}{\sqrt\frac{150*10^{6}}{0.25*10^{-6}}}[/tex]

Q = 0.24

Final answer:

The maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz in an RLC series circuit with the given values of R, L, and C. The quality factor of the circuit is approximately 57.74.

Explanation:

In an RLC series circuit, the maximum power dissipation in the resistor is achieved at the resonant frequency, which is given by the formula:

fr = 1 / (2π √LC)

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the resonant frequency:

fr = 1 / (2π √(0.15 x 0.00000025))

fr ≈ 1175.5 Hz

Therefore, the maximum power dissipation in the resistor occurs at a frequency of approximately 1175.5 Hz.

The quality factor (Q) of the circuit is a measure of its damping ability. It is given by the formula:

Q = R √C / L

Substituting the given values:

R = 100Ω, L = 150mH (or 0.15H), and C = 0.25μF (or 0.00000025F), we can calculate the quality factor:

Q = 100 √(0.00000025) / 0.15

Q ≈ 57.74

Therefore, the quality factor of the circuit is approximately 57.74.

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of box[tex]A =24\times35=840\ cm[/tex]

Drag coefficient = 0.80

Density of air [tex]\rho= 1.2\times kg/m^3[/tex]

We need to calculate the terminal velocity

Using formula of drag force

[tex]F_{d} = \dfrac{1}{2}\rho v^2Ac[/tex]

Where,

[tex]\rho[/tex] = density of air

A = area

C= coefficient of drag

Put the value into the formula

[tex]78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80[/tex]

[tex]v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}[/tex]

[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}[/tex]

[tex]v=137.68\ m/s[/tex]

Hence, The terminal speed of his is 137.68 m/s.

The terminal speed of the skydiver with dimensions as a rectangular box as he falls feet first is 137.68 m/s.

Terminal speed of a body is the maximum speed, which is achieved by the object when it fall through a fluid.

In the case of terminal velocity, the force of gravity becomes equal to the sum of the drag force and buoyancy force due to fluid on body.

Terminal velocity can be find out as,

[tex]v=\sqrt{\dfrac{2mg}{\rho AC_d}}[/tex]

Here, (m) is the mass, (g) is gravitational force, ([tex]\rho[/tex]) is the density of fluid, (A) is the project area and ([tex]C_d[/tex]) is the drag coefficients.

It is given that, the mass of the skydiver is 78 kg The dimensions of the skydiver is s 24 cm × 35 cm × 170 cm.

The coefficient of drag is 0.80 and the density of air is 1.2 kg/m³.

Put the values in the above formula as,

[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times0.24\times0.35\times 0.8}}\\v=137.68\rm m/s[/tex]

Thus the terminal speed of the skydiver as he falls feet first is 137.68 m/s.

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Answer:

The acceleration of the block is 6.35 m/s².

Explanation:

It is given that,

A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N

Mass of the block, m = 5 kg

We need to find the acceleration of the block. From the attached figure, it is clear that.

[tex]F_x=ma_x[/tex]

[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]

[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]

[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]

[tex]a_x=6.35\ m/s^2[/tex]

So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.

The given data in the problem,

F is the applied force =65 N

Θ is the angle of inclined plane=30°

m is the mass of the block= 5 kg

We need to find the acceleration of the block in the x-direction

[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]

Hence the magnitude of the acceleration of the block will be 6.35 m/sec².

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Answer:

-423 m³/s

Explanation:

Volume of a cone is:

V = ⅓ π r² h

Given r = 2h:

V = ⅓ π (2h)² h

V = ⁴/₃ π h³

Taking derivative with respect to time:

dV/dt = 4π h² dh/dt

Given h = 1010 cm and dh/dt = 33 cm/s:

dV/dt = 4π (1010 cm)² (33 cm/s)

dV/dt ≈ 4.23×10⁸ cm³/s

dV/dt ≈ 423 m³/s

The pile is growing at 423 m³/s, so the bin is draining at -423 m³/s.

The rate at which the sand is leaving the bin at that instant is [tex]423\times 10^{6} cm^{3}/s[/tex].

It is given that the radius of a conical bin is two times its height and at the instant when the height of the bin is [tex]1010cm[/tex], the height of the pile increases at a rate of [tex]33 cm/s[/tex].

The formula for the volume of a cone is given as,

[tex]V=\frac{1}{3}\pi r^{2}h[/tex]

Substitute [tex]r=2h[/tex] as per the question

[tex]V=\frac{4}{3}\pi h^{3}[/tex]

This is the volume of the conical bin.

To find the rate of change in the volume of the bin, differentiate the expression for volume w.r.t. time using the chain rule as follows,

[tex]\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}[/tex]

[tex]\frac{dV}{dt}=\frac{4}{3}\pi (3h^{2}) \times \frac{dh}{dt}\\\\\frac{dV}{dt}=4\pi h^{2} \times \frac{dh}{dt}\\[/tex]

Now, according to the question, at [tex]h=1010cm[/tex], [tex]\frac{dh}{dt}=33[/tex].

Substituting these values, the rate at which the sand is leaving the bin is,

[tex]\frac{dV}{dt}=4\pi (1010)^{2} \times 33\\\frac{dV}{dt}=423,025,503.90235\\\frac{dV}{dt}=423\times10^{6}cm^{3}/s[/tex]

So, the rate at which the sand leaves the conical bin at the given instant is [tex]423\times10^{6}cm^{3}/s[/tex]

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Answer:

v = 0.025 m/s

Explanation:

Given that the voltage is

[tex]V = 5000 x^2[/tex]

now at x = 0

[tex]V_1 = 0 Volts[/tex]

also we have at x = 8 cm

[tex]V_2 = 5000(0.08)^2 = 32 Volts[/tex]

now change in potential energy of the charge is given as

[tex]\Delta U = q\Delta V[/tex]

[tex]\Delta U = (10 \times 10^{-9})(32 - 0)[/tex]

now by mechanical energy conservation law

[tex]\frac{1}{2}mv^2 - 0 = 3.2 \times 10^{-7}[/tex]

[tex]\frac{1}{2}(1 \times 10^{-3})v^2 = 3.2 \times 10^{-7}[/tex]

[tex]v = 0.025 m/s[/tex]

The maximum speed of the particle in arrangement of source charges produces 32 volts electric potential is 0.025 meter per second.

Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.

It can be given as,

[tex]U=qV[/tex]

Here, (q) is the charge and (V) is the electric potential difference.

Given infroamtion-

The electric potential producers by the arrangement of source charges is given by,

[tex]V=5000x^2[/tex]

The mass of the particle is 1.0 gram.

The charge of the particles 10 nC.

As, the electric potential producers by the arrangement of source charges is given by,

[tex]V=5000x^2[/tex]

At the x equal to 8 cm or 0.08 m, the equation become,

[tex]V=5000(0.08)^2\\V=32\rm V[/tex]

Thus the potential difference at the is 32 volts.

The electric potential energy of the particle is,

[tex]U=10\times10^{-9}\times32\\U=3.2\times10^{-7}\rm j[/tex]

Now the electric potential energy is equal to the kinetic energy of the particle. Thus,

[tex]\dfrac{1}{2}\times0.001\times v^2=3.2\times10^{-7}\\v=0.025\rm m/s[/tex]

Thus the maximum speed of particle is 0.025 meter per second.

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Answer:

B= [tex]\sqrt{65}[/tex] ≅8.06

Explanation:

Using the Pythagorean theorem:

[tex]C^{2}[/tex]= [tex]A^{2}[/tex] + [tex]B^{2}[/tex]

where C represents the length of the hypotenuse and A and B the lengths of the triangle's other two sides, we can find out the lenght of B assuming the value of the hypotenuse being 9 and A being 4.

[tex]9^{2}[/tex]=[tex]4^{2}[/tex] + [tex]B^{2}[/tex]

81= 16+ [tex]B^{2}[/tex]

81-16= [tex]B^{2}[/tex]

B= [tex]\sqrt{65}[/tex] ≅8.06

The length of B is equal to 8.06 units

Data given;

To solve this question, we have to use the formula of finding resultant vectors

Since it's a right-angle triangle, let's use Pythagoras' theorem

[tex]C^2=A^2 + B^2\\9^2 = 4^2 + B^2\\b^2 = 9^2 - 4^2\\b^2 = 65\\b = \sqrt{65}\\b = 8.06[/tex]

From the calculation above, the length of B is equal to 8.06.

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You haven't told us anything about the detectors being used.  We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.

All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector.  That is, assume . . .

(double the number of photons) ===> (detect the source in half the time) .

-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.

So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.

-- So we'd expect your instrument to detect the same star in

(119.5 min) / (12.96) = 9.22 minutes  .

We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.

If your telescope has diameter 0.18 metres, for 9.22 minutes, you need to expose for.

An optical telescope is one that collects and sharply concentrates light. Mostly from the visible parts of the spectrum. That is to make a magnified image for close inspection, to take a picture, or you might say to get information from an electronic sensor image.

A reflecting telescope, sometimes called a reflector, is one that creates images by reflecting light off either a single curved mirror or a group of mirrors. Sir Isaac Newton created the reflecting telescope inside the 17th century as a replacement for the refracting one.

your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.

(119.5 min) / (12.96) = 9.22 minutes

Therefore, for 9.22 minutes, you need to expose for.

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Answer:

Magnetic field, B = 0.0045 T            

Explanation:

It is given that,

Speed of the proton, [tex]v=1.5\times 10^7\ m/s[/tex]

Acceleration of the proton, [tex]a=0.66\times 10^{13}\ m/s^2[/tex]

Charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]

The magnetic force is balanced by the force due to the acceleration of the proton as :

[tex]qvB=ma[/tex]

[tex]B=\dfrac{ma}{qv}[/tex]

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.66\times 10^{13}\ m/s^2}{1.6\times 10^{-19}\ C\times 1.5\times 10^7\ m/s}[/tex]

B = 0.0045 T

So, the magnitude of magnetic field on the proton is 0.0045 T. Hence, this is the required solution.

Answer:

1)  The acceleration of the car = 1.17 m/s²

2) The time required to reach 55 mi/h = 7.59 seconds.

Explanation:

1)  Initial speed of motorist, u = 35 mph = 15.56 m/s

   Final speed of motorist, v = 55 mph = 24.44 m/s

   Distance traveled, s = 500ft = 152.4 m

   We have v² = u² + 2as

                  24.44² = 15.56² + 2 x a x 152.4

                  a = 1.17 m/s²

  The acceleration of the car = 1.17 m/s²  

2) We have v = u + at

                   24.44 = 15.56 + 1.17 x t

                           t = 7.59 s

   The time required to reach 55 mi/h = 7.59 seconds.  

Answer:

(a) 2.23 m/s

(b) 0.9 m

Explanation:

h = 1 m, t = 0.1 second

horizontal component of initial velocity, ux = 2 m/s

vertical component of initial velocity, uy = 0  

(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

The horizontal component of velocity remains constant as in this direction, acceleration is zero.

vx = ux = 2 m/s

Use first equation of motion in Y axis direction.

vy = uy + g t

vy = 0 + 9.8 x 0.1 = 0.98 m/s

Resultant velocity after 0.1 second

v^2 = vx^2 + vy^2

v^2 = 2^2 + 0.98^2

v = 2.23 m/s

(b) Let it takes time t to land.

Use second equation of motion along Y axis

h = uy t + 1/2 g t^2

1 = 0 + 1/2 x 9.8 x t^2

t = 0.45 second

Let it lan at a distance x.

so, x = ux x t

x = 2 x 0.45 = 0.9 m

It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.

The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.

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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.

Orbit of Venus Around the Sun-

The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.

Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.

Answer:

3260.33 J

Explanation:

[tex]n [/tex]  = number of rods = 5

[tex]L [/tex]  = length of each rod = 0.715 m

[tex]m [/tex]  = mass of rod = 2.51 kg

[tex]I [/tex]  = total moment of inertia

Total moment of inertia is given as

[tex]I = \frac{nmL^{2}}{3}[/tex]

[tex]I = \frac{(5)(2.51)(0.715)^{2}}{3}[/tex]

[tex]I [/tex] = 2.14 kgm²

[tex]w [/tex]  = angular speed = 527 rpm = 55.2 rad/s

Rotational kinetic energy is given as

E = (0.5) [tex]I [/tex]  ([tex]w [/tex] )²

E = (0.5) (2.14) (55.2)²

E = 3260.33 J

Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

Answer:

c) Elastic Modulus

Explanation

As we know that when deformation is under elastic limit then stress applied to the given material is proportional to the strain developed in it

So here we can say that since they both are directly proportional to each other so the proportionality constant here is known as Modulus of elasticity.

So we can say it is given as

[tex]stress = E (strain)[/tex]

so now if we draw a graph between between stress and strain then it must be a straight line and the the slope of this straight line is given as

[tex]Slope = \frac{Stress}{Strain}[/tex]

So here correct answer will be

c) Elastic Modulus

Final answer:

The slope of the stress-strain curve in the elastic deformation region is known as the Elastic Modulus or Young's Modulus, which measures the stiffness of the material. The correct answer to the given question is 'c. Elastic Modulus'.

Explanation:

The slope of the stress-strain curve in the elastic deformation region represents the relationship between stress and strain under elastic conditions. This slope is in fact the Elastic Modulus, which is also known as Young's Modulus when it's in tension or compression. It serves as a measure of the stiffness or rigidity of the material, indicating how much stress is required to achieve a certain amount of strain.

In the context of a stress-strain curve, the plastic modulus is associated with plastic deformation, not the elastic region. Poisson's ratio is another material property that describes the ratio of transverse strain to axial strain, and is not the slope of the curve. Hence, the correct answer to the question is 'c. Elastic Modulus'.

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

[tex]x(t)=(0.28\ m)\cos[(8\ rad/s)t][/tex]....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

[tex]y=A\omega \cos\omega t[/tex]

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

[tex]F_{g}[/tex] - [tex]F_{d}[/tex] - [tex]T[/tex] = 0

7000 - 1800 - [tex]T[/tex] = 0

[tex]T[/tex] = 5200 N

[tex]T[/tex] = 5.2 x 10³ N

Part B)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

[tex]T[/tex]  - [tex]F_{g}[/tex] - [tex]F_{d}[/tex] = 0

[tex]T[/tex] - 7000 - 1800  = 0

[tex]T[/tex] = 8800 N

[tex]T[/tex] = 8.8 x 10³ N

Answer:

The new force F' will be same of the original force F.

Explanation:

Given that,

Charges = 0.3

We need to calculate the force between the charges

Suppose that the distance between the charges is r.

The force between the charges

[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]

[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]

If the charges are doubled, and the distance between them increased by 100%.

So, The charges are 0.6 and the distance is 2r.

Then,

The force between the charges

[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]

[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]

[tex]F'=F[/tex]

Hence, The new force F' will be same of the original force F.

Answer:

Force, F = 16814.95 N

Explanation:

It is given that,

Mass of space station, m = 2010 kg

Altitude, [tex]d=5.35\times 10^5\ m[/tex]

Mass of earth, [tex]m=5.98\times 10^{24} kg[/tex]

Mean radius of earth, [tex]r=6.37\times 10^6\ m[/tex]

Magnitude of force is given by :

[tex]F=G\dfrac{Mm}{R^2}[/tex]

R = r + d

[tex]R=6.37\times 10^6\ m+5.35\times 10^5\ m=6905000\ m[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{2010\ kg\times 5.98\times 10^{24} kg}{(6905000\ m)^2}[/tex]

F = 16814.95 N

So, the force between the space station and the Earth is 16814.95 N. Hence, this is the required solution.

Explanation:

It is given that,

Mass of first ball, m₁ = 2 kg

Mass of other ball, m₂ = 3.5 kg

Velocities of both balls, u = 0.9 m/s

(1) The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. We need to find the final velocity of second ball. Applying the conservation of momentum as :

[tex]2\ kg\times 0.9-3.5\ kg\times 0.9\ m/s=-2\ kg\times 0.9\ m/s+3.5v[/tex]

v is the final velocity of heavier ball.

v = 0.128 m/s

or

v = 0.13 m/s

Initial kinetic energy, [tex]E_i=\dfrac{1}{2}\times (2\ kg+3.5\ kg)\times (0.9\ m/s)^2=2.23\ J[/tex]

Final kinetic energy, [tex]E_f=\dfrac{1}{2}\times 2\ kg\times (0.9\ m/s)^2+\dfrac{1}{2}\times 3.5\ kg\times (0.13\ m/s)^2=0.84\ J[/tex]

Lost in kinetic energy, [tex]\Delta KE=0.84-2.23=-1.39\ J[/tex]

Hence, this is the required solution.

Answer:

The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².

Explanation:

Given that,

Angular velocity of optical disk= 1045 rad/s

Angular velocity of particular disk = 646.1 rad/s

Time = 0.234 sec

We need to calculate the average angular acceleration

Using formula of  angular acceleration

[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]

Where, [tex]\omega_{f}[/tex] = final angular velocity

[tex]\omega_{i}[/tex] = Initial angular velocity

t = time

Put the value in the equation

[tex]\alpha = \dfrac{0-646.1}{0.234}[/tex]

[tex]\alpha= -2761.1\ rad/s^2[/tex]

Negative sign shows the angular deceleration.

Hence, The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².

Explanation:

It is given that,

Mass of SUV, m = 900 kg

It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :

F = m a

Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity

As SUV is travelling at a constant speed. This gives acceleration of it as zero.

So, F = 0

So, the total force acting on SUV is 0 N. Hence, this is the required solution.

Answer:

The total force acting on the vehicle is 0 N.

Explanation:

Given that,

Mass of SUV = 900 kg

Velocity = 1.44 m/s

SUV is travelling at a constant speed of 1.44 m/s due north.

We need to calculate the force on the vehicle.

Using newton's second law

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

We know that,

The acceleration is the first derivative of the velocity of the particle.

[tex]a = \dfrac{dv}{dt}[/tex]

Now,put the value of a in equation (I)

[tex]F=m\dfrac{dv}{dt}[/tex]

SUV is traveling at a constant speed it means acceleration is zero.

So, The force will be zero.

Hence, The total force acting on the vehicle is 0 N.

Answer:

21828 m

Explanation:

[tex]v_{ground}[/tex] = speed of sound through ground = 6 km/s = 6000 m/s

[tex]v_{air}[/tex] = speed of sound through air = 343 m/s

t = time taken for the vibrations to arrive

t' = time taken for sound to arrive = t + 60

d = distance of the point of explosion

time taken for the vibrations to arrive is given as

[tex]t = \frac{d}{v_{ground}}[/tex]                            eq-1

time taken for the sound to arrive is given as

[tex]t' = \frac{d}{v_{air}}[/tex]

[tex]t + 60 = \frac{d}{v_{air}}[/tex]

using eq-1

[tex]\frac{d}{v_{ground}} + 60 = \frac{d}{v_{air}}[/tex]

[tex]\frac{d}{6000} + 60 = \frac{d}{343}[/tex]

d = 21828 m

Answer:

The average power is 619 W.

(B) is correct option

Explanation:

Given that,

Weight = 62.0 kg

Height = 4.28 m

Time t = 4.20 s

We need to calculate the work done

Work done by woman

[tex]W=mgh[/tex]

Where,

m = mass

g = acceleration due to gravity

t = time

Put the value into the formula

[tex]W=62\times9.8\times 4.28[/tex]

[tex]W=2600.528 J[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{W}{t}[/tex]

Where,

W = work done

t = time

Put the value in to the formula

[tex]P=\dfrac{2600.528}{4.20}[/tex]

[tex]P=619\ W[/tex]

Hence, The average power is 619 W.

The average power supplied by the woman is calculated by first determining the work done against gravity to move up the stairs, and then dividing this by the time taken. After performing these calculations, the most accurate answer is approximately 629 W.

The question is asking to calculate the average power supplied by a woman while running up the stairs. Power is defined as the rate at which work is done. In this scenario, the work is the woman's movement against the gravitational force, which is calculated using the formula W = mgh, where m is mass, g is the acceleration due to gravity and h is the height of the stairs. Substituting the given values, we get W = 62.0 Kg * 9.8 m/s² * 4.28 m = 2650.368 Joules. Power is calculated by dividing the work done by the time taken, so P = W/t = 2650.368 J / 4.20 s = 630.8 Watts, which is closest to option (d) 629 W.

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Answer: false

Explanation: the air will increase if it is compressed by an adiabatic compressor

Explanation:

A satellite orbiting the Earth is a characteristic example of the uniform circular motion, where the direction of the velocity vector [tex]\vec{V}[/tex] is perpendicular to the radius [tex]r[/tex] of the trajectory.  Hence, the velocity is changing although the speed is constant.

On the other hand,  acceleration [tex]\vec{a}[/tex] is directed toward the center of the circumference (that is why it is called centripetal acceleration).  

Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:  

[tex]\vec{F}=m.\vec{a}[/tex]  

Therefore, the net force on the satellite resulting from its circular motion points towards the center of the circle (where the Earth is in the Earth-satellite system).

Answer:

The speed of the ambulance is 4.66 m/s.

Explanation:

Given that,

The siren emitting a whine at 1570 Hz

The cyclist pedaling a bike at 2.45 m/s

The cyclist hears a frequency of 1560 Hz

We know that,

Speed of sound wave

[tex]v = 343\ m/s[/tex]

We calculate the speed of the ambulance

Using Doppler effect,

[tex]f'=f\times\dfrac{v+v_{o}}{v+v_{s}}[/tex]

Where,

[tex]f' [/tex]= frequency of ambulance siren

[tex]f [/tex]= cyclist hears the frequency

[tex]v_{s}[/tex]=speed of source

[tex]v_{v}[/tex]= speed of observer

Put the value in to the formula

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

[tex]v_{s}=1570\times\dfrac{343-2.45}{1560}-343[/tex]

[tex]v_{s}=4.66\ m/s[/tex]

Hence, The speed of the ambulance is 4.66 m/s.

Answer:

a) There are 100 centimeters in 1 meter.

b) [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Explanation:

a) We have the conversion

         1 m = 100 cm

   So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

    [tex]1cm=\frac{1}{2.54}inch[/tex]

   [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]