A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a mass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?

To develop this problem it is necessary to apply the concepts related to kinetic energy and gravitational potential energy.

By conserving energy we know that

[tex]PE = KE[/tex]

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

Where,

m = mass

v = Velocity

G = Gravitational universal constant

M = Mass of Spherical asteroid

m = mass of object

r = Radius

Mass of a Sphere can be expressed as,

[tex]M= \rho* (\frac{4}{3}\pi r^3 )[/tex]

Replacing we have that,

[tex]\frac{1}{2}*2.8^2 - 6.67*10^{-11}*\frac{4\pi*2000*r^3}{3r} = 0[/tex]

[tex]6.67*10^{-11}*(\frac{4\pi}3{}2000*r^2)=\frac{1}{2}*2.8^2[/tex]

[tex]r = \sqrt{\frac{1}{2}*\frac{2.8^2}{6.67*10^{-11}*2000*4/3*\pi}}[/tex]

[tex]r=2648 m[/tex]

Therefore the diameter is 5296 m.

b) Applying the concept of gravitational force and centripetal force we have to

[tex]F_g = F_c[/tex]

[tex]\frac{G M m}{r^2} = \frac{m v^2}{r}[/tex]

[tex]\frac{G M}{r} = v^2[/tex]

[tex]6.67*10^{-11}*\frac{(\frac{4\pi}{3}2000*r^3)}{r} = 2.8^2[/tex]

[tex]r=3746 m[/tex]

Therefore the diameter would be 7492 m

The diameter of the largest asteroid from which a froghopper could jump free can be calculated using the escape velocity formula. Similarly, for the froghopper to enter a circular orbit, its horizontal speed needs to equal the required orbital speed, calculated using the gravitational field strength at the surface of the asteroid.

To find the diameter of the largest spherical asteroid from which a froghopper could jump free, we need to understand the escape speed, the speed necessary to overcome an object's gravitational field. The formula for escape velocity is √(2*G*M/R), where G is the gravitational constant, M is the mass of the asteroid, and R is its radius (half of the diameter).

In this case, we can solve the formula for the radius, R = 2*G*M/v², where v is the takeoff speed of a froghopper (2.8 m/s). The mass of the spherical asteroid can be calculated from the given density (2.0 g/cm³ or 2000 kg/m³) and the volume of a sphere (4/3*π*R³). Substituting the mass into the formula will provide the radius, and hence diameter, of the asteroid.

For a froghopper to enter a circular orbit just above the surface of the asteroid, its horizontal speed must equal the circular orbital speed, which depends on the gravitational field strength at the surface of the asteroid. It's determined by the formula v = √(G*M/R), which can be rearranged to solve for the required diameter of the asteroid.

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Answer:

T= 1.71×10^{-3} sec= 1.71 mili sec

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

Explanation:

First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)=  5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C

a) I(t) = -ωqosin(ωt+φ)

⇒Io= ωqo

⇒ω= Io/qo

also we know that T= 2π/ω

⇒T= [tex]2\pi\frac{q_0}{I_0}[/tex]

now putting the values we get

= [tex]2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}[/tex]

= 1.71×10^{-3} sec

b) note that the time [tex]t_{fc}[/tex] it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is

[tex]t_{fc}= \frac{T}{4}[/tex]

[tex]t_{fc}= \frac{ 1.71\times10^{-3} }{4}[/tex]

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

[tex]F = ma[/tex]

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{11*10^3}{80}[/tex]

[tex]a = 137.5m/s[/tex]

From the cinematic equations of motion we know that

[tex]v_f^2-v_i^2 = 2ax[/tex]

Where,

[tex]v_f =[/tex]Final velocity

[tex]v_i =[/tex]Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

[tex]v_f^2-v_i^2 = 2ax[/tex]

[tex]0-v_i^2 = 2(-g)x[/tex]

[tex]v_i =\sqrt{2gx}[/tex]

[tex]v_i = \sqrt{2*9.8*4.8}[/tex]

[tex]v_i = 9.69m/s[/tex]

From the equation of motion where acceleration is equal to the velocity in function of time we have

[tex]a = \frac{v_i}{t}[/tex]

[tex]t = \frac{v_i}{a}[/tex]

[tex]t =\frac{9.69}{137.5}[/tex]

[tex]t = 0.0705s[/tex]

Therefore the time required is 0.0705s

The minimum time for the mass to come to rest at the end of the fall is about 0.071 s

[tex]\texttt{ }[/tex]

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

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

[tex]\texttt{ }[/tex]

[tex]\large {\boxed {F = \Delta (mv) \div t }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

v = Velocity of Object ( m/s )

t = Time Taken ( s )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of climber = m = 80 kg

height of fall = h = 4.8 m

net force = ∑F = 11 k N = 11000 N

Asked:

minimum time = t = ?

Solution:

FIrstly , we could find the initial velocity of climber as he caught by the rope:

[tex]v^2 = u^2 + 2gh[/tex]

[tex]v^2 = 0^2 + 2(9.8)(4.8)[/tex]

[tex]v^2 = 94.08[/tex]

[tex]v = \frac{28}{3}\sqrt{5} \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

Next , we will use Newton's Law of Motion to calculate the minimum time:

[tex]\Sigma F = \Delta p \div t[/tex]

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]11000 = 80 (\frac{28}{3}\sqrt{5}) \div t[/tex]

[tex]t = 80 (\frac{28}{3}\sqrt{5}) \div 11000[/tex]

[tex]t \approx 0.071 \texttt{ s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer:

.09 mm

Explanation:

In case of diffraction by single slit with width a the width of central maxima is given below

width = 2 λD/a

where λ is wavelength of light , D is distance of screen , a is width o slit

substituting the given values

14 x 10⁻³ = [tex]\frac{2\times530\times10^{-9}\times1.2}{a}[/tex]

a = .09 mm

b) If a is greater , width of central maxima will be less wide.

Answer:

c) Equal to R

b) the width of the resonance

Explanation: In general the impedance Z of an electrc circuit is:

Z = R + jX

Now when the circuit is capacitive, the above mentioned relation become  Z = R + 1/jwc  where ( w = 2πf  )

And when the circuit is inductive Z becomes

Z  =  R + j wl

Resonance  condition implies that the reactance created by capacitors are equal at the inductances produced by inductors, in other words the circuit will behaves as it were resistive.

The impedance will be equal to R

The Q factor is[

Q = X/R  in which X is the module of either the capacitive or inductive reactance.

Q has an inverse relation with the band width.  

Answer:[tex]a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )[/tex]

Explanation:

Given

Electric Field [tex]\vec{E}=2\hat{i}+5.4\hat{j}[/tex]

[tex]\vec{B}=0.4\hat{k}\ T[/tex]

velocity [tex]\vec{v}=8\hat{i} m/s[/tex]

mass of electron [tex]m=9.1\times 10^{-31} kg[/tex]

Force on a charge Particle moving in Magnetic Field

[tex]F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ][/tex]

[tex]a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}[/tex]

[tex]a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}[/tex]  

[tex]a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2[/tex]

The Lorentz force law is used to calculate the force exerted on an electron in an electromagnetic field. This force is then divided by the mass of the electron to calculate the acceleration.

The acceleration of an electron moving in both electric and magnetic fields can be found using the Lorentz force law, which states that the force exerted on a charged particle in an electromagnetic field is the vector sum of the electric and magnetic forces. The formula is F = q(E + v × B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field. The acceleration a is then calculated as the force divided by the mass m of the electron, which is known to be 9.11 x 10^-31 kg.

First, calculate the force F due to the electric field E (since F = qE) and the magnetic field B (since F = qvBsinθ). Using the given values, E = (2.00î + 5.40ĵ) V/m, v = 8.0î m/s, and B = 0.400k T, we find that Fe = qE = -(1.60 x 10^-19 C)(2.00î + 5.40ĵ) N (the charge of an electron is -1.60 x 10^-19 C), and Fb = qvBsinθ = - (1.60 x 10^-19 C)(8.0î m/s)(0.400 T), since the angle θ between v and B is 90 degrees. Then, the total force F is the vector sum of Fe and Fb.

Finally, calculate the acceleration a by dividing the total force F by the mass m of the electron to get a = F/m.

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Answer:

B) AC current can be used in transformers while DC current cannot.

Explanation:

AC current is used for long distance transportation of electrical energy with amplifying the voltage and reducing the current in the conductors so that there is minimum loss of energy in the form of heat.

The relation between heat energy and electrical current is given by Joule's law as:

[tex]Q=i^2.R.t[/tex]

where:

The above mentioned variation  of current and voltage in the transmission lines of AC is achieved by a device called transformer. It consist of a rectangular shaped steel core common to two of the insulated wire winding of which one side acts as input (primary coil) and the other one acts as secondary (output).

AC i.e. alternating current can only be used in the transformers because they have the continuously varying amplitude which in turn provides the change in flux with respect to the time inducing an emf in the  nearby secondary coil of the transformer without any physically moving parts.

The alternating nature of AC current aloows changes in magnet fields to be used to transform electricity and make it more efficient to transport over long distances. AC current can be used in transfromers while DC current cannot.

The question is from the domain of Physics, where it deals with the behavior of gases subjected to temperature changes, as per Gay-Lussac's Law. It can be solved using the formula P1/T1 = P2/T2 after converting Celsius to Kelvin.

The subject of this question is Physics, and it is suitable for students at the High School level. The question deals with the concept of gases and their behavior when subjected to changes in temperature, specifically within the context of the ideal gas law and Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant. To find the new absolute pressure p2 after the gas is heated from 20.0°C to 40.0°C in a rigid container, we can use Gay-Lussac's Law in the form of P1/T1 = P2/T2, assuming the amount of gas and the volume remains constant. Since temperatures must be in Kelvin for gas law calculations, they will be converted first: T1 = 293.15 K and T2 = 313.15 K. The new pressure p2 can then be calculated by multiplying p1 by the ratio of T2 to T1.

Answer:

66.4767 m

Explanation:

[tex]L_0[/tex] = Original length of power line = 145 km

[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]

Initial temperature = 13.7°C

Final temperature = 42°C

Change in length of a material is given by

[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=1.62\times 10^{-5}\times 145000\times (42-13.7)\\\Rightarrow \Delta=66.4767\ m[/tex]

The change in length will be 66.4767 m

To solve this problem we resort to the concept of potential energy which is given by the equation

[tex]PE = mgh[/tex]

Where,

m= mass

g= Gravitational acceleration

h = Height

The height is given in the form of a component, that is, it is given the length that is 5 m and the angle of 37 degrees, therefore the height would be

[tex]h=Lsin\theta[/tex]

[tex]h = 5*sin37[/tex]

[tex]h = 3m[/tex]

Applying the potential energy formula we have to

[tex]PE = mgh[/tex]

[tex]PE = F_g(h)[/tex]

[tex]PE = 40N*3m[/tex]

[tex]PE = 120J[/tex]

Therefore the net change in potential energy of the crate is 120J.

The net change in potential energy of the crate is 148.3 N.m.

To calculate the net change in potential energy of the crate, we can use the equation: ∆PE = (mgh) +fric, where ∆PE is the change in potential energy, m is the mass of the crate, g is the acceleration due to gravity, h is the height of the inclined plane, and fric is the work done by friction. In this case, the work done by friction is given by the formula: fric = Ffric * d * cos(θ), where Ffric is the force of friction, d is the length of the inclined plane, and θ is the angle of the inclined plane with respect to the horizontal.

Plugging in the values, we have: ∆PE = (40 N * 5.0 m * sin(37°)) + (10 N * 5.0 m * cos(37°)). Evaluating the trigonometric functions, we get: ∆PE = (107.8 N·m) + (40.5 N·m) = 148.3 N·m.

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Answer:

B  blue

Explanation:

This is an exercise of photoelectric effect where electrons are removed from the metal by the incident radiation, the explanation of this phenomenon was given by Einstein, they assume that the light is formed by quanta and these collide with the electrons promoting their transfer; The equation that describes the process is

       [tex]K_{max}[/tex] = h f - Φ

Where[tex]K_{max}[/tex]is the kinetic energy, maximize the electrons started, h the plan constant, f the frequency and Φ the work function of the metal (link energy)

The radiation with the lowest energy that an electron can start

      h f = Φ

Let's analyze the energy of the light rays with the wavelength

    c = λ f

    f = c / λ

λ red = 700 10⁻⁹ m

    f = 3 108/700 10⁻⁹

    f₁ = 4.29 10¹⁴ Hz

Yellow  λ = 600 10⁻⁹ m

    f₂ = 3 108/600 10⁻⁹

    f₂ = 5 10¹⁴ Hz

λ blue = 450 10⁻⁹ m

    f₃ = 6.67 10¹⁴ Hz

In the initial metal with a work function fi1 yellow light was needed, to start the electrons, when analyzing the initial equation if the work function increases Φ₂> Φ₁ more energy is needed to take out the electrons from the atom  (E₂>E₁)

light of greater energy (frequency) than yellow is needed so that light is needed to go BLUE

The kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.

To find the change in kinetic energy of the spaceship as it moves from position R1 = 7300 km to position R1 = 6700 km in the gravitational field of Earth, we can use the gravitational potential energy and the fact that the mechanical energy is conserved. The change in kinetic energy is equal to the negative of the change in potential energy.

The gravitational potential energy (U) is given by the formula:

U = -G * (m1 * m2) / R

Where:

- G is the universal gravitational constant (approximately 6.674 × 10^(-11) N·(m/kg)^2).

- m1 is the mass of the Earth (approximately 5.972 × 10^24 kg).

- m2 is the mass of the spaceship (8600 kg in this case).

- R is the distance from the centre of the Earth.

Now, let's calculate the potential energy at R1 = 7300 km and R2 = 6700 km:

For R1 = 7300 km:

R1 = 7300 km = 7,300,000 meters

U1 = -G * (m1 * m2) / R1

U1 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 7,300,000 m

U1 ≈ -5.125 × 10^9 J

For R2 = 6700 km:

R2 = 6700 km = 6,700,000 meters

U2 = -G * (m1 * m2) / R2

U2 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 6,700,000 m

U2 ≈ -5.504 × 10^9 J

Now, let's find the change in potential energy (ΔU) as the spaceship moves from R1 to R2:

ΔU = U2 - U1

ΔU ≈ (-5.504 × 10^9 J) - (-5.125 × 10^9 J)

ΔU ≈ -0.379 × 10^9 J

So, the change in potential energy is approximately -0.379 × 10^9 Joules.

The change in kinetic energy is equal to the negative of the change in potential energy, so:

Change in kinetic energy = -ΔU

Change in kinetic energy ≈ 0.379 × 10^9 J

Therefore, the kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.

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Answer:

Estimation: year 1110.

Explanation:

We need to know how much time it takes to move 100 arcseconds if it moves at 0.11 arcsecond per year. Similarly to any velocity equation [tex]v=\frac{d}{t}[/tex], where in our case the distances are angular, we will obtain the time by doing:

[tex]t=\frac{d}{v}=\frac{100arc}{0.11arc/year}=909 years[/tex]

Which, considering from 2019, the explosion ought to have been observed around 1110 (in reality it was observed by Chinese astronomers in 1054).

The explosion of the Crab Nebula was estimated to have been observed around 1054 AD based on the given information.

To estimate the year in which the explosion of the Crab Nebula was observed, we can use the information provided. The gas blobs that are now 100 arcseconds from the center of the nebula are moving away from the center by about 0.11 arcsecond per year. By dividing the distance of 100 arcseconds by the rate of motion of 0.11 arcsecond per year, we can estimate that it would have taken approximately 909 years for the gas blobs to reach their current position. Therefore, the explosion of the Crab Nebula would have been observed around 1054 AD.

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To solve the problem it is necessary to apply the concepts related to wavelength and error calculation.

The calculation of the error is usually defined by

[tex]z = \frac{\lambda_o-\lambda_r}{\lambda_r}[/tex]

Where

[tex]\lambda_o =[/tex] Wavelength from the observer

[tex]\lambda_r =[/tex] Wavelength from the rest

The previous formula is exactly equal to,

[tex]z=  \frac{\lambda_o}{\lambda_r}-1[/tex]

[tex](z+1) = \frac{\lambda_o}{\lambda_r}[/tex]

[tex]\frac{\lambda_o}{\lambda_r}=(10+1)[/tex]

[tex]\lambda_o=11 \lambda_r[/tex]

Therefore the observed wavelength will be 11 times longer that the emitted wavelength.

If we make the comparison for the ultraviolet region (from 10nm to 400nm) we get that,

[tex]\lambda_o=11 \lambda_r[/tex]

[tex]\lambda_o=11 (\frac{10nm*400nm}}{2})[/tex]

[tex]\lambda_o = 2255nm[/tex]

That is the infrared region of electromagnetic spectra.

The wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.

The redshift (z) corresponds to a change in the way in which the frequency of light photon waves is observed in the spectroscope as a function of the relative velocity between the emitting source and the observing receiver (a measure of distance, of according to cosmological expansion, based on recessional velocity).

A decrease in frequency occurs as the distance from the source galaxy increases, causing the light ray to be captured as a standard color. The deviation or distance measurement is represented by the letter z.

With this information, we can conclude that the wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.

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Answer:204.9 N

Explanation:

Given

Radius [tex]r=3.4 in.[/tex]

Density of iron weight [tex]\rho =0.28 pound/in.^3[/tex]

Volume of sphere[tex]=\frac{4\pi r^3}{3}[/tex]

[tex]V=\frac{4\pi (3.4)^3}{3}[/tex]

[tex]V=164.65\ in.^3[/tex]

mass of sphere [tex]m=\rho \cdot V[/tex]

[tex]m=0.28\times 164.65=46.10 Pounds\approx 20.91 kg[/tex]

[tex]weight=mg=20.91\times 9.81=204.91\ N\ or\ 46.063353\ Pound-Force [/tex]                  

The weigh of the sphere depend on its mass. The weigh of the sphere is 204.722 N.

Mass is defined as the amount of matter contained in a physical body.

Given that the radius r of the sphere is 3.4 inches and the density [tex]\rho[/tex] of the iron is 0.28 pounds per cubic inch.

The volume of the sphere is calculated as below.

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

[tex]V = \dfrac {4}{3} \times 3.14\times (3.4)^3[/tex]

[tex]V = 164.55 \;\rm cube\;Inches[/tex]

Now the mass of the sphere is calculated as given below.

[tex]m = \rho \times V[/tex]

[tex]m = 0.28 \times 164.55[/tex]

[tex]m = 46.074 \;\rm Pounds[/tex]

We know that 1 pound is equal to 0.453592 kg. So,

[tex]m = 46.074 \times 0.453592 \;\rm kg[/tex]

[tex]m = 20.89 \;\rm kg[/tex]

The weight of the sphere is given below.

[tex]W = mg[/tex]

Where g is the gravitational acceleration.

[tex]W = 20.89 \times 9.8[/tex]

[tex]W = 204.722 \;\rm N[/tex]

Hence we can conclude that the weigh of the sphere is 204.722 N.

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To solve this problem it is necessary to apply the concepts related to the conservation of the moment and the conservation of energy.

The conservation of energy implies that potential energy is transformed into kinetic energy or vice versa, depending on the process, in this way

[tex]KE = PE[/tex]

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

Where,

m = mass

v = velocity

h = height

g = gravitational acceleration

Re-arrange to find v,

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

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

[tex]v = \sqrt{2gh_1}[/tex]

Replacing with our values we have that

[tex]v = \sqrt{2gh_1}[/tex]

[tex]v = \sqrt{2(9.8)(3.9)}[/tex]

[tex]v = 8.74m/s[/tex]

At this point we can find the final speed through the conservation of the momentum, that is

[tex]mv = (m+M)V[/tex]

Here,

m = mass of the first person

M = Mass of the second person

v = Initial velocity

V = Final Velocity

Replacing,

[tex]mv = (m+M)V[/tex]

[tex](85)(8.74) = (85+55)V[/tex]

[tex]V = \frac{(85)(8.74) }{(85+55)}[/tex]

[tex]V = 5.30m/s[/tex]

Applying energy conservation again, but in this case using the values of the final state we have to

[tex]\frac{1}{2}(m+M)V^2 = (m+M)gh_2[/tex]

[tex]\frac{1}{2}V^2 = gh_2[/tex]

[tex]h_2 = \frac{V^2}{2g}[/tex]

[tex]h_2 = \frac{5.3^2}{2*9.8}[/tex]

[tex]h_2 = 1.433m[/tex]

Therefore the maximum height that they reach after their upward swing is 1.433m

Answer:

A)[tex]0.306k[/tex]

B)[tex]0.1325k[/tex]

C)[tex]v_{s_{f}} =7.65(-i)+17.35(-j)[/tex]

D)[tex]w=41.64 rads^{-1}[/tex]

Explanation:

Given:

To find the initial angular momentum about origin which is the 50th mark of the metre scale (It's COM) :                                                                                                                          angular momentum [tex]L[/tex]=[tex]mv[/tex]x[tex]r[/tex]                                      where,                                                                                                                       v  - is the velocity of the puck perpendicular to the radial vector

r  -  is the radius vector

∴

A) [tex]L_{i} =  m_{h}r*v_{h_{i} }\\L_{i}= 0.17*\frac{50-30}{100} (-i)*9(-j)\\L_{i} = 0.306 k[/tex]

B) after collision , it moves 30° from original path;

   and it's speed = [tex]\frac{9}{2} =4.5ms^{-1}[/tex];

   ∴the perpendicular velocity [tex]v_{per}[/tex] = 4.5[tex]cos30[/tex] = [tex]2.25\sqrt{3}[/tex][tex]ms^{-1}[/tex]

⇒[tex]L=m_{h}r*v=0.17*0.2(-i)*2.25\sqrt{3}(-j) \\L= 0.1325 k[/tex]

C) since the net external force on the system is zero , the total momentum of the system can be conserved .

thus ,

[tex]m_{s}v_{s_{i}}+m_{h}v_{h_{i}}=m_{s}v_{s_{f}}+m_hv_{h_{f}}\\0+0.17*9(-j)=0.05*v_{s_{f}}+0.17*(2.25(i)+2.25\sqrt{3}(-j))\\[/tex]

solving this we get,

⇒[tex]v_{s_{f}} =7.65(-i)+17.35(-j)[/tex]

D) since there is no external torque about the system ,the angular momentum can be conserved.

[tex]L_{h_{i}}= L_{h_{f}} + Iw[/tex]

where ,

[tex]w[/tex] is the angular velocity of the stick.

[tex]I[/tex] is the moment of inertia of the stick about COM :

[tex]I =\frac{m_{s}l^{2}}{12} \\m_{s}=0.05kg\\l=1m\\I = 0.004167 kgm^{2}[/tex]

∴

⇒[tex]0.306k=0.1325k+0.004167w\\w=41.64 rads^{-1}[/tex]

A) The angular momentum of the hockey puck before the collision is 0.459 kg.m^2/s. B) The angular velocity of the puck after the collision is 15 rad/s. C) The velocity of the stick's center of mass after the collision is 3.475 m/s.

A) The angular momentum of an object is given by the equation:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Since the meter stick is at rest, its angular velocity is zero, and therefore its angular momentum is also zero.

On the other hand, the angular momentum of the hockey puck before the collision can be calculated using the equation:

L = mvr

where m is the mass of the puck, v is its linear velocity, and r is the distance from its axis of rotation.

Using the given values, the angular momentum of the hockey puck before the collision is 0.17 kg * 9 m/s * 0.3m = 0.459 kg.m^2/s.

B) After the collision, the puck is deflected at an angle of 30 degrees and its linear velocity is halved.

The angular velocity of the puck after the collision can be calculated using the equation:

ω = v'/r'

where v' is the new linear velocity of the puck and r' is the new distance from its axis of rotation.

Since the linear velocity of the puck is halved, its new linear velocity is 4.5 m/s.

Assuming the radius of rotation remains the same, the angular velocity of the puck after the collision is 4.5 m/s / 0.3m = 15 rad/s.

C) The velocity of the stick's center of mass after the collision can be calculated using the conservation of linear momentum.

Before the collision, the momentum of the system is zero, since the meter stick is at rest.

After the collision, the momentum of the system is the linear momentum of the puck, which can be calculated using the equation:

p = mv

where m is the mass of the puck and v is its linear velocity.

Using the given values, the linear momentum of the puck after the collision is 0.17 kg * 4.5 m/s = 0.765 kg.m/s.

Since the meter stick and the puck move together after the collision, their velocities are the same and equal to 0.765 kg.m/s divided by the total mass of the system, which is 0.05 kg + 0.17 kg = 0.22 kg.

Therefore, the velocity of the stick's center of mass after the collision is 0.765 kg.m/s / 0.22 kg = 3.475 m/s.

D) The angular velocity of the stick can be calculated using the equation:

ω = v/r

where v is the linear velocity of the stick's center of mass and r is the radius of the stick.

Using the given values, the linear velocity of the stick's center of mass after the collision is 3.475 m/s.

Assuming the radius remains the same, the angular velocity of the stick is 3.475 m/s / 0.3m = 11.583 rad/s.

Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

 distance = speed x time

[tex]t_1 = \dfrac{d}{s}[/tex]

[tex]t_1 = \dfrac{4}{30}[/tex]

     t₁= 0.1333 hr

time from another 4 km trip

[tex]t_2 = \dfrac{d}{s}[/tex]

[tex]t_2= \dfrac{4}{50}[/tex]

     t₂ = 0.08 hr

average speed

     [tex]s= \dfrac{d}{t}[/tex]  

     [tex]s= \dfrac{8}{0.1333 + 0.08}[/tex]  

     [tex]s= \dfrac{8}{0.2133}[/tex]  

            s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

[tex]\lambda = \frac{v}{f}[/tex]

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

[tex]\lambda = \frac{192}{320}[/tex]

[tex]\lambda = 0.6m[/tex]

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

[tex]N = \frac{L}{\lambda}[/tex]

[tex]N = \frac{3.6}{0.6}[/tex]

[tex]N = 6[/tex]

Therefore the number of wavelengths of the wave fit on the string is 6.

To solve this problem it is necessary to apply the concepts related to Pressure, Strength and Area.

We know by definition that Pressure is the amount of Force expressed per unit area, that is,

[tex]P = \frac{F}{A}[/tex]

Where,

F = Force

A = Cross-sectional Area

The net pressure on the bottle would be given by the difference between the internal pressure and the atmospheric pressure, therefore

[tex]P_{net} = P_{in}-P_{atm}[/tex]

[tex]P_{net} = 2.5atm-1atm[/tex]

[tex]P_{net} = 1.5atm (\frac{101325Pa}{1atm})[/tex]

[tex]P_{net} = 151987.5Pa[/tex]

The given radio is,

[tex]r = 3cm^2[/tex]

Hence the Cross-sectional Area would be

[tex]A= \pi r^2[/tex]

[tex]A = \pi(3*10^{-2})^2[/tex]

[tex]A = 2.827*10^{-3}m^2[/tex]

Applying the equation for Pressure we have that

[tex]P = \frac{F}{A}[/tex]

[tex]151987.5= \frac{F}{2.827*10^{-3}}[/tex]

[tex]F = 429.66N[/tex]

Therefore the frictional force on the cork due to the neck of the bottle is 429.66N.

Answer:

3. F = m g

Explanation:

The centripetal force the person is exerting on the wall is;

                                         [tex]f = \frac{mv^{2}}{r}[/tex]

The maximum force due to  static friction is f = µ R ,

where R is  the normal force exerted by  the wall on the person. For the person to be held up against the wall, the maximal  friction force must be larger than the force of  gravity mg.

The actual force is now equal to or less than the maximum µR

                                                 f ≤ fmax = µR

When the angular velocity increase beyond the minimum angular velocity ωmin that is required  to hold the person against the wall,  the frictional force does to increase any more. Therefore, f still equals mg regardless of  the maximum frictional force.  

The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two.

The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two:

Ffriction = mω²r

Where Ffriction is the frictional force, m is the mass of the person, ω is the angular velocity, and r is the distance between the person and the axis of rotation (which is the radius of the cylinder in this problem). Therefore, the correct answer is option 9, F = 3mg.

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Answer:

[tex]\epsilon = 0.018 V[/tex]

Explanation:

given,

Area of rectangular loop = 0.15 m²

Magnetic field = B = 0.20 T

angle between magnetic field and the normal to the plane = π/2

increasing at the rate = 0.60 rad/s

magnitude of emf induced = ?

induce emf through the loop

   [tex]\epsilon = -\dfrac{d\phi}{dt}[/tex]

      Φ = BA cos θ

   [tex]\epsilon = -\dfrac{BAd(cos \theta)}{dt}[/tex]

   [tex]\epsilon = -BA (-sin \theta)\dfrac{\theta)}{dt}[/tex]

   [tex]\epsilon = BA(sin \theta)\dfrac{\theta)}{dt}[/tex]

now substituting all the given values

   [tex]\epsilon = 0.2 \times 0.15 (sin 90^0)\times 0.6[/tex]

   [tex]\epsilon = 0.018 V[/tex]

the magnitude of induced emf  is equal to [tex]\epsilon = 0.018 V[/tex]

Final answer:

The magnitude of the induced emf in the loop is 0.18 V.

Explanation:

The magnitude of the emf induced in a rectangular loop can be calculated using the formula ε = (dB/dt)(A cos θ), where ε is the induced emf, dB/dt is the rate of change of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the plane of the loop.

In this case, the area of the loop is given as 0.15 m2 and the magnetic field B = 0.20 T. The angle θ is given as π/2 radians and is increasing at a rate of 0.60 rad/s.

Plugging these values into the formula, we get ε = (0.20 T)(0.15 m2)(0.60 rad/s)(cos π/2) = 0.18 V.

Answer:

99.019 kPa

Explanation:

[tex]P_a[/tex] = Atmospheric pressure = 100 kPa

[tex]\rho[/tex] = Density of water in manometer = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Difference in height = 100 mm

As the pressure in the test section is lower than the pressure of the atmosphere it will be subtracted.

The absolute pressure in the test section is given by

[tex]P_{ab}=P_a-\rho gh\\\Rightarrow P_{ab}=100\times 10^3-1000\times 9.81\times 100\times 10^{-3}\\\Rightarrow P_{ab}=99019\ Pa=99.019\ kPa[/tex]

The absolute pressure in the test section is 99.019 kPa

Answer:[tex]5.67\times 10^{3} m/s[/tex]

Explanation:

Given

mass of satellite [tex]m=155 kg[/tex]

Satellite is orbiting 5995 km above Earth surface

mass of Earth [tex]M=5.97\times 10^{24} kg[/tex]

Radius of Earth [tex]R=6370 km[/tex]

here [tex]r=R+5995=6370+5995=12,365 km[/tex]

Gravitational Force will Provide the centripetal Force

[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]

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

[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{12365\times 10^{3}}}[/tex]

[tex]v=\sqrt{32.203\times 10^6}[/tex]

[tex]v=5.67\times 10^3m/s=5674.7 km/s[/tex]            

Answer:

Mass of the planet, [tex]M=9.59\times 10^{24}\ kg[/tex]

Explanation:

Given that,

Mass of the object, m = 0.5 kg

Weight of the object, W = 20 N

Radius of the planet, [tex]r=4\times 10^6\ m[/tex]

The weight of an object is given by :

W = mg

g is the acceleration due to gravity on the surface of Planet

[tex]g=\dfrac{W}{m}[/tex]

[tex]g=\dfrac{20}{0.5}[/tex]

[tex]g=40\ m/s^2[/tex]

The expression for g is given by :

[tex]g=\dfrac{GM}{r^2}[/tex]

[tex]M=\dfrac{r^2g}{G}[/tex]

[tex]M=\dfrac{(4\times 10^6)^2\times 40}{6.67\times 10^{-11}}[/tex]              

[tex]M=9.59\times 10^{24}\ kg[/tex]

So, the mass of the planet X is [tex]9.59\times 10^{24}\ kg[/tex]. Hence, this is the required solution.                                

The free-fall acceleration (g) on Planet X is 40 m/s². The estimated mass of Planet X, derived using Newton's law of universal gravitation and provided data, is 9.57 * 10⁵⁴ kg.

The free-fall acceleration, or the acceleration due to gravity, can be found by using the equation F (force which is the weight of the object) = m (mass) * g (acceleration due to gravity). Given the weight as 20 N and the mass is 0.50 kg, we can rearrange the equation to find g = F/m = 20 N /0.50 kg = 40 m/s².

To determine the mass of Planet X (M), we use Newton's law of universal gravitation: F = G * (m * M) / r², where F is the force (the weight of the object), G is the universal gravitational constant, m is the object's mass, and r is the radius of the planet. We already know g (acceleration due to gravity), which replaces F/m in the equation, thus becoming g = G * M / r². After rearranging the formula to solve for M, M = g*r²/G = 40 m/s² * (4.0 * 10⁶ m)² / (6.67 * 10^-11 N*m²/kg²) = 9.57 * 10⁵⁴ kg.

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Answer:

3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]

Explanation:

β(dB)=10 × [tex]log_{10}(\frac{I}{I_{0} })[/tex]

[tex]I_{0}[/tex]=[tex]10^{-12}[/tex] W/[tex]m^{2}[/tex]

β=55 dB

Therefore plugging into the equation the values,

55=10 [tex]log_{10}(\frac{I}{ [tex]10^{-12}[/tex]})[/tex]

5.5= [tex]log_{10}(\frac{I}{ [tex]10^{-12}[/tex]})[/tex]

[tex]10^{5.5}[/tex]= [tex]\frac{I}{10^{-12} }[/tex]

316227.76×[tex]10^{-12}[/tex]= I

I= 3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]

Answer:

18.2 cm

Explanation:

The kinetic energy of the block

= .5 x 3.49 x 10⁻² x 10.3²

= 1.85 J

If x be the compression created in the spring due to this kinetic energy

1.85 = 1/2 k x²

= .5 x 111 x²

x² = 3.3 x 10⁻²

x = 0.182 m .

= 18.2 cm

This will be the amplitude of oscillation under SHM.

Answer:

A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Explanation:

Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is

    P =[tex]P_{atm}[/tex] + ρ g y

Where [tex]P_{atm}[/tex] is the atmospheric pressure, ρ  the water density, and 'y' the depth measured from the surface.

Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth

Now we can examine the claims

A) To true. State agreement or with the equation above

B) False. Pressure changes with atmospheric pressure

C) False. It's the opposite

D) False. They are directly proportional

A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Hydrostatic pressure is pressure caused by the weight of a liquid.

The weight of a liquid is affected by the force of gravity.

If a liquid is placed in a container, the higher the liquid content in the container, the heavier the liquid content is and the greater the liquid pressure at the bottom of the container.

The hydrostatic pressure of a liquid can be formulated:

[tex]\large{\boxed{\bold {P_h ~ = ~ \rho.g.h}}[/tex]

Ph = hydrostatic pressure (N / m², Pa)

ρ = density of liquid (kg / m³)

g = acceleration due to gravity (m / s²)

h = height / depth of liquid surface (m)

If the container is open, then the atmospheric pressure (P₀) can be entered into the equation.

[tex]\large{\boxed {\bold {P_h ~ = ~ P_o ~ + ~ \rho.g.h}}[/tex]

The magnitude of P₀ is usually equal to = 1.01.10⁵ Pascal (Pa = N / m²) = 1 atm

For example submarines  :

The deeper a submarine is, the greater the hydrostatic pressure it experiences, so that the hull / wall of the submarine is made thick to withstand that pressure.

As a storm approaches, the atmospheric pressure drops

Because the hydrostatic pressure is proportional to the atmospheric pressure, the pressure at the bottom of the pool will also be reduced

Isaac Newton's investigations of gravity

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Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

[tex]F_n = \rho A (v - u)^2 sin \theta[/tex]

v is the velocity of jet

Q = A V

[tex]v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}[/tex]

[tex]v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}[/tex]

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

[tex]F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0[/tex]

F n = 2295 N

now force along x direction

[tex]F_x = F_n sin 60^0[/tex]

[tex]F_x = 2295 \times sin 60^0[/tex]

[tex]F_x = 1987.52\ N[/tex]

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

The power produced by the stream is; 40.5 kW

We are given;

Constant speed of cart on right side; u = 2 m/s

Diameter of nozzle; d = 50 mm = 0.05 m

Discharge flow through nozzle; Q = 0.04 m³/s

Formula for the normal force acting on the cart is;

Fₙ = ρA(v - u)²sin θ

Let us find v the speed of the jet from;

Q = Av

v = Q/A

v = 0.04/(π * 0.05²/4)

v = 20.37 m/s

From online research about this question, the angle of inclination is 60°. Thus;

Fₙ = ρA(v - u)²sin θ

Fₙ = 1000 * (π * 0.05²/4) * (20.37 - 2)sin 60

Fₙ = 2295 N

Force along the x-direction will be;

Fₓ = Fₙ * sin 60

Fₓ = 2295 * sin 60

Fₓ = 1987.52 N

Power of the Cart is;

P = Fₓ * v

P = 1987.52 * 20.37

P = 40.5 kW

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Answer:

The distance h above the victim's arm where the level of the blood in the transfusion bottle should be located is 0.254 m

Explanation:

given information:

density, ρ = 1060 [tex]kg/m^{3}[/tex]

viscosity, η = 4 x [tex]10^{-3}[/tex] Pa.s

needle length, L = 3 cm = 0.03 m

radius, r = 0.25 mm = 0.00025 m

volume flow rate, Q = 4.70 x [tex]10^{-8}[/tex]  [tex]m^{3} /s[/tex]

according to  Poiseuille’s law

Q = (π[tex]r^{4}[/tex]ΔP)/8ηL

ΔP = (8QηL)/(π[tex]r^{4}[/tex])

     = 8 (4.70 x [tex]10^{-8}[/tex])(4 x [tex]10^{-3}[/tex])(0.03)/(π[tex]0.00025^{4}[/tex])

     = 3676.71

Now we can calculate the distance h

ΔP = ρ g h

h = ΔP / ρ g

  = 3676.71/(1060)(9.8)

  = 0.354 m