What is an extrasolar planet?
a. A planet that is considered an "extra," in that it was not needed for the formation of its solar system.
b. A planet that is extra large compared to what we'd expect.
c. A planet that is larger than the Sun.
d. A planet that orbits a star that is not our own Sun

Answer:

Mutual inductance will be [tex]M=1.8\times 10^{-3}Hennry[/tex]

Explanation:

We have given induced current [tex]\Delta i=5A[/tex]

Time is given as [tex]t=1ms=10^{-3}sec[/tex]

We have to find the mutual inductance between the coils

Induced emf is given as e = 9 volt

We know that induced emf is given by

[tex]e=M\frac{\Delta i}{\Delta t}[/tex]

[tex]9=M\times \frac{5}{10^{-3}}[/tex]

[tex]M=1.8\times 10^{-3}Hennry[/tex]

Answer:

The answer is to insulate Earth's surface temperature.

Explanation:

Clouds inhibit the outflow of terrestrial radiation. This acts to insulate Earth's surface temperature, keeping it warmer at night and cooler in the day.

Final answer:

Using the given formula and solving for the constant using the temperature 10 minutes after pouring, the temperature of the coffee 30 minutes after it was poured is found to be 60 degrees Fahrenheit.

Explanation:

The question asks us to determine the temperature of coffee 30 minutes after it was poured, given the formula F = 120(2–at) + 60, where F is in degrees Fahrenheit and t is the time in minutes since the coffee was poured. Since the temperature of the coffee 10 minutes after it was poured was 120 degrees Fahrenheit, we first need to find the value of the constant a using the given formula. Substituting F with 120 and t with 10 gives us 120 = 120(2-10a) + 60. Solving for a, we find that a is equal to 0.1. Now, to find the temperature of the coffee 30 minutes after it was poured, we substitute t with 30 in the formula, getting F = 120(2-0.1*30) + 60, which simplifies to F = 60 degrees Fahrenheit.

Answer:

1. Rotating 2. Revolving

Explanation:

Just did it on edge 2021

Answer:

positive reinforcer; negative reinforcer

Explanation:

positive reinforcer; negative reinforcer

Receiving delicious food can be seen as a positive reinforcer while escaping electric shock can be seen as a negative reinforcer.

Answer:1.742 J

Explanation:

Given

mass of block [tex]m=2.2 kg[/tex]

[tex]k=825 N/m[/tex]

Compression in spring [tex]x=0.065 m[/tex]

Elastic Potential Energy of the Spring-mass system is given by

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

where [tex]k=spring\ constant[/tex]

[tex]x=compression\ in\ spring[/tex]

[tex]E=\frac{1}{2}\times 825\times (0.065)^2[/tex]

[tex]E=412.5\times 42.25\times 10^{-4}[/tex]

[tex]E=1.742 J[/tex]                          

Final answer:

The elastic potential energy of the block-spring system with a spring constant of 825 N/m compressed by 0.0650 m is approximately 1.74 J.

Explanation:

To determine the elastic potential energy of the block-spring system, the formula for elastic potential energy (EPE) in a compressed or stretched spring is used, which is EPE = (1/2)kx2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant k is 825 N/m and the compression x is 0.0650 m.

Using the formula:

EPE = (1/2)(825 N/m)(0.0650 m)2

EPE = (1/2)(825)(0.004225)

EPE = (1/2)(3.481125)

EPE = 1.7405625 Joules

Therefore, the elastic potential energy of the block-spring system when the spring is compressed by 0.0650 m is approximately 1.74 J.

Explanation:

Given that, Speed of the automobile, v = 55 km/h = 15.27 m/s

Diameter of the tire, d = 77 cm

Radius, r = 0.385 m

(a) Let [tex]\omega[/tex] is the angular speed of the tires about their axles.

The relation between the linear speed and the angular speed is given by :

[tex]v=r\times \omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

[tex]\omega=\dfrac{15.27\ m/s}{0.385\ m}[/tex]

[tex]\omega=39.66\ rad/s[/tex]

(b) Number of revolution,

[tex]\theta=38\ rev=238.76\ radian[/tex]

Final angular speed of the car, [tex]\omega_f=0[/tex]

Initial angular speed, [tex]\omega_i=39.66\ rad/s[/tex]

Let [tex]\alpha[/tex] is the angular acceleration of the car. Using third equation of rotational kinematics as :

[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]

[tex]\alpha =\dfrac{-\omega_i^2}{2\theta}[/tex]

[tex]\alpha =\dfrac{-(39.66)^2}{2\times 238.76}[/tex]

[tex]\theta=-3.29\ rad/s^2[/tex]

(c) Let d is the distance covered by the car during the braking. It is given by :

[tex]d=\theta\times r[/tex]

[tex]d=238.76\times 0.385[/tex]

d = 91.92 meters

Answer:

Neither Technician A nor Technician B.

Explanation:

chrome is very expensive and we do need to check the ring before reassembling engine.

The solution to this problem requires the understanding of the concepts of specific heat and heat of fusion. It involves three parts: heating up the ice to the melting point, melting the ice, and as the melted ice is absorbed by the water, it cools down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.

The subject of this question falls under Physics, specifically in the topic of Thermodynamics. The problem involves the concept of specific heat, and heat of fusion. We need to account for three parts: heating the ice from -5°C to 0°C, melting the ice at 0°C, and the absorbed heat by the water from the melting ice.

1. Firstly, calculate the energy needed to heat the ice from -5°C to 0°C (phase change has not yet happened). Use Q = m*c*ΔT, where m is the mass of ice, c is the specific heat of ice, and ΔT is the change in temperature.

2. Secondly, calculate the energy required to melt the ice. This time, use Q = m*Lf, where m is the mass of ice and Lf is the heat of fusion.

3. Lastly, as the melted ice (now at 0°C) is absorbed by the water, it will cool down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.

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The speed of the mass when t = 3 s in simple harmonic motion is approximately 6.283 m/s.

The speed of the mass in simple harmonic motion can be determined by using the formula v = ωA sin(ωt), where v is the speed, ω is the angular frequency, A is the amplitude, and t is the time. In this case, the frequency of the motion is given as 4.0 Hz, which corresponds to an angular frequency of ω = 2πf = 2π(4.0 Hz) = 8π rad/s. The amplitude is given as 4.0 cm, which corresponds to A = 0.04 m. Plugging these values into the formula, we get:

v = (8π rad/s)(0.04 m) sin((8π rad/s)(3 s))

v ≈ 6.283 m/s

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The speed of the mass when[tex]\( t = 3 \) s is \( v = 8\pi \)[/tex] cm/s.

To find the speed of the mass at any given time \( t \) during its simple harmonic motion, we can use the following equation for velocity in SHM:

[tex]\[ v(t) = -A\omega \sin(\omega t + \phi) \][/tex]using the relation [tex]\( \omega = 2\pi f \).[/tex]Substituting the given frequency:

[tex]\[ \omega = 2\pi (4.0 \, \text{Hz}) = 8\pi \, \text{rad/s} \][/tex]

 The phase constant \( \phi \) is 0 because the timer is started when the mass is at its maximum displacement, which corresponds to \[tex]( x = A \)[/tex]when [tex]\( t = 0 \).[/tex]

 Now, we can calculate the velocity at [tex]\( t = 3 \)[/tex] s:

[tex]\[ v(3) = -A\omega \sin(\omega \cdot 3 + \phi) \][/tex]

[tex]\[ v(3) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi \, \text{rad/s} \cdot 3 \, \text{s} + 0) \][/tex]

[tex]\[ v(3) = -32\pi \, \text{cm/s} \sin(24\pi \, \text{rad}) \][/tex]

Since \[tex]( \sin(24\pi \, \text{rad}) = 0 \)[/tex] (because [tex]\( 24\pi \)[/tex] is a multiple of[tex]\( 2\pi \)[/tex]), the velocity at [tex]\( t = 3 \)[/tex] s is:

[tex]\[ v(3) = -32\pi \, \text{cm/s} \cdot 0 = 0 \, \text{cm/s} \][/tex]

 However, this result corresponds to the instantaneous speed at a point where the mass changes direction, and thus its speed is momentarily zero. To find the speed at a point where the mass is actually moving, we need to consider the time just before or after this point of maximum displacement.

 Now, we calculate the velocity at [tex]( t = 3 - \Delta t \):[/tex]

[tex]\[ v(3 - \Delta t) = -A\omega \sin(\omega (3 - \Delta t) + \phi) \][/tex]

[tex]\[ v(3 - \Delta t) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi (3 - \frac{1}{8\pi}) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(24\pi - 1) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(-1) \, \text{rad}) \][/tex]

 Since [tex]\( \sin(-1) \approx -1 \)[/tex] for a very small angle in radians:

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \cdot (-1) \][/tex]

[tex]\[ v(3 - \Delta t) = 32\pi \, \text{cm/s} \][/tex]

 Therefore, just before the mass reaches its maximum displacement at [tex]\( t = 3 \) s[/tex], its speed is [tex]\( 32\pi \)[/tex] cm/s.

Answer:

Explanation:

Let the velocity of rocket case and payload after the separation be v₁ and v₂ respectively. v₂ will be greater because payload has less mass so it will be fired with greater speed .

v₂ - v₁ = 910

Applying law of conservation of momentum

( 250 + 100 ) x 7700 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 ( 910 +v₁ )

v₁ = 7440 m /s

v₂ = 8350 m /s

Total kinetic energy before firing

= 1/2 ( 250 + 100 ) x 7700²

= 1.037575 x 10¹⁰ J

Total kinetic energy after firing

= 1/2 ( 250 x 7440² + 100 x 8350² )

= 1.0405325 x 10¹⁰ J

The kinetic energy has been increased due to addition of energy generated in firing or explosion which separated the parts or due to release of energy from compressed spring.

Answer:

[tex]\theta = 20.98 degree[/tex]

Explanation:

As we know that the speed of the sound is given as

[tex]v = 332 + 0.6 t[/tex]

now at t = 273 k = 0 degree

[tex]v = 332 m/s[/tex]

so we have

[tex]a sin\theta = N\lambda[/tex]

[tex]a sin\theta = N(\frac{v_1}{f})[/tex]

now when temperature is changed to 313 K we have

[tex]t = 313 - 273 = 40 degree[/tex]

now we have

[tex]v = 332 + (0.6)(40)[/tex]

[tex]v_2 = 356 m/s[/tex]

[tex]a sin\theta' = N(\frac{v_2}{f})[/tex]

now from two equations we have

[tex]\frac{sin19.5}{sin\theta} = \frac{332}{356}[/tex]

so we have

[tex]sin\theta = 0.358[/tex]

[tex]\theta = 20.98 degree[/tex]

The diffraction angle of sound from a loudspeaker changes with temperature because the speed of sound varies with temperature. The higher the temperature, the faster the sound travels, resulting in a longer wavelength and, consequently, a larger diffraction angle.

The question involves the concept of wave diffraction, specifically as it pertains to sound waves exiting a diffraction horn loudspeaker. In this scenario, the diffraction angle θ of sound is observed to change with temperature, due to the dependence of the speed of sound on air temperature. To find the new diffraction angle when the temperature rises from 273 K to 313 K, we need to understand that the speed of sound in air increases with temperature. The formula v = 331.4 + 0.6T (with T in ℃) can be used to calculate the speed of sound at different temperatures, but here we'll use Kelvin and recall that v = 331.4 + 0.6(T - 273) to scale it correctly.

At 273 K, the speed of sound is approximately 331.4 m/s, and at 313 K, it is higher. Assuming the frequency of the sound remains constant, and using the relationship v = fλ (where v is the speed of sound, f is the frequency, and λ is the wavelength), the wavelength will increase with the speed of sound. According to the diffraction formula for a single slit, sin(θ) = mλ/d (where m is the order of the minimum, λ is the wavelength, and d is the width of the slit), if wavelength λ increases while d stays the same, the angle θ must also increase to satisfy the equation for the first minimum (m=1). Thus, the diffraction angle on a summer day when the temperature is higher would be larger than 19.5°.

Answer:

U / K

Explanation:

The energy stored in the capacitor is given by

[tex]U=\frac{q^{2}}{2C}[/tex]

where, q be the charge and C be the capacitance of the capacitor.

If a dielctric is inserted beteen the plates of capacitor having dielctric constant K, the new capacitance is C' = KC

The charge will remain same

The new energy stored is

[tex]U'=\frac{q^{2}}{2C'}[/tex]

[tex]U'=\frac{q^{2}}{2KC}[/tex]

U' = U / K

So, the energy reduces by the factor of K.

The new energy stored in the capacitor is U/K. Hence, option(e) is correct.

To determine the energy stored in a capacitor after inserting a dielectric, recognize the following points:

When a dielectric is inserted and the capacitor is disconnected from any power source, the charge Q remains constant.

However, the capacitance C increases by a factor of K due to the dielectric. The energy stored in a capacitor(U) is given by:

Since the capacitance becomes KC, the new energy stored will be:

Given the original energy U = Q²/(2C), we can relate the new energy U' to the old energy U:

Hence, the energy stored in the capacitor after inserting the dielectric is decreased by a factor of K, resulting in the new energy U' = U/K.

complete question:

A charged capacitor stores energy . Without connecting this capacitor to anything, dielectric having dielectric constant is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the capacitor now store?

a) U/2K

b) KU

c) 2KU

d) U

e) U/K

Answer:

[tex]x=0.0049\ m= 4.9\ mm[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

Explanation:

Given:

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

[tex]f=k.x[/tex]

[tex]\mu_s.N=k.x[/tex]

where:

N = m.g = the normal reaction force acting on the body under steady state.

[tex]0.1\times (9.8\times 0.75)=150\times x[/tex]

[tex]x=0.0049\ m= 4.9\ mm[/tex]

(b)

Now, according to the question:

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

[tex]U=\frac{1}{2} k.A^2[/tex]

This energy will be equal to the work done by the kinetic friction to stop it.

[tex]U=F_k.d[/tex]

[tex]\frac{1}{2} k.A^2=\mu_k.N.d[/tex]

[tex]0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

is the total distance does it travel before stopping.

To find the maximum distance the spring can be stretched without moving the mass, calculate the force of static friction and equate it to the spring force.

(a) How far can the spring be stretched without moving the mass?

To find the maximum distance the spring can be stretched without moving the mass, we need to calculate the force of static friction. The force of static friction can be calculated using the equation:

Fs = μs * m * g

where μs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.

Once we have the force of static friction, we can equate it to the spring force:

Fs = k * x

where k is the force constant of the spring and x is the maximum distance the spring is stretched without moving the mass.

Answer:

seismograph or seismometer.

Explanation:

Seismograph is an instrument used to detect and record earthquakes.

It has a mass attached to a fixed base. During an earthquake, the base moves and the mass does not. The motion of the base with respect to the mass is commonly transformed into an electrical voltage. The electrical voltage is recorded. This record is proportional to the motion of the seismometer mass relative to the earth, but it can be mathematically converted to a record of the absolute motion of the ground.

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex] (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

[tex]F=kx[/tex]

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

[tex]mg=kx[/tex]

and using [tex]g=9.8 m/s^2[/tex] and [tex]x=0.0177 m[/tex], we find

[tex]k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m[/tex]

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

[tex]m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg[/tex]

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Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p)  -

[tex]p_{i}[/tex]=[tex]p_{f}[/tex]

System initially is at rest

∴[tex]p_{i}[/tex]=0

∴ From the above 2 equations

[tex]p_{f}[/tex]=0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let [tex]m_{1}[/tex] and [tex]m_{2}[/tex] be the mass of the person and the book respectively and [tex]v_{1}[/tex] and [tex]v_{2}[/tex] be the final velocities of the person and book respectively.

∴[tex]p_{f}[/tex]=[tex]m_{1}[/tex][tex]v_{1}[/tex]+[tex]m_{2}[/tex][tex]v_{2}[/tex]=0

From the question ,

[tex]m_{1}[/tex] = 74.9 kg

[tex]m_{2}[/tex] = 2.44 kg

[tex]v_{2}[/tex] = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × [tex]v_{1}[/tex] )+ (2.44×8.25) = 0

∴[tex]v_{1}[/tex]  = - 0.2687 m/s (Negative sign suggests that the motion of  the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

Answer:12.12 m/s

Explanation:

Given

mass of child [tex]m=49 kg[/tex]

height of hill [tex]h=7.5 m[/tex]

inclination [tex]\theta =26^{\circ}[/tex]

Conserving Energy at top and bottom Point of hill

Potential Energy at Top =Kinetic Energy at bottom

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

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

[tex]v=\sqrt{2\times 9.8\times 7.5}[/tex]

[tex]v=\sqrt{147}[/tex]

[tex]v=12.12 m/s[/tex]

Answer:

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Explanation:

The final speed of the child-sled system is determined by means of the Principle of Energy Conservation:

[tex]U_{1} + K_{1} = U_{2} + K_{2}[/tex]

[tex]K_{2} = (U_{1}-U_{2})+K_{1}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot \Delta h[/tex]

[tex]v = \sqrt{2\cdot g \cdot \Delta h}[/tex]

[tex]v = \sqrt{2\cdot\left(9.807\,\frac{m}{s^{2}} \right)\cdot (7.50\,m)}[/tex]

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Answer:

Explanation:

a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s.

F = M x 48

Mass M = F / 48

a )

When force = 2F and mass = M

Acceleration = force / mass

= 2F /F/48

= 48 X 2 = 96 cm/s²

b )

When force = F and mass = 2M

Acceleration = force / mass

= F /2F/48

=  24 cm/s²

c )

When force = 2F and mass = 2M

Acceleration = force / mass

= 2F /2F/48

=  48  cm/s

d )

When force = 2F and mass = 4M

Acceleration = force / mass

= 2F /4F/48

=  24  cm/s

e)

When force = 4F and mass = 2M

Acceleration = force / mass

= 4F / 2M

= 4F / 2 F/48

= 48 x 2

Acceleration  = 96 cm/s²

Final answer:

The acceleration of carts with different masses acted upon by varying forces can be calculated using Newton's second law.

Explanation:

Mass is a fundamental factor in determining acceleration according to Newton's second law, which states force (F) = mass (m) x acceleration (a). When a cart with mass M is acted upon by a net force of 2F, the acceleration will be 96 cm/s/s (48 cm/s/s x 2). Similarly, for a mass of 2M acted upon by force F, the acceleration will be 24 cm/s/s (48 cm/s/s / 2). When a mass of 2M is subject to 2F, the acceleration will be 96 cm/s/s, and for a mass of 4M with a force of 2F, the acceleration will be 24 cm/s/s. Finally, when a mass of 2M experiences 4F, the acceleration will be 192 cm/s/s.

Answer:

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

Explanation:

The intermolecular forces are interactions between particles. And decrease as we go from solid >liquid >gas. And on gases we have weak intermolecular forces.

The most common 3 types of intermolecular forces occurs between neutral molecules are known as Van der Walls forces (Dipole-Dipole,Hydrogen bonding, London Dispersion forces).

If the intermodelucar attraction forces increases we have:

1. Decreasing vapor pressure (pressure of the vapor that is in equilibrium with the liquid)

2. Increasing boiling point (temperature at which the vapor pressure is equal to the pressur exerted on the surface of the liquid)

3. Increasing surface tension (resistance of a liquid to spread out and increase the surface area)

4. Increasing viscosity (resistance of a liquid to flow)

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

Answer:

See explanation

Explanation:

To do this, you need to use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path so, you can write the expression like this:

1/2mv1² + mgh1 = (1/2)mv2² + mgh2

Where:

v1 = 0 because it's released from rest

h1 = 18.6 m

v2 = speed we want to solve.

h2 = height at point A. In this case, you are not providing the picture or data, so, I'm going to suppose a theorical data to solve this. Let's say h2 it's 12 m.

Now, let's replace the data in the above expression (assuming h2 = 12 m). Also, remember that we don't have the mass of the bead, but we don't need it to solve it, because it's simplified by the equation, therefore the final expression is:

1/2v1² + gh1 = 1/2v2² + gh2  

Replacing the data we have:

1/2*(0) + 9.8*18.6 = 1/2v² + 9.8*12

182.28 = 117.6 + 1/2v²

182.28 - 117.6 = v²/2

64.68 * 2 = v²

v = √129.36

v = 11.37 m/s

Now, remember that you are not providing the picture to see exactly the value of height at point A. With that picture, just replace the value in this procedure, and you'll get an accurate result.

Answer:

[tex]Re=1992.24[/tex]

Explanation:

Given:

vertical height of oil coming out of pipe, [tex]h=25\ m[/tex]

diameter of pipe, [tex]d=0.1\ m[/tex]

length of pipe, [tex]l=50\ m[/tex]

density of oil, [tex]\rho = 900\ kg.m^{-3}[/tex]

viscosity of oil, [tex]\mu=1\ Pa.s[/tex]

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

Using the equation of motion:

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

where:

v = final velocity

u = initial velocity

Putting the respective values:

[tex]0^2=u^2-2\times 9.8\times 25[/tex]

[tex]u=22.136\ m.s^{-1}[/tex]

For Reynold's no. we have the relation as:

[tex]Re=\frac{\rho.u.d}{\mu}[/tex]

[tex]Re=\frac{900\times 22.136\times 0.1}{1}[/tex]

[tex]Re=1992.24[/tex]

Answer:

D = 71.20 km

Explanation:

In this case, let's get the exercise by parts.

Part 1 would be the first displacement which is 30 km. However, it's displacing north to east 30°, so in this case, if we trace a line of this, it should be a diagonal. So we need to get the value of this displacement in the x and y axis.

Dx1 = 30 cos 30° = 25.98 km

Dy1 = 30 sin 30° = 15 km

Doing the same thing but in the second travel we have:

Dx2 = 70 cos 50° = 45 km

Dy2 = 70 sin 50° = 53.62 km

Now we need to know how was the displacement in both axis. In travel 2, the plane is moving north to west, so, in the x axis, is moving in the opposite direction, therefore:

Dx = -45 + 25.98 = -19.02 km

In the y axis, is moving upward so:

Dy = 53.62 + 15 = 68.62 km

Finally to get the resultant displacement:

D = √Dx² + Dy²

Replacing:

D = √(-19.02)² + (68.62)²

D = √5070.46

D = 71.20 km

Horizontal component of vector A: 5.0 cm

Vertical component of vector A: 8.7 cm

Explanation:

The horizontal and vertical components of a vector on the cartesian plane are calculated as follow:

[tex]v_x = v cos \theta\\v_y = v sin \theta[/tex]

where

[tex]v_x[/tex] is the horizontal component

[tex]v_y[/tex] is the vertical component

v is the magnitude of the vector

[tex]\theta[/tex] is the angle between the direction of the vector and x-axis

In this problem, we have

v = 10.0 cm is the magnitude of vector A

[tex]\theta=60.0^{\circ}[/tex] is the angle of its direction with the x-axis

Applying the equations, we find

[tex]v_x = (10.0)(cos 60)=5.0 cm[/tex]

[tex]v_y = (10.0)(sin 60)=8.7 cm[/tex]

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Explanation: edmuntum answer

Answer and Explanation:

Tropical storm:

The point at which the tropical depression intensifies and can sustain a maximum wind speed in the range of 39-73 mph, it is termed as a tropical storm.

Hurricane:

A hurricane is that type of tropical cyclone which includes with it high speed winds and thunderstorms and the intensity of the sustained wind speed is 74 mph or more than this.

Hurricanes are the most intense tropical cyclones.

The only difference between the tropical storm and hurricane is in the intensity.

A tropical storm becomes a hurricane when its sustained wind speeds exceed 74 miles per hour, with both systems featuring low pressure centers and heavy rains, but hurricanes pose a greater threat due to their higher wind speeds.

The primary difference between a tropical storm and a hurricane lies in their wind speeds. When a storm's sustained winds reach between 39 to 73 miles per hour, it is classified as a tropical storm. Once those winds exceed 74 miles per hour, the storm is then classified as a hurricane. Both types of storm systems form over warm ocean waters, typically warmer than 80 °F and involve low pressure centers, strong winds, and heavy rains. However, a hurricane's increased wind speed is what gives it the potential to cause significantly more damage upon making landfall.

The Coriolis force is responsible for the cyclonic rotation of these storms—with hurricanes in the Northern Hemisphere rotating counterclockwise and those in the Southern Hemisphere rotating clockwise. The terms hurricane, typhoon, and tropical storm are regional names for these cyclones. Regardless of the name, these storms are dangerous due to the combination of high winds, heavy rains, and consequential risks such as flooding and structural damage.

Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

Answer:

[tex]U_k[/tex] = 0.3731

Explanation:

First we will identify the important data of the question.

M = 2420 kg

V = 14 m/s

d = 26.8 m

[tex]U_k =[/tex] ?

So, we will use the law of the conservation of energy, it says that:

[tex]E_i - E_f = W_f[/tex]

therefore:

[tex]E_i = \frac{1}{2}MV^2\\E_f = 0\\W_f = F_kd[/tex]

where [tex]F_k[/tex] is the friction force

Replacing on the first equation, we get:

[tex]\frac{1}{2}MV^2 -0 = F_kd[/tex]

[tex]F_k[/tex] is also equal to [tex]U_kN[/tex]

where N is the normal force and Uk is the coefficient of kinetic friction.

solving the equation:

[tex]\frac{1}{2}(2420)(14)^2 = U_kN(26.8)[/tex]

Before solve for [tex]U_k[/tex] we need to know the value of N, so we use the law of newton as:

∑[tex]F_y[/tex] = N - (2420)(9.8m/s) = 0

N = 23716

Finally, just solve for [tex]U_k[/tex] as:

[tex]U_ k = \frac{\frac{1}{2}(2420)(14)^2 }{(26.8)(23716)}[/tex]

[tex]U_k[/tex] = 0.3731

Final answer:

To find the coefficient of kinetic friction on a wet road, we use the work-energy principle to set the work done by friction equal to the car's initial kinetic energy. By substituting and simplifying the given values, we calculate the coefficient as approximately 0.746.

Explanation:

The question asks to find the coefficient of kinetic friction between the tires of a car and a wet road, given that the car, which has a mass of 2420.0 kg and was initially moving at a constant speed of 14.0 m/s, slides to a halt in a distance of 26.8 m after the driver slams on the brakes. To solve for the coefficient of kinetic friction (μk), we will use the work-energy principle which states that the work done by the friction force equals the change in kinetic energy of the car since no other forces do work.

The equation is as follows:

Ff × d = ½ m × vi2 - ½ m × vf2

Where Ff is the friction force, d is the distance, m is the mass, vi is the initial velocity, and vf is the final velocity (which is 0 since the car stops).

Since Ff = μk × N, and N = m × g (where N is the normal force and g is the acceleration due to gravity), we have:

μk = (½ m × vi2)/ (m × g × d)

Plugging in the known values: μk = [(½ × 2420.0 kg × (14.0 m/s)2)] / [2420.0 kg × 9.8 m/s2 × 26.8 m]

This simplifies to: μk = (2420.0 kg × 196.0 m2/s2) / (2420.0 kg × 9.8 m/s2 × 26.8 m)

We can cancel the mass from both the numerator and denominator, leaving us with:

μk = (196.0 m2/s2) / (9.8 m/s2 × 26.8 m)

Thus, the coefficient of kinetic friction is:

μk = 196.0 / 262.64 = 0.746 (rounded to three decimal places)

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

Answer:

[tex]fraction = 0.11[/tex]

Explanation:

Linear kinetic energy of the bicycle is given as

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

[tex]K_1 = \frac{1}{2}(13) v^2[/tex]

[tex]K_1 = 6.5 v^2[/tex]

Now rotational kinetic energy of the wheels

[tex]K_2 = 2(\frac{1}{2}(I)(\omega^2))[/tex]

[tex]K_2 = (mR^2)(\frac{v^2}{R^2})[/tex]

[tex]K_2 = mv^2[/tex]

[tex]K_2 = 3.1 v^2[/tex]

now kinetic energy of the rider is given as

[tex]K_3 = \frac{1}{2}Mv^2[/tex]

[tex]K_3 = \frac{1}{2}(38) v^2 [/tex]

[tex]K_3 = 19 v^2[/tex]

So we have

[tex]fraction = \frac{K_2}{K_1 + K_2 + K_3}[/tex]

[tex]fraction = \frac{3.1 v^2}{6.5 v^2 + 3.1 v^2 + 19 v^2}[/tex]

[tex]fraction = 0.11[/tex]

Answer:

166 666 666.7 years

Explanation:

We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.

We are also told that the continents are now 5000 km = 5 000 000 m apart

So to calculate the time it took for them to be this far apart

t = distance/speed

t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years