The end point of a spring vibrates with a period of 2.1 seconds when a mass m is attached to it. When this mass is increased by 6.810×101 kg, the period is found to be 3.4 seconds. Find the value of m.

Answer:

a) -1.57 rad/s²

b) 800 revolutions

Explanation:

a)In here you can use the equations of velocity as if it were a linear movement. In this case:

wf = wo + at

wo is the innitial angular velocity, that we can get this value using the fact that a revolution is 2π so:

wo = 20 * 2π = 125.66 rad/s

We have the time of 80 seconds, and the final angular speed is zero, because it's going to a rest so:

0 = 125.66 + 80a

a = -125.66 / 80

a = -1.57 rad/s²

b) In this part, we will use the following expression:

Ф = Фo + wo*t + 1/2 at²

But as this it's coming to rest then:

Ф = 1/2at²

solving we have:

Ф = 0.5 * (-1.57)*(80)²

Ф = 5,024 rad

Ф = 5024 / 2π

Ф = 800 revolutions

(a) The angular acceleration of the grinding wheel is approximately [tex]-1.57 \, \text{rad/s}^2[/tex]. (b) The grinding wheel made about 800 revolutions during the time it was coming to rest.

To solve the problem of the grinding wheel, we need to find the angular acceleration and the number of revolutions during its deceleration.

First, we convert the initial angular velocity from revolutions per second to radians per second:
[tex]\omega_0 = 20.0 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 40\pi \, \text{rad/s} \approx 125.66 \, \text{rad/s}[/tex]

Angular acceleration [tex]\alpha[/tex] can be calculated using the formula:
[tex]\alpha = \frac{\omega_f - \omega_0}{t}[/tex]
where [tex]\omega_f = 0 \, \text{rad/s}[/tex] (final angular velocity, when the wheel comes to rest).

Substituting in the values:
[tex]\alpha = \frac{0 - 125.66}{80.0} = \frac{-125.66}{80.0} \approx -1.57 \, \text{rad/s}^2[/tex]

To find how many revolutions the wheel made while coming to rest, we can use the formula for angular displacement [tex]\theta[/tex]:
[tex]\theta = \omega_0 t + \frac{1}{2} \alpha t^2[/tex]

Substituting the values:
[tex]\theta = (125.66 \, \text{rad/s}) (80.0 \, ext{s}) + \frac{1}{2} (-1.57) (80.0)^2[/tex]
[tex]\theta = 10052.8 \, \text{rad} - \frac{1}{2} (1.57) (6400)\approx 10052.8 - 5024\approx 5028.8 \, \text{rad}[/tex]

Now convert radians to revolutions:
[tex]\text{Revolutions} = \frac{\theta}{2\pi} = \frac{5028.8}{2\pi} \approx 799.8 \, ext{rev} \approx 800 \, ext{rev}[/tex]

The flux of F through the curved surface of the cylinder is [tex]\( 18\pi \).[/tex]  

The flux of the vector field F = xi + yj + zk  through the curved surface of the cylinder [tex]\( x^2 + y^2 = 9 \),[/tex] bounded below by the plane [tex]\( x + y + z = 2 \)[/tex] and above by the plane [tex]\( x + y + z = 4 \),[/tex] and oriented away from the z-axis can be calculated using the surface integral formula.

The flux [tex]\( \Phi \)[/tex]is given by:

[tex]\[ \Phi = \iint_S \mathbf{F} \cdot d\mathbf{S} \][/tex]

Where [tex]\( d\mathbf{S} \)[/tex] is the outward-pointing normal vector to the surface [tex]\( S \).[/tex]

First, let's parameterize the surface. Since the cylinder is described by [tex]\( x^2 + y^2 = 9 \)[/tex], we can use cylindrical coordinates:

[tex]\[ x = 3\cos \theta \]\[ y = 3\sin \theta \]\[ z = z \][/tex]

The normal vector [tex]\( d\mathbf{S} \)[/tex] can be calculated using the cross product of the partial derivatives of [tex]\( \mathbf{r}(\theta, z) = (3\cos \theta, 3\sin \theta, z) \)[/tex] with respect to [tex]\( \theta \) and \( z \):[/tex]

[tex]\[ \frac{\partial \mathbf{r}}{\partial \theta} = (-3\sin \theta, 3\cos \theta, 0) \]\[ \frac{\partial \mathbf{r}}{\partial z} = (0, 0, 1) \][/tex]

Taking the cross product:

[tex]\[ d\mathbf{S} = \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial z} \right| d\theta dz = 3 \, d\theta dz \][/tex]

Now, calculate F.ds:

[tex]\[ \mathbf{F} \cdot d\mathbf{S} = (3\cos \theta, 3\sin \theta, z) \cdot (0, 0, 3) = 3z \, d\theta dz \][/tex]

The bounds of integration for [tex]\( \theta \) are \( 0 \) to \( 2\pi \)[/tex] since we want to cover the entire curved surface of the cylinder. For [tex]\( z \),[/tex] the bounds are from [tex]\( 2 \) to \( 4 \)[/tex] since the surface is bounded below by [tex]\( x + y + z = 2 \)[/tex] and above by [tex]\( x + y + z = 4 \).[/tex]

Now integrate:

[tex]\[ \Phi = \int_0^{2\pi} \int_2^4 3z \, dz \, d\theta \]\[ \Phi = \int_0^{2\pi} \left[ \frac{3}{2}z^2 \right]_{z=2}^{z=4} \, d\theta \]\[ \Phi = \int_0^{2\pi} \frac{3}{2} \cdot 12 \, d\theta \]\[ \Phi = 18\pi \][/tex]

So, the flux of F through the curved surface of the cylinder is [tex]\( 18\pi \).[/tex]

Complete Question:
Compute the flux of F⃗ =xi⃗ +yj⃗ +zk⃗ through just the curved surface of the cylinder x2+y2=9 bounded below by the plane x+y+z=2, above by the plane x+y+z=4, and oriented away from the z-axis

Answer:

Stoma represents the interface between the environment and the plant, helping to obtain the necessary CO2.

Explanation:

Stoma are groups of two or more specialized epidermal cells whose function is to regulate gas exchange and perspiration.

The frequency or density varies widely from a few tens to thousands per mm2, due to the influence of environmental factors, leaf morphology and genetic composition.

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

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

Where:

[tex]F[/tex] is the gravitational force between Earth and Moon

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant

[tex]M=5.972(10)^{24} kg[/tex] is the mass of the Earth

[tex]m=7.349(10)^{22} kg[/tex] is the mass of the Moon

[tex]r=3.9(10)^{8} m[/tex] is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth a circular orbit, the Earth exercts a centripetal force on the moon, which is equal to [tex]F[/tex]:

[tex]F=m.a_{C}[/tex] (2)

Where [tex]a_{C}[/tex] is the centripetal acceleration given by:

[tex]a_{C}=\frac{V^{2}}{r}[/tex] (3)

Being [tex]V[/tex] the orbital velocity of the moon

Making (1)=(2):

[tex]m.a_{C}=G\frac{Mm}{r^{2}}[/tex] (4)

Simplifying:

[tex]a_{C}=G\frac{M}{r^{2}}[/tex] (5)

Making (5)=(3):

[tex]\frac{V^{2}}{r}=G\frac{M}{r^{2}}[/tex] (6)

Finding [tex]V[/tex]:

[tex]V=\sqrt{\frac{GM}{r}}[/tex] (7)

[tex]V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}[/tex] (8)

Finally:

[tex]V=1010.92 m/s[/tex]

To find the moon's speed, we use Newton's law of universal gravitation and the formula for centripetal acceleration, substituting the Earth's mass and Earth-Moon distance to solve for the moon's orbital speed.

Calculating the Speed of the Moon from Earth-Moon Distance

To find the speed of the moon in its orbit using Newton's law of universal gravitation, we apply the formula for gravitational force, F, which is given by F = (G * m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of two objects, and r is the distance between their centers. For the Earth and Moon, we use Earth's mass (m2) and the Earth-Moon distance (r).

The moon's orbital speed (v) can be related to the gravitational force (F) since the force providing the moon's centripetal acceleration (a_c) required for circular motion is the same as the gravitational force exerted by Earth. That is, a_c = (v^2)/r. Setting the expression for gravitational force equal to mass times centripetal acceleration (F = m * a_c), and solving for v, we find v = sqrt(G * m2 / r).

We are given the distance between the Earth and Moon (r = 3.9 × 10^8 m) and the Earth's mass. Substituting these values into the equation, we can solve for the orbital speed of the Moon.

Answer:

Option (B)

Explanation:

A rockfall is defined as a type of mass movement process, where the small and large fragments of rocks are weathered and fall down from the top of the cliff under the influence of gravity. It is a process that takes place in a dry condition. This means that it does not depend on the moisture content that could be present in the rock. The rock particles fall down by bouncing and rolling. The base of this rockfall area forms a collection of smaller to larger size rock fragments. This process mainly takes place in the region where the temperature of the area is very high and experiences comparatively less amount of rainfall.

Thus, this type of mass movement is the fastest of all and is not dependent on the amount of moisture content.

Thus, the correct answer is option (B).

The option that gives the type of mass movement that is the fastest and depends upon moist or saturated conditions the least is;

Option B; Rockfalls

To answer this question, let us analyze each of the options.

Option A; Solifluction; This occurs when is the soil becomes saturated and then the wet soil starts to flow down a slope. This alongside creep are the slowest types of mass movement.

Option B; Rockfall; These occur when rock fragments that could be tiny or huge fall down from cliffs that are steep. Thus, this is the fastest type of mass movement as it is not dependent on moist or saturated conditions.

Option C; Rockslides; These are also known as landslides and they occur when numerous amounts of loose rocks get to join with soils and suddenly fall down from a slope.

Option D; Soil Creep; This means a slow movement of soil material down a slope under the Influence of gravity.

Option E; Talus Creep; This is a very slow movement of individual rock fragments individually down a slope.

Looking at the analysis of the options above, we can say that the fastest form of mass movement is Rockfalls.

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Explanation:

Taking the incident light to be traveling in the + x-direction so that it is at normal incidence to the left side of the film(referred to as the "Front side"). This means the beam transmitted into the liquid is essentially as strong as the incident beam.

Almost all the light that is reflected off the back surface will get through the front surface. (But only 2.78% gets re-reflected off the the front surface back to the right) this means that there are two beams reflected to the - x-direction, one from the front surface and one from the back, and these beams are of almost equal intensity.

Hope this is helpful. Thanks

Answer:

a)TE=0.0245 J

b)A = 0.0128 m

c)V=0.57 m/s

Explanation:

Given that

m = 0.150 kg

K= 300 N/m

x= 0.012 ,v= 0.2 m/s

The velocity of the toy at any point given as

[tex]v=\omega\sqrt{A^2-x^2}[/tex]

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

[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]

[tex]0.2=\sqrt{\dfrac{300}{0.15}}\times \sqrt{A^2-0.012^2}[/tex]

2 x 10⁻⁵ = A² - 0.000144

A=0.0128 m

Amplitude ,A = 0.0128 m

The total energy TE

[tex]TE=\dfrac{1}{2}KA^2[/tex]

[tex]TE=\dfrac{1}{2}300\times 0.0128^2[/tex]

TE=0.0245 J

The maximum speed

[tex]V=\omega A[/tex]

[tex]V=\sqrt{\dfrac{K}{m}}\times A[/tex]

[tex]V=\sqrt{\dfrac{300}{0.15}}\times 0.0128[/tex]

V=0.57 m/s

(a) The total energy of the toy is 0.0246 J (b) The amplitude of its motion is 0.0128 m (c) The maximum speed it can attain is 0.573 m/s.

To solve the given problem involving a toy undergoing simple harmonic motion (SHM) on a horizontal spring, let's go step-by-step:

(a) The total energy (E) in SHM is the sum of kinetic energy (KE) and potential energy (PE) at any point.

Given the mass (m) is 0.150 kg, spring constant (k) is 300.0 N/m, displacement (x) is 0.0120 m, and speed (v) is 0.200 m/s, we calculate:

Kinetic Energy: [tex]KE = (\frac{1}{2})mv^2 = (\frac{1}{2})(0.150 kg)(0.200 m/s)^2[/tex] = 0.003 J

Potential Energy:[tex]PE = (\frac{1}{2})kx^2 = (\frac{1}{2})(300.0 N/m)(0.0120 m)^2[/tex] = 0.0216 J

Total Energy: E = KE + PE = 0.003 J + 0.0216 J = 0.0246 J

Amplitude: The total energy in SHM is also equal to the potential energy at maximum displacement (amplitude A). Therefore, using [tex]E = (\frac{1}{2})kA^2[/tex], we solve for A:

[tex]0.0246 J = (\frac{1}{2})(300.0 N/m)A^2[/tex]

[tex]A^2 = (\frac{0.0246 J}{150 N/m}) = 0.000164 m^2[/tex]

[tex]A = \sqrt{0.000164 m^2} = 0.0128 m[/tex]

(b) Maximum Speed: The maximum speed (vm)) occurs at the equilibrium position.

Using [tex]E = (\frac{1}{2})mv^2_m[/tex], we solve for vm:

[tex]0.0246 J = (\frac{1}{2})(0.150 kg)v^2_max[/tex]

[tex]v^2_m = (\frac{0.0246 J}{0.075 kg}) = 0.328 m^2/s^2[/tex]

[tex]v_m = \sqrt{0.328 m^2/s^2} = 0.573 m/s[/tex]

Therefore, the total energy of the system is 0.0246 J, the amplitude of oscillation is 0.0128 m, and the maximum speed is 0.573 m/s.

The magnitude of the frictional force on the crate is 20 N.

The magnitude of the frictional force on the crate can be determined using the equation fs = μsN, where fs is the frictional force, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the crate, which is 40 N. Therefore, the frictional force is fs = (0.5)(40 N) = 20 N.

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Answer:

498.67 seconds, wich is 8.31 minutes

Explanation:

We can calculate it because we know the distance between the sun and the earth, and the speed at which light travels.

The distance from the sun to the earth is:

[tex]d=149.6 x10^9m[/tex]

and the speed of light:

[tex]v=3x10^8m/s[/tex]

this the tie it takes for light of the sun to reach earth can be calculated by the next equation:

[tex]t=\frac{d}{v}[/tex]

Substituting the v and d values:

[tex]t=\frac{149.6 x10^9m}{3x10^8m/s}=498.67s[/tex]

This is the anwer: 498.67s

if we want it in minutes we just divide by 60:

[tex]498.67s/60=8.31 minutes[/tex]

Answer:

Q = 230.36 C

Explanation:

Given

R = 3.54 Ω

t = 30'45" = (30')*(60"/1') + 45" = 1845 s

V = 442 mV = 442*10⁻³V

Q = ?

We can use Ohm's Law in order to get I as follows

V = I*R   ⇒    I = V / R

⇒    I = 442*10⁻³V / 3.54 Ω = 0.1248 A

Finally we use the formula

I = Q / t     ⇒    Q = I*t = (0.1248 A)(1845 s)

⇒    Q = 230.36 C

Answer:

6.46*10^-11m

Explanation:

We can use the equation of the center of the mass of two atoms

[tex]Xcm=\frac{mcxc+moxo}{mc+mo}[/tex]

If we take the origin at the center of the molecule carbon monoxice, xc will be 0, so

[tex]Xcm=\frac{moxo}{mc+mo}[/tex]=[tex]\frac{xo}{mc/mo+1}[/tex]

[tex]Xcm=\frac{1.13.10^-3}{0.750mo/mo+1}[/tex]=6.46*10^-11m

In an isochoric process, where volume is constant and the gas is cooled, no work is done by the gas. Therefore, the answer is zero.

In the thermodynamics, if a given amount of a gas is maintained at a constant volume and is cooled, the whole process occurs in what is known as an isochoric process. In an isochoric process, the volume is held constant which means the gas does no work because there isn't any volume change. Therefore, work done by the gas in this case is zero, hence, the answer is (c) zero. Work = Change in Internal Energy (ΔU) - Heat (Q)

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The volume of the monatomic gas is kept constant. When the volume is constant, the work done by the gas is zero because work is done when there is a change in volume. Therefore, the correct answer to this physics question is c) zero.

Considering the situation, we can note that the volume of the monatomic gas is kept constant. When the volume is constant, the work done by the gas is zero because work is done by a gas when it expands or contracts, which involves a change in volume. The only change occurring here is the cooling of the gas by 50K, that is achieved by removing 400J of energy from the gas, but this does not involve any work done by the gas itself.

Therefore, the correct answer to this physics question is c) zero. This answer can be found using the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transferred to the system, minus the work done by the system. In this case, there is no work done by the system since the volume is constant.

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Answer:

[tex]F_n = mg - \frac{mv^2}{R}[/tex]

Explanation:

As we know that when an object moves in a circle with uniform speed then the force required by the object in moving the circular path is known as centripetal force.

This force is always towards the center of the circle and points towards it

This force is the sum of all forces towards the center

so we have

[tex]mg - F-n = F_c[/tex]

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

so we have

[tex]mg - F_n = \frac{mv^2}{R}[/tex]

[tex]F_n = mg - \frac{mv^2}{R}[/tex]

Answer:

a)

[tex]mv l[/tex]

b)

[tex]\frac{M }{(M + m)}[/tex]

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

a)

[tex]m[/tex] = mass of the bullet

[tex]v[/tex] = velocity of the bullet before collision

[tex]r[/tex] = distance of the line of motion of bullet from pivot = [tex]l[/tex]

[tex]L[/tex] = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

[tex]L = m v r[/tex]

[tex]L = mv l[/tex]

b)

[tex]V[/tex] = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

[tex](M + m) V l = m v l\\V =\frac{mv}{(M + m)}[/tex]

[tex]K_{o}[/tex] = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

[tex]K_{o} = (0.5) m v^{2}[/tex]

[tex]K_{f}[/tex] = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

[tex]K_{f} = (0.5) (M + m) V^{2}[/tex]

Fraction of original kinetic energylost is given as

Fraction = [tex]\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}[/tex]

Fraction = [tex]\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}[/tex]

Fraction = [tex]\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}[/tex]

The question is about the physics principle of Conservation of Angular Momentum in a completely inelastic collision involving a bullet and a block attached to a rod. After the collision, their total angular momentum remains constant due to no external torques acting on the system. This allows the analysis of their post-collision movement.

The subject of this question is Physics and it deals with the concept of Conservation of Angular Momentum during a collision. When the bullet travels horizontally and hits the wooden block, it becomes embedded in it. In this scenario, the system of the block and bullet then starts moving as a single unit. This event is typically referred to as a completely inelastic collision.

The conservation of angular momentum principle states that the total angular momentum of a system remains constant if no external torques act on it. Here, the bullet-block system doesn't experience any external torques. Therefore, their total angular momentum before the collision is equivalent to their total angular momentum after the collision. This principle allows us to analyze the movement of the bullet-block system after the collision.

It's worth noting that in this setup, the block, the rod, and the position at which the bullet hits the block together form a simple physical pendulum. Therefore, the post-collision dynamics of the system could also involve oscillatory movements, which relate to the aspects of pendulum physics.

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Answer:

They would be the same

Explanation:

Since there's no external force, or the external net force acting on the satellite is 0. By the law of conservation of momentum, the momentum before and after the explosion must be the same.

Even if the satellite is stationary prior to the explosion, its momentum is 0. After explosion it sends many of its pieces in all direction with different non 0 momentum. The total momentum by account of their direction would sum up to be 0.

Final answer:

The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion.

Explanation:

The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion. This is because linear momentum is conserved in an explosion. Although the distribution of momentum among the individual pieces may change, the total momentum remains the same. In this case, when the satellite explodes, the momentum of the satellite is redistributed among the thousands of pieces that are scattered in all directions.

Answer:

d = 25.51 m

Explanation:

the law of the conservation of energy says that:

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

where [tex]E_i[/tex] is the inicial energy, [tex]E_f[/tex] is the final energy and [tex]W_f[/tex] is the work of the friction.

so:

[tex]E_i[/tex] = [tex]\frac{1}{2} MV^2[/tex]

[tex]E_f = 0[/tex]

where M is the mass and V the velocity.

also,

[tex]W_f = U_kNd[/tex]

where [tex]U_k[/tex] is the coefficient of kinetic frictio, N is the normal force and d is the distance.

therefore:

[tex]\frac{1}{2}MV^2=U_kNd[/tex]

also, N is equal to the mass of the hockey puck multiplicated by the gravity.

replacing:

[tex]\frac{1}{2}m(5)^2=(0.05)(m(9.8))(d)[/tex]

canceling the m:

[tex]\frac{1}{2}5^2=0.05(9.8)(d)[/tex]

solving for d:

[tex]d = \frac{\frac{1}{2}5^2 }{0.05(9.8)}[/tex]

d = 25.51 m

The distance which the hockey puck slide before coming to rest is equal to  25.51 meters.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].

To find how far (distance) the hockey puck slide before coming to rest, we would use the law of conservation of energy:

According to the law of conservation of energy:

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

The final kinetic energy of the hockey puck is zero (0) because it came to rest or stop.

[tex]K.E_i - 0 = W_f\\\\K.E_i = W_f\\\\\frac{1}{2}mv_i^2 = umgd\\\\\frac{1}{2}v_i^2 = ugd\\\\v_i^2 = 2ugd\\\\d = \frac{v_i^2}{2ug}[/tex]

Substituting the given parameters into the formula, we have;

Distance, d = 25.51 meters

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Answer:

Height reached by the ball, h = 3.57 meters

Explanation:

It is given that,

Mass of the disk, m = 42 kg

Diameter of the disk, d = 3.2 m

Radius, r = 1.6 m

Angular speed of the disk, [tex]\omega=4.27\ rad/s[/tex]

The kinetic energy of the disk is equal to its potential energy. Using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2=mgh[/tex]

[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}(mr^2/2)\omega^2=mgh[/tex]

[tex]\dfrac{1}{2}v^2+\dfrac{1}{2}(r^2/2)\omega^2=gh[/tex]

[tex]\dfrac{1}{2}(r\omega)^2+\dfrac{1}{2}(r^2/2)\omega^2=gh[/tex]

[tex]\dfrac{1}{2}(1.6\times 4.27)^2+\dfrac{1}{2}(1.6^2/2)\times 4.27^2=gh[/tex]

[tex]h=\dfrac{35.0071}{9.8}[/tex]

h = 3.57 meters

So, the solid disk will reach to a height of 3.57 meters.

The height attained by the disk on the hill is 3.57 m.

The given parameters;

The height attained by the disk is calculated by applying the law of conservation of energy as follows;

[tex]\frac{1}{2}mv^2 + \frac{1}{2}I \omega ^2 = mgh\\\\mv^2 + I \omega ^2 = 2mgh\\\\mv^2 + (\frac{mr^2}{2} ) \omega ^2 = 2mgh\\\\v^2 + \frac{1}{2} r^2 \omega ^2 = 2gh\\\\(\omega r)^2 + 0.5(\omega r )^2 = 2gh\\\\1.5 (\omega r)^2 = 2gh\\\\h = \frac{1.5 (\omega r)^2}{2g} \\\\h = \frac{1.5 \times (4.27 \times 1.6)^2 }{2\times 9.8} \\\\h = 3.57 \ m[/tex]

Thus, the height attained by the disk on the hill is 3.57 m.

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Final answer:

To calculate the pressure after a temperature change, use the equation PV = nRT, and simplifying it to P1/T1 = P2/T2. Rearranging the equation, the final pressure (P2) can be calculated as P2 = P1 * (T2 / T1). In this case, the pressure after the temperature has risen to 35.0ºC is approximately 7.33 × 10^5 Pa.

Explanation:

To calculate the pressure after the temperature change, we can use the equation PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature. Since the volume and moles remain constant in this scenario, we can simplify the equation to P1/T1 = P2/T2, where P1 and T1 represent the initial pressure and temperature, and P2 and T2 represent the final pressure and temperature. Rearranging the equation, we get P2 = P1 * (T2 / T1). Therefore, the pressure after the temperature has risen to 35.0ºC can be calculated as:

P2 = 7.00 × 10^5 Pa * (35.0 + 273.15) / (21.0 + 273.15)

P2 = 7.00 × 10^5 Pa * 308.15 / 294.15

P2 = 7.33 × 10^5 Pa

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

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

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

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

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

[tex]k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m[/tex]

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Answer:

Explanation:

On the rod , three forces are acting .

1 ) its weight  from the middle point .

2 ) Tension in the wire in the horizontal direction

3 ) reaction force by the hinge which balances the other two forces

Now taking moment of forces acting on the rod about the hinge

and equating opposite moment

3.7 x 9.8 x .6 x cos 25 = T x .51 ( T is tension in the horizontal wire)

T = 38.6 N

a )

The horizontal component of the force exerted on the rod by the hinge

will balance the tension in the wire acting in horizontal direction so it will be equal to

38.6 N

b )  The vertical component of the force exerted on the rod by the hinge

will balance the weight of rod so it will be equal to

3.7 x 9.8 N

= 36.26. N

=

(a) The horizontal component of the force exerted on the rod by the hi1ge is 38.67 N.

(b) The vertical component of the force exerted on the rod by the h1nge is 36.26N.

The tension in the wire is calculated by applying the following formula as shown below;

We will apply the principle of moment, sum of clockwise moment is equal to sum of anti clockwise moment.

(a) The horizontal component of the force exerted on the rod by the hi1ge is calculated as;

taking moment of forces acting on the rod about the h1nge;

3.7 x 9.8 x .6 x cos 25 = T x .51

where;

3.7 kg x 9.8 m/s² x 0.6 m x cos 25 = Tₓ x .51 m

19.72 = 0.51Tₓ

Tₓ = 19.72 / 0.51

Tₓ = 38.67 N

(b) The vertical component of the force exerted on the rod by the h1nge will balance the weight of rod so it will be equal to;

Ty = 3.7 x 9.8 N

Ty = 36.26N

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Answer:

d = 5.8 m

Explanation:

Principle of work and energy

ΔE = Wf

ΔE = Ef-Ei

ΔE : mechanical energy change

Wf : Work done by kinetic friction force

Ef : final mechanical energy

Ei : initial mechanical energy

K =(1/2 )mv² :  Kinetic energy

U= mgh   :Potential energy

m: mass (kg)

v : speed (m/s)

h: high (m)

Data

vi= 5.00 m/s

vf=0

hi=0

hf=0

μk=0.220

Kinetic friction force

fk = μk* FN

FN = W  : normal force (N)

W = m*g : weight  (N)

FN= 9.8*m (N)

fk =0.220*9.8*m

fk = (2.156)*m    Equation (1)

Work done by the  kinetic friction force

Wf = -fk*d  (J)  Equation (1)

d: distance traveled by force

Principle of work and energy to the skier

ΔE = Ef-Ei

Ef = Kf + Uf = 0

Ei = Kf + Uf = (1/2)(m)(5)² + 0

ΔE = -(1/2)(m)(5)²

ΔE = Wf

0- (1/2)(m)(5)² = -fk*d

We replace fk = (2.156)*(m )of the equation (1)

-(1/2)(m)(5)²= -(2.156)*(m)*d

We divide by (-m ) both sides of the equation

(1/2)(5)²=  (2.156)*d

12.5=  ( 2.156)*d

d = (12.5) / ( 2.156)

d = 5.8 m

To calculate the distance a skier travels before stopping, we equate the skier's initial kinetic energy to the work done by friction. The force exerted by friction is calculated using the given coefficient of kinetic friction and the skier's weight. The distance traveled can be calculated using this force with the initial speed of the skier.

The problem you've posed can be solved using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. We can use the principle of conservation of energy here. Initially, the skier has kinetic energy and no potential energy.

When she stops, all her kinetic energy has been transferred into frictional work, hence her kinetic energy becomes 0. Kinetic energy can be defined as (1/2)mv^2 and the work done by friction is force times distance (Fd).

Force exerted by friction (F) can be calculated using the formula F = μN where μ corresponds to the coefficient of kinetic friction given as 0.220 and N is the normal force and in this case, equals to the weight of the skier (assuming the skier is on a horizontal plane).

To find the distance (d) travelled before she comes to a stop, we equate the skier's initial kinetic energy to the work done by friction, giving us (1/2)mv^2 = μmgd. The mass of the skier (m) cancels out, and the distance comes out to be d = v^2/(2μg).

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Answer:

infinity

Explanation:

Given that

Resistance = R

Resistivity = ρ

Length = L

Diameter = d

The resistance of wire R given as

[tex]R=\rho\dfrac{L}{A}[/tex]

A=Area

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

Now by putting the value of A

[tex]R=\rho\dfrac{L}{\dfrac{\pi d^2}{4}}[/tex]

[tex]R=\rho\dfrac{4L}{\pi d^2}[/tex]

When d tends to infinity then d² will also tends to infinity.

So when d tends to zero then the resistance tends to infinity.

Therefore answer is ---

infinity

Answer:

d

Explanation:

Answer: D. 40,000Joules

Explanation:

Heat capacity is defined as the heat required to change the state of a substance.

Simce the substance in question is a block of ice (solid), the heat needed to cause it to melt is Latent heat H = mL

m is the mass of the substance

L is the latent heat of fusion of water

H= 1 × 334,000

= 334,000Joules

Heat required to raise the temperature of the water to 10°C

Q = mCƦ

where C is specific heat capacity of water and Ʀ is change in temperature

Q = 1 × 4000 ×(10-0)

Q= 40,000Joules

Since the block already melts, the heat required will just be

40,000Joules (D)

Final answer:

To raise the temperature of 1 kg of melted ice water to 10°C, the heat energy required is calculated using the specific heat capacity of water (4,000 J/kg-°C) and the formula Q = mcΔT, resulting in 40,000 J.

Explanation:

The question is asking about the amount of heat energy required to raise the temperature of 1 kg of water (from melted ice) to 10 °C. After the ice melts, the water at 0 °C needs to be heated to 10 °C. Using the specific heat capacity of water, which is 4,000 J/kg-°C for this calculation, we can find the amount of energy needed with the formula:

Q = mcΔT

where Q is the heat energy, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values:

Q = (1 kg)(4,000 J/kg-°C)(10 °C - 0 °C)

Q = (1 kg)(4,000 J/kg-°C)(10 °C)

Q = 40,000 J

Therefore, the correct answer is D. 40,000 J.

The statement is correct. Recrystallization refers to any chemical, physical, or biological changes that occur after sediments are deposited, whereas burial and lithification occur when unconsolidated sediments become sedimentary rocks.

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Answer:

Size and charge and speed

Explanation:

Gel electrophoresis  is method designed for the separation of DNA molecules. The are also used to separate other micromolecules like RNA and protein, etc. This separation takes place based on their size and charge. The molecule travels through the gel in all different directions but their speed are different and thus they are separated.

Gel electrophoresis is a technique used to separate DNA fragments based on their size.

Gel electrophoresis is a technique used to separate DNA fragments based on their size. It involves loading DNA samples onto a gel matrix and applying an electric current. Smaller DNA fragments move faster through the gel, resulting in bands at specific distances from the top of the gel. Molecular weight standard samples can be run alongside the DNA fragments to provide a size comparison. The separated DNA fragments can then be visualized by staining the gel with a DNA-specific dye.

Gel electrophoresis is a fundamental technique used in molecular biology, genetics, and forensics for various purposes, including DNA fingerprinting, DNA sequencing, and the analysis of PCR products. It provides a visual representation of DNA fragment sizes, enabling researchers to draw conclusions about genetic material and study genetic variations.

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Answer:

Lone pairs cause bond angles to deviate away from the ideal bond angles

Explanation:

Bonded electrons are stabilized and clustered between the bonding electrons meaning they are much closer together. Non-bonding electrons however are not being shared between any atoms which allows them to roam a little further spreading the charge density over a larger space and therefore interfering with what would be an expected bond angle

Answer:

Not possible.

Explanation:

Exoplanets are an active area of modern research. The article states that the outermost planet (Planet C) goes all the way around its star in less time than the innermost planet (Planet A). Which is not possible as it will violet Kepler's third law of planetary motion.  Which says that the square of orbital period of a planet is proportional to the cube of its semi-major axis of its orbit.

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

Explanation:

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen.  At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution.  In the event, that the net torque is zero, at that point angular momentum is steady or saved.  

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework.  Be that as it may, we likewise double the moment of inertia. Since [tex]L=I \times \omega[/tex], the turning rate of the two-puck framework must stay unaltered.

The answer involves comparing the centripetal force exerted by the turntable on the coin to the static and kinetic friction forces that resist the coin's motion.

The situation involves a coin on a turntable where rotation is causing a force. The physics principles at play here are centripetal force and the forces of static and kinetic friction. The centripetal force needed to keep the coin rotating can be calculated using the equation F = mrω², where m is the mass of the coin, r is the distance from the coin to the center of the rotation, and ω is the angular speed of the rotation.

The static friction force is what keeps the coin from sliding as the turntable speeds up. Static friction can be calculated using the equation fs = µsmg, where µs is the coefficient of static friction, m is the mass, and g is acceleration due to gravity. If the force of static friction is less than the centripetal force, the coin will start to slide, and kinetic friction will come into play. Kinetic friction can be calculated with f = µkmg.

By calculating these forces, we can compare and determine if the centrifugal force from the turntable's acceleration would overcome the friction between the coin and the turntable, causing the coin to slide off.

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The main important characteristic that distinguishes meteorites and terrestrial rocks is the presence of fusion crust and the presence of iron metal alloys.

Meteorites are usually covered with dark, pitted crust resulting from their fiery passage through the atmosphere. Terrestrial rock usually has a metal content and can able to attract the magnet.

Meteorite has rare earth element such as iridium, and terrestrial rocks do not have rare earth elements. Meteorite has a dark crust from burning Earth's atmosphere. A fusion crust is present only in meteorites whereas terrestrial rock doesn't.

Meteorites have high metal content and it has different isotope ratios of the particular elements. The terrestrial rocks do not have isotope and metal content.

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Final answer:

A meteorite is distinguished from a terrestrial rock primarily by its composition and origin, with meteorites often containing metallic elements and coming from asteroids. Primitive and differentiated are two main categories of meteorites, providing insight into the early solar system and respective parent bodies’ structures.

Explanation:

The characteristic that distinguishes a meteorite from a terrestrial rock is mainly its origin and composition. Meteorites are typically fragments from asteroids and are often rich in metallic elements, unlike many terrestrial rocks. A key distinction between meteoritic and terrestrial rocks can be made on the basis of whether they are primitive or differentiated. Primitive meteorites are composed of materials that have not been altered by heat or pressure since their formation and provide insights into the early solar system. On the other hand, differentiated meteorites are remnants of larger parent bodies that experienced molten states, allowing materials to separate by density, similar to the process on Earth but with different conditions. Examples of differentiated meteorites include irons and stony-irons, which come from the metal cores or mantle-core boundaries of their parent bodies.

Typical stony meteorites closely represent the terrestrial crust or mantle, while typical iron meteorites have compositions akin to the Earth's core. Scientists use various methods, including chemical and mineralogical analysis, to distinguish meteorites from terrestrial material. Lunar meteorites, for instance, can be identified by their unique chemical properties compared to Earth rocks. By studying meteorites, scientists gain valuable knowledge about the history and formation of the solar system.