A ball whose mass is 1.8 kg is suspended from a spring whose stiffness is 8.5 N/m. The ball oscillates up and down with an amplitude of 13 cm. (a) What is the angular frequency

Answer:separate

Explanation:

To determine the magnetic force on a straight wire carrying a current in a uniform magnetic field, we can use the formula for the magnetic force:

F = I * L * B * sin(θ)

where:

F is the magnetic force,

I is the current in the wire,

L is the length of the wire,

B is the magnitude of the magnetic field, and

θ is the angle between the wire and the magnetic field.

In this case, the values are:

I = 30 A (current in the wire)

L = 2.0 m (length of the wire)

B = 55 mT = 0.055 T (magnitude of the magnetic field)

θ = 20° (angle between the wire and the magnetic field)

Substituting the values into the formula:

F = 30 A * 2.0 m * 0.055 T * sin(20°)

Calculating sin(20°):

F = 30 A * 2.0 m * 0.055 T * 0.3420

F ≈ 1.5714 N

Therefore, the magnetic force on the 2.0-meter length of wire carrying a current of 30 A in a region with a uniform magnetic field of magnitude 55 mT and at an angle of 20° away from the wire is approximately 1.5714 N.

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Answer:

Explanation:

It's graph A because the pressure in the tire is increasing as the amount of air going into it increases. B says the pressure drops exponentially as air goes in, and C says that the pressure stays the same as air goes in. Pressure in a tire increases proportionally to the amount of air in it.

Newton's First Law of Motion states "An object continues to be in its state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force".

The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

The speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.

K.E = P.E

¹/₂mv² = mgh

v = √2gh

where;

v(2) = √[(2 x 9.8)(39 - 0)]

v(2) = 28 m/s

v(3) = √[(2 x 9.8)(39 -  26)]

v(3) = 16 m/s

v(4) = √[(2 x 9.8)(39 - 14)]

v(4) = 22 m/s

Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

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The work done (W) in lifting the crate of bananas is equal to the force applied (F) times the distance moved (d). Therefore, W = F * d. Since we know the work done (W) and the distance moved (d), we can solve for the force applied (F).


W = 40 W

d = 5 m


F = W / d

F = 40 W / 5 m

F = 8 N


Therefore, the force applied is 8 N and the work done is 40 J.

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Answer:

change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

Explanation:

For this exercise we can use the conservation of angular momentum.

Initial. Before the crash

        L₀ = I w₀

final. After the crash

         L_f = I w_f + p r

where the moment is

        p = mv

As the system is formed by the two bodies, the forces during the impact are internal, therefore the angular momentum is conserved

        L₀ = L_f

        I w₀ = I w_f + m v r

        w_f = w₀ - \(m \frac{ v \ r }{I}\)

We can approximate the Earth to a sphere, so its angular momentum is

       I = 2/5 M r²

we substitute

         w_f = w₀ - \(\frac{5 \ m \ v}{2 \ M \ r}\)

We can find the angular velocity of the Earth with the duration of a spin which is the period of one day

        w₀ = 2π / T

        T = 24 h (3600 s / 1h) = 86 400 s

         w₀ = 2π / 86400

         w₀ = 7.272 10⁻⁵ rad / s

let's calculate

         w_f = 7.27 10⁻⁵ - \(\frac{5 \ 250 \ 10^{13} \ 33.0 \ 10^{3} }{ 2 \ 5.98 \ 10^{24} \ 6.37 10^{6} }\)

         w_f = 7.272 10⁻⁵ - 1.0829 10⁻¹³

         w_f = 7.27199999 10⁻⁵

this change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

The car is going at speed of 3.82 m/s at the end of the breaking period and the car traveled the distance from the start is 25.33 m.

The rate of change in an object's velocity concerning time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

According to the question, the given values are :

Time, t₁ = 4.9 s

Time, t₂ = 1.60 s

Acceleration, a₁ = 1.4 m/s²

Acceleration, a₂ = -1.9 m/s²

Initial Velocity, u = 0 m/s

(a)

By the Equation of motion,

v = u + a₁t₁

v = 0 + (1.4)(4.9)

v = 6.86 m/s.

Now, the velocity of the car at the end of the breaking period :

v' = v + a₂t₂

v' = 6.86 + (-1.9)(1.60)

v' = 3.82 m/s.

(b)

s₁ = ut₁ + 1/2 a₁t₁²

s₁ = 0 + 1/2 (1.4)(4.9)²

s₁ = 16.8 m.

Now for the decelerated motion :

s₂ = vt₂+1/2 a₂t₂²

s₂= 6.86(1.60) + 1/2 (-1.9)(1.60)²

s₂ = 8.53 m.

So, total distance covered from the start, D :

D = s₁ + s₂

D = 16.8 + 8.53

D = 25.33 m.

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Answer:

Explanation:

The pressure is calculated by dividing the force by the area:

P = F / A = 160 N / 24 cm² = 6.67 N/cm²

The uncertainty in the calculation of pressure can be determined by propagating the uncertainties in the force and area measurements. According to the formula for uncertainty propagation, the fractional uncertainty in the pressure calculation is:

ΔP / P = √((ΔF / F)^2 + (ΔA / A)^2)

Plugging in the values:

ΔP / P = √((6 N / 160 N)^2 + (3 cm² / 24 cm²)^2)

ΔP / P = √(0.0375) = 0.195

So the fractional uncertainty in the pressure calculation is 0.195 or 19.5%.

To find the absolute uncertainty in the pressure, multiply the fractional uncertainty by the pressure:

ΔP = P * ΔP / P = 6.67 N/cm² * 0.195 = 1.29 N/cm²

So, the pressure is 6.67 ± 1.29 N/cm².

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\(\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}} = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N\)

The force,by the charge q₂ on the q₃;

\(\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N\)

The net force is the sum of the two forces;

\(\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N\)

Hence, the net force on q₃ will be 17.51 N.

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I think its -88 on acellus

The normal force on the block is approximately 20.05 N.

Force describes the interaction between objects or between an object and its environment. It causes a change in the motion of the body it acts upon.

The normal force, which counteracts the force of gravity dragging the block downward, is the force generated by the slope acting perpendicular to its surface. As the incline has no friction, there is no force acting parallel to its surface.

To ascertain the parts of the force of gravity pulling on the block, we can apply trigonometry. There are two parts to the force of gravity: one that is parallel to the incline's surface and the other that is perpendicular to it. The weight of the block, mg, where m is its mass and g is its gravitational acceleration, is equal to the component of gravity perpendicular to the inclination. mg sin θ, where is the angle of the incline, is the component of gravity that is parallel to the incline.

To calculate the acceleration of the block moving down the incline, we can apply Newton's second law, F = ma. The component of gravity parallel to the inclination, or mg sin θ, represents the net force exerted on the block. As a result, we have:

\(mg sin(theta) = ma\)

To solve for a, we obtain:

\(a = g sin(theta)\)

Substituting the given values, we get:

\(a = 9.8 m/s^2 * sin(27°)\) ≈ \(4.69 m/s^2\)

Now that we know the normal force acting on the block, we can use Newton's second law once more. The component of gravity's force perpendicular to the incline is equal in magnitude to the normal force and moves in the opposite direction. As a result, we have:

\(mg cos(theta) = N\)

Inputting the values provided yields:

\(N = 2.3 kg * 9.8 m/s^2 * cos(27°)\) ≈ \(20.05 N\)

Therefore, the normal force on the block is approximately 20.05 N.

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The correct answer is option (a)

Protons and neutrons are the particles in the atom that accounts for the most mass of the atom.

Answer:

4.05×10⁶ N.

Explanation:

The following data were obtained from the question:

Breadth (B) = 4 m

Length (L) = 10 m

Standard pressure (P) = 101325 Nm¯²

Force (F) =..?

Next, we shall determine the area of the wall. This can be obtained as follow:

Area (A) = length (L) × breadth (B)

A = 10 × 4

A = 40 m²

Finally, we shall determine the force exerted on one side of the wall as follow:

Standard pressure (P) = 101325 Nm¯²

Area (A) = 40 m²

Force (F) =.?

Pressure (P) = Force (F) /Area (A)

101325 = F/40

Cross multiply

F = 101325 × 40

F = 4.05×10⁶ N.

Therefore, a force of 4.05×10⁶ N was exerted on one side of the wall.

Option A) 3.9 km is the correct answer. the magnitude of Rhea's displacement vector is approximately 3.92 km.

In order to find out the magnitude of Rhea's displacement vector, we have to add up all of the displacement vectors.

Then we can use the Pythagorean theorem to calculate the magnitude of the resultant vector.

Since Rhea is first driving north for 2.4 km and then west for 3.1 km, we can represent her displacement vectors as follows: Δx = 0 km and Δy = 2.4 km for the first vector, and Δx = -3.1 km and Δy = 0 km for the second vector.

We can then add these vectors together by adding their components: Δx = 0 km + (-3.1 km) = -3.1 km and Δy = 2.4 km + 0 km = 2.4 km.

This gives us a resultant vector of -3.1 km east and 2.4 km north.

Using the Pythagorean theorem, we can find the magnitude of this vector: \(\sqrt{(\(-3.1 km)^{2} + (2.4 km)^{2} ) } = \sqrt{(9.61 + 5.76) km^{2} } = \sqrt{15.37 km^{2} } \approx 3.92 km.\)

Therefore, the magnitude of Rhea's displacement vector is approximately 3.92 km.

Therefore, option A) 3.9 km is the correct answer.

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Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

If the number of revolutions is 300 and the paired poles are 50 , then the frequency would be 15 kHz, therefore the correct answer is option A.

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,

F = 1 / T

As given in the problem, we have to calculate frequency if the number of revolutions is 300 and the paired poles are 50.

F = 300 × 50

  = 1500 kHz

Thus, If the number of revolutions is 300 and the paired poles are 50, then the frequency would be 15 kHz, therefore the correct answer is option A.

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Answer:

B.Focused,warm

Explanation:

when the rays are focused more energy an area recieves and it is warmer there.

In a DC generator, the generated electromotive force (emf) is directly proportional to the rotational speed of the generator's armature and the strength of the magnetic field within the generator.

This relationship is described by the equation for the generated emf in a DC generator:

Emf = Φ * N * A * Z / 60

Where:

Emf is the generated electromotive force (in volts),

Φ is the magnetic flux density (in Weber/meter^2\(meter^2\) or Tesla),

N is the number of turns in the armature winding,

A is the effective area of the armature coil (in square meters),

Z is the total number of armature conductors, and

60 is a constant representing the conversion from seconds to minutes.

From this equation, we can see that the generated emf is directly proportional to the magnetic flux density (Φ) and the product of the number of turns (N), effective area (A), and the total number of armature conductors (Z). This means that increasing any of these factors will result in a higher generated emf.

The magnetic flux density (Φ) can be increased by using stronger permanent magnets or increasing the strength of the field windings in the generator.

The number of turns (N) and the effective area (A) are design parameters and can be optimized for a specific generator. Increasing the number of turns or the effective area will result in a higher generated emf.

Similarly, the total number of armature conductors (Z) can be increased to enhance the generated emf.

By controlling and optimizing these factors, the generated emf in a DC generator can be increased, resulting in higher electrical output. However, it is important to note that there are practical limits to these factors based on the design and construction of the generator.

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A ) The acceleration of two crates = 1.4 m / s²

B ) The magnitude of force that each create exerts on other = 389.17 N

C ) If the crates are reversed, the acceleration = 1.4 m / s²

a ) ∑ \(F_{x}\) = m a

F = 610 N

m1 = 70 kg

m2 = 123 kg

m = m1 + m2 = 193 kg

μ = 0.18

g = 9.8 m / s²

\(F_{k}\) = μ N

\(F_{k}\) = μ m g

\(F_{k}\) = 0.18 * 193 * 9.8

\(F_{k}\) = 340.45 N

∑ \(F_{x}\) = m a

F - \(F_{k}\) = m a

610 - 340.45 = 193 a

a = 269.55 / 193

a = 1.4 m / s²

b ) Let the force exerted on crate 1 by crate 2 be F12.

\(F_{k2}\) = μ m2 g

\(F_{k2}\) = 0.18 * 123 * 9.8

\(F_{k2}\) = 216.97 N

∑ \(F_{x}\) = m a

F12 - \(F_{k2}\) = m2 a

F12 - 216.97 = 123 * 1.4

F12 = 172.2 + 216.97

F12 = 389.17 N

The force exerted on crate 2 by crate 1 will be equal in magnitude but will be opposite in direction.

F21 = - 389.17 N

c ) Eventhough the crates are reversed the masses, the applied force and the coefficient of kinetic friction are same. So the acceleration will be same.

a = 1.4 m / s²

d ) Let the force exerted on crate 1 by crate 2 be F12.

\(F_{k2}\) = μ m2 g

\(F_{k2}\) = 0.18 * 70 * 9.8

\(F_{k2}\) = 123.48 N

∑ \(F_{x}\) = m a

F12 - \(F_{k2}\) = m2 a

F12 - 123.48 = 70 * 1.4

F12 = 98 + 123.48

F12 = 221.48 N

The force exerted on crate 2 by crate 1 will be equal in magnitude but will be opposite in direction.

F21 = - 221.48 N

Therefore,

A ) The acceleration of two crates = 1.4 m / s²

B ) The magnitude of force that each create exerts on other = 389.17 N

C ) If the crates are reversed, the acceleration = 1.4 m / s²

D ) If the crates are reversed , determine the magnitude of the force that each crate exerts on the other = 221.48 N

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Air can do work when it exerts a force on an object and causes it to undergo displacement. The ability of air to do work is evident in various phenomena, such as wind pushing sails, fans moving objects, and air pressure powering pneumatic systems.

Air can do work through its ability to exert a force over a distance. Work is defined as the transfer of energy that occurs when a force is applied to an object and it undergoes displacement in the direction of the force. When air is in motion, it possesses kinetic energy and can exert a force on objects in its path, thus performing work.

To understand how air can do work, we can consider the example of a moving fan. When a fan is turned on, the blades start to rotate, creating a flow of air. As the air moves, it carries kinetic energy. When the moving air encounters an object, such as a piece of paper, the air molecules collide with the paper's surface and exert a force on it. This force causes the paper to move and displaces it from its initial position.

The work done by the air can be calculated using the equation:

Work = Force * Distance * cos(θ)

Where Force is the magnitude of the force exerted by the air, Distance is the displacement of the object, and θ is the angle between the direction of the force and the displacement.

In the case of air doing work on an object, the force exerted by the air is perpendicular to the direction of motion, resulting in θ = 90 degrees. Since cos(90) = 0, the equation simplifies to:

Work = Force * Distance * 0

Therefore, the work done by the air on the object is zero when the force exerted by the air is perpendicular to the displacement.

However, if the force exerted by the air is not perpendicular to the displacement, such as when blowing air at an angle to move an object, then work is performed. The air exerts a force on the object and causes it to move in the direction of the force, resulting in the transfer of energy.

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Answer:

c.8

Explanation:

Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

One notable athlete who has been associated with the use of performance-enhancing drugs (PEDs) is the American professional cyclist Lance Armstrong.

Armstrong gained worldwide recognition for his unprecedented seven consecutive victories in the Tour de France from 1999 to 2005. However, his remarkable achievements were tarnished when it was revealed that he had engaged in systematic doping throughout his career.

In 2012, after years of denial, Armstrong finally admitted to using banned substances, including erythropoietin (EPO), testosterone, corticosteroids, and blood transfusions, to enhance his performance. These substances boosted his endurance and oxygen-carrying capacity, providing him with an unfair advantage over his competitors. Armstrong's confession came after substantial evidence, including testimonies from teammates and extensive investigations, exposed his involvement in one of the most elaborate and sophisticated doping schemes in sports history.

Following his admission, Armstrong was stripped of his Tour de France titles and received a lifetime ban from professional cycling. The revelations surrounding his drug use had a profound impact on the sport, shaking its credibility and raising concerns about the prevalence of doping in cycling.

Armstrong's story serves as a cautionary tale, highlighting the ethical and moral dilemmas associated with doping in sports. His case underscores the importance of maintaining the integrity of athletic competition, the significance of stringent anti-doping measures, and the need for education and awareness regarding the risks and consequences of performance-enhancing substances.

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45º angle will result in the trebuchet’s furthest thrown projectile.

It is acceptable to infer from the results shown in the demonstration on Interactive Physics that as the weight of a counterweight on a trebuchet rises, so too should the projectile's range. Except for the fact that the distance did not rise linearly but rather more quadratically, the results support the theory. The results showed that the distance rose as the weight of the trebuchet's counterweight was increased in steps of 5 kilograms, starting at 20 kilograms and ending at 200 kilograms.

Each time, the projectiles were launched from the trebuchet at a 45-degree angle, and their distances typically followed the equation -8.1551E-4x2 +.304388x + 8.12756 (where x is the mass of the counterweight). The graph was thought to be more quadratic than linear because gravity has more time to work against the projectile and pull it down to the earth the longer it is in the air. Therefore, as additional mass is applied and the projectile is in the air for a longer period of time, the projectile distances would not grow as quickly. These findings back up Newton's Third Law of Motion as well as earlier, historical investigations.

Thus, Thomas should launch the trebuchet at a 45º angle to get the farthest thrown projectile.

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Answer:

A and B.

Explanation:

Answer C is wrong - there are no tall trees underwater

Answer D is wrong, there are mountain chains underwater due to plate tectonics.

Explanation:

The height of the rise of liquid with capillary tube is given by the formula as follows :

\(h=\dfrac{2S\cos\theta}{r\rho g}\)

Where

r is radius

It is clear that the height of the rise of liquid is inversely proportional to the radius of the capillary tube.

If the radius of the capillary tube is doubled, it means the height of rise of liquid with capillary tube become half.

Answer:

There are 5 general layers; organism, population, community, ecosystem, and biosphere.

Explanation:

Hope this helps!

The horizontal distance the fish travels before hitting the water is independent of the height the fish is dropped from, and the speed at which the pelican is traveling. The horizontal distance is determined only by the time the fish is in the air, which can be found using the equation:

Distance = Speed * Time

To find the time, we use the equation:

Time = sqrt(2 * Distance / Acceleration due to gravity)

Where distance is the height the fish is dropped from (5.4m) and the acceleration due to gravity is 9.8 m/s^2

Time = sqrt(2 * 5.4 / 9.8) = sqrt(2 * 5.4 / 9.8) = sqrt(1.08) = 1.03s

So the fish is in the air for 1.03s and the horizontal distance it travels is given by:

Distance = Speed * Time = 5.0m/s * 1.03s = 5.15m

So the fish will travel horizontally 5.15m before hitting the water.

Answer:

\(\mathbf{\alpha = 25.88 \ rad/s^2 }\)

Explanation:

Consider the force due to friction:

\(F = \mu_kN -------- (1)\)

where;

N = normal reaction

\(\mu_k\) = coefficient of kinetic friction.

Via the horizontal direction, the forces of equilibrium are:

\(\sum F_x = \sum(F_x) _{eff} \\ \\ N- F_{AB} \ cos \theta = 0 \\ \\ F_{AB} \ cos \theta = N ------ (2)\)

where;

\(\theta\) = angle of AB is associated with the horizontal;

\(F_{AB}\) = force exerted by AB

Let's take a look at the equilibrium of forces along the vertical direction.

\(\sum F_y = \sum (F_y)_{eff} \\ \\ F+F_{AB} sin \theta - W = 0\\ \\ \mu_k N + F_{AB} sin \theta -mg =0 \\ \\ F_{AB} sin \theta = mg - \mu_kN-------(3)\)

By dividing (3) by (2), we get:

\(\dfrac{F_{AB} sin \theta }{F_{AB} cos \theta} = \dfrac{mg - \mu_k N}{N} \\ \\ tan \theta = \dfrac{mg-\mu_kN}{N} \\ \\ Ntan \theta = mg - \mu_kN \\ \\ N(tan \theta + \mu_k ) = mg \\ \\ N = \dfrac{mg}{tan \theta + \mu_k}\)

By replacing the obtained value of N into:

\(F = \mu_k N\)

we have:

\(F = \mu_k (\dfrac{mg}{tan \theta + \mu_k}) \\ \\ \\ F = \dfrac{mg \mu_k}{tan \theta + \mu_k}\)

Moment about center A can be expressed as:

\(\sum M_A = \sum (M_A) _{eff} \\ \\ Fr = I \alpha\\ \\ \alpha = \dfrac{Fr}{I}\)

where;

\(\alpha\) = angular acceleration of the disc

I = mass moment of inertia

r = radius of disc

By replacing F and I; ∝ becomes:

\(\alpha = \dfrac{\bigg( \dfrac{mg \mu_k}{tan \theta + \mu_k} \bigg) r}{\dfrac{1}{2}mr^2}\)

\(\alpha = \dfrac{2g}{r} \dfrac{\mu_k}{tan \ \theta + \mu_k}------- (4)\)

where;

g = 9.8 m/s²

r = 0.18 m

\(\mu_k = 0.54\)

θ = 60°

\(\alpha = \dfrac{2(9.8)}{0.18} (\dfrac{0.54}{tan \ 60+ 0.54})\)

\(\mathbf{\alpha = 25.88 \ rad/s^2 }\)