A rope exerts a 35N force on an object at an angle of 12N degrees above the horizontal. What horizontal and vertical components of the force?

Answer:

a c d

Explanation:

yeet

Answer:

a c d

Explanation:

The force constant of the spring is 104.18 N/m given a 7.00-kg object with a period of 2.60 s.

When an object is hung from a spring and set into vertical oscillations, the force constant of the spring can be determined using the formula that relates it to the mass of the object and the period of oscillation. In this problem, the mass of the object is 7.00 kg, and the period of oscillation is 2.60 s. Substituting these values into the formula yields a force constant of 104.18 N/m. The force constant is a measure of how stiff the spring is, and it relates the force exerted by the spring to the displacement of the object from its equilibrium position.

The force constant of a spring can be calculated using the formula:

k = (4π²m) / T²

where k is the force constant of the spring, m is the mass of the object attached to the spring, and T is the period of oscillation.

In this problem, we are given:

m = 7.00 kg (the mass of the object)

T = 2.60 s (the period of oscillation)

Substituting the given values into the formula, we get:

k = (4π² * 7.00 kg) / (2.60 s)²

= 104.18 N/m

Therefore, the force constant of the spring is 104.18 N/m.

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The complete question is:

A 7.00−kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 2.60 s. Find the force constant of the spring.

The equation of the motion of this object is  \(\mathbf{y = 3 cos \Big [ \dfrac{ \pi}{2}(t -2) \Big ] }\)

An oscillation in a waveform is a periodic set of vibrations an object in the medium produces.

From  the parameters given:

The angular velocity can be computed by using the formula:

\(\mathbf{\omega = \dfrac{2 \pi}{T}}\)

\(\mathbf{\omega = \dfrac{2 \pi}{4.0}}\)

The amplitude A from the oscillation is:

\(\mathbf{\Big|A \Big| = \Big|\dfrac{3 - (-3)}{2} \Big| = 3}\)

It implies that the vertical shift (D) is:

\(\mathbf{\dfrac{minimum + maximum}{2} = \dfrac{3+(-3)}{2} = 0}\)

However, by using the formula of the general form of a sinusoidal equation, we have:

\(\mathbf{y = A cos \Big [ \dfrac{2 \pi}{4}(t -2) \Big ] + 0}\)

\(\mathbf{y = 3 cos \Big [ \dfrac{ \pi}{2}(t -2) \Big ] }\)

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The graph of the planet's gravitational potential energy as it completes one orbit would be in the shape of a curve, with the highest point corresponding to the farthest distance from the center of the gravitational field and the lowest point corresponding to the closest distance.

The gravitational potential energy of a planet is directly related to its distance from the center of the gravitational field. As the planet moves closer to the center, its gravitational potential energy decreases, and as it moves farther away, the potential energy increases. Therefore, the graph would show a curve that starts at a maximum point when the planet is at its farthest distance from the gravitational source (e.g., the Sun), and gradually decreases as the planet moves closer during its orbit.

At the point closest to the gravitational source, the graph would reach its minimum value, representing the lowest gravitational potential energy. This occurs when the planet is at its perihelion (closest point to the Sun) in the case of a solar system orbit. From there, the graph would again rise as the planet moves away from the gravitational source and reaches its maximum potential energy at the farthest point of the orbit, known as the aphelion.

The specific shape of the graph will depend on the eccentricity of the planet's orbit. For a circular orbit, the graph would form a symmetrical curve, whereas for an elliptical orbit, the curve would be elongated and asymmetrical.

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Answer:

The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m

Explanation:

Given that,

First charge = 40 μC

Second charge = 80 μC

Distance between the two particles = 2.0 m

Kinetic energy = 16 J

We need to calculate the distance separating the two particles when the moving particle is momentarily stopped

Using conservation of energy

\(K.E+\dfrac{kq_{1}q_{2}}{d}=\dfrac{kq_{1}q_{2}}{x}+K.E\)

Put the value into the formula

\(16+\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{2}=\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{x}+0\)

\(16+14.4=\dfrac{28.8}{x}\)

\(30.4x=28.8\)

\(x=\dfrac{28.8}{30.4}\)

\(x=0.947\ m\)

Hence, The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m

The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m.

The given parameters;

The kinetic energy is zero at the instant the moving particle is momentarily stopped.

The distance separating the two particles when the moving particle is momentarily stopped is calculated as follows;

\(K.E_1 \ + W_1 = K.E_2 + W_2\\\\K.E_1 + \frac{kq_1q_2}{x_1} = K.E_2 + \frac{kq_1q_2}{x_2} \\\\16 \ + \ \frac{(9\times 10^9)\times (40\times 10^{-6})\times (80\times 10^{-6})}{2} = 0 \ + \ \frac{(9\times 10^9)\times (40\times 10^{-6})\times (80\times 10^{-6})}{x_2} \\\\30.4= \frac{28.8}{x_2} \\\\x_2 = \frac{28.8}{30.4} \\\\x_2 = 0.947 \ m\)

Thus, the distance separating the two particles when the moving particle is momentarily stopped is 0.947 m.

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Answer:

The answer is B

Explanation:

Only things that are moving have kinetic energy, therefore B is the only correct answer

Uniform circular motion is a motion that is uniform in terms of speed and circular in terms of its path.

Uniform means constant speed and its path is circular.

Thus, the statement is true.

The moment of inertia about the rotation axis for the given playground toy, with two children sitting opposite each other, is approximately 145.35 kg·m².

To determine the moment of inertia about the rotation axis for the given playground toy, we need to consider the contributions from the seats and the two children.

Given:

Mass of each seat = 6.4 kg

Length of the rods (r) = 1.5 m

Mass of the first child (m₁)= 16 kg

Mass of the second child (m₂) = 23 kg

The moment of inertia of each seat can be calculated using the formula for the moment of inertia of a point mass about an axis:

\(I_{seat} = m_{seat times} r^2\)

For each seat, the moment of inertia is:

\(I_{seat} = 6.4 kg times (1.5 m)^2= 14.4 kg\cdot m^2\)

Now, to calculate the moment of inertia contributed by the children, we need to consider that the children are located opposite each other. Assuming the axis of rotation passes through the center of mass of the children-seats system, the moment of inertia for each child is:

\(I_{child} = m_{child times} r^2\)

For the first child (m₁):

\(I_1 = 16 kg times (1.5 m)^2 = 36 kgm^2\)

For the second child (m₂):

\(I_2 = 23 kg times (1.5 m)^2 = 51.75 kgm^2\)

Finally, we can calculate the total moment of inertia by summing the contributions from the seats and the children:

Total moment of inertia =\(4 times I_{seat} + I_1 + I_2\)

= \(4 times (14.4 kgm^2) + 36 kgm^2 + 51.75 kgm^2\)

= \(57.6 kgm^2 + 36 kgm^2 + 51.75 kgm^2\)

= \(145.35 kgm^2\)

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When a 1.5 kg mass is hung from the spring, the spring will stretch approximately 9.6 cm.

To determine how much the spring stretches when a 1.5 kg mass is hung from it, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it.

The formula for Hooke's Law is:

F = k.x,

where F is the force applied to the spring, k is the spring constant, and x is the extension or stretch of the spring.

In this case, we know that the spring stretches 6.4 cm (or 0.064 m) when a 1.0 kg mass is hung from it. Let's denote this extension as x1 and the mass as m₁:

x₁= 0.064 m

m₁ = 1.0 kg

Now, we need to find the stretch of the spring when a 1.5 kg mass is hung from it. Let's denote this extension as x₂ and the mass as m₂:

m₂ = 1.5 kg

Since the force applied to the spring is proportional to the stretch, we can set up the following ratio:

F₁ / F₂ = x₁ / x₂,

where F₁ is the force when the 1.0 kg mass is hung, and F₂ is the force when the 1.5 kg mass is hung.

Since the force is proportional to mass (F = m x g, where g is the acceleration due to gravity), we can rewrite the ratio as:

(m₁  x g) / (m₂ x g) = x₁ / x₂,

Cancelling out the g (acceleration due to gravity) and substituting the known values:

(1.0 kg) / (1.5 kg) = 0.064 m / x₂.

Now we can solve for x₂:

x₂ = (0.064 m) x (1.5 kg) / (1.0 kg)

x₂ = 0.096 m or 9.6 cm.

Therefore, when a 1.5 kg mass is hung from the spring, the spring will stretch approximately 9.6 cm.

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Answer:

20000000000000hz only for this

The tension in the string supporting the rod and the attached weights is 7.11 N.

The tension in the string supporting the rod and the attached weights is the sum of the weights supported by the strings.

T = (m1 + m2)g

where;

0.4lb + 1.2lb = 1.6lb = 0.725 kg

T = 0.725 x 9.8

T = 7.11 N

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b. Farthest from the axis of rotation.

Torque is the rotational equivalent of force and depends on both the magnitude of the force and its distance from the axis of rotation. The torque (τ) can be calculated using the formula:

Torque = Force × Distance.

The greater the distance between the force and the axis of rotation, the greater the torque produced. This is because the lever arm acts as a moment arm, and the perpendicular distance from the axis of rotation to the line of action of the force determines the lever arm's effectiveness in generating torque.

By applying the force farthest from the axis of rotation, the lever arm's effective length is maximized, resulting in the highest torque. Therefore, option b, farthest from the axis of rotation, is the correct choice for producing the most torque.

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Answer:

\(16 * 10^{-7}\) F

Explanation:

The ability of a capacitor to store an electric charge (Q) per unit of voltage (V) across its plates of a capacitor is called capacitance. SI unit of capacitance is Farad.

It can be measured by using the formula C=Q/V.

Q = 4.0 x 10^-10 C

V = 250.0 V

C = Q / V

   = \(\frac{4.0 * 10^{-10} }{250}\)

   = \(16 * 10^{-7}\) F

Hence, the capacitance is \(16 * 10^{-7}\) F.

Answer:

Explanation:

1.6 X 10^-12 F

Answer:

B. They travel at different speeds through the Earth.

Explanation:

Answer:

true

Explanation:

Because the more mass you have, the faster the speed and the greater its potential energy, which is a storage energy.

Answer:

15 meters/second

Explanation:

Mortgage company would be the least likely place for Tom and Mary to obtain a loan.

A mortgage is a loan used to finance the acquisition or upkeep of real estate, vacant land, or other kinds of rental properties. The borrower and lender both agree to repay the loan over a period of time, typically in a series of consistent instalments that are separated into principal and interest. For loans, the property acts as insurance.

Borrowers should submit loan applications to their preferred lenders and meet a number of requirements, such as minimum credit ratings and prepayment history.

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Kim's policy remained in force for a certain number of days even though she forgot to pay the premium. the provision that allows this is called grace period provision

If the policyowner fails to make the premium payments due to some issue or reason, the insurance company will not immediately cancel the policy , whereas the company will wait for few days if no action seen than only they will take action on that account .

The grace period provision allots a specifically designated amount of time in which the policyowner has to make the required premium payments after the stipulated due date.

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Answer:

Explanation:

We are given the following the following

Initial speed u = 0.5m/s

initial distance S1 = 15cm = 0.15m

Final distance = 0m

Required

Final speed v

Using the equation of motion;

v² = u²+2gS

v² = 0.5²+2(9.8)(0.15)

v² = 0.25+2.94

v² = 3.19

v = √3.19

v = 1.786m/s

Hence  its speed when it reaches the bottom is 1.786m/s

Answer:

Req= 220 ohm

Explanation:

In series combination just add the resistances.

Req=R1+Req(23)

Req=100+120

Req= 220 ohm

Answer:

A increases

Explanation:

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The bouncing of the ball backward is a result of the conservation of energy and momentum. When the ball hits the target, it exerts a force on the target, which causes the target to move. When the ball hits the target, it experiences an equal and opposite reaction force, which causes it to bounce backward.

As a result, the ball bounces backward due to the reaction force. This reaction force is equal in magnitude but opposite in direction to the force that the ball exerts on the target. Therefore, when the ball hits the target, it experiences an equal and opposite reaction force, which causes it to bounce backward.

The bouncing of the ball backward is a result of the conservation of energy and momentum. When the ball hits the target, it exerts a force on the target, which causes the target to move. However, this force is also exerted on the ball, and as a result, the ball also moves.

This is because the ball and the target are interacting with each other and exchanging energy and momentum. In order to understand this phenomenon more clearly, it is helpful to think about the properties of the ball and the target.

The ball is made of rubber, which means that it has a high coefficient of restitution. This means that when the ball hits a surface, it bounces back with a lot of energy.The target, on the other hand, is usually made of a less elastic material such as wood or metal. This means that it does not absorb the energy of the ball as well as a rubber ball would. As a result, when the ball hits the target, it bounces back with more force than it would if it hit a rubber surface.

This is why the ball bounces backward when it hits the target. The force that the ball exerts on the target is equal and opposite to the force that the target exerts on the ball. Therefore, when the ball hits the target, it experiences an equal and opposite reaction force, which causes it to bounce backward.

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The momentum flow rate through the pipe carrying liquid ammonia is 1 × \(10^6\) kg·m/s.

The flow rate of momentum (Ṁ) through the pipe can be calculated by multiplying the mass flow rate (ṁ) by the velocity (v). The speed can be determined using the equation v = ṁ / (ρA), where ρ is the density of the liquid ammonia and A is the pipe's cross-sectional area.

Given:

ṁ = 100 kg/s

A = 0.01 m²

Assuming the density (ρ) of liquid ammonia is 700 kg/m³, we can calculate the velocity (v):

v = ṁ / (ρA)

v = 100 kg/s / (700 kg/m³ × 0.01 m²)

v = 10000 m/s

Now, we can calculate the flow rate of momentum (Ṁ):

Ṁ = ṁv

Ṁ = 100 kg/s × 10000 m/s

Ṁ = 1 × \(10^6\) kg·m/s

Therefore, the momentum flow rate through the pipe is 1 × \(10^6\) kg·m/s.

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If you went to a fireworks show in Atlanta you would see the fireworks explode before you heard them go BOOM. However if astronauts are

watching the same fireworks show from space, they would see them explode, but never hear them because  Sound waves travel slower than light waves and they cannot travel through a vacuum. Hence option C is correct.

Sound waves are a form of energy transmission method that uses adiabatic loading and unloading to move across a material. Acoustic pressure, particle velocity, particle displacement, and acoustic intensity are all important parameters for defining acoustic waves. Acoustic waves have a particular acoustic velocity that relies on the medium through which they move. Acoustic waves include audible sound from a speaker (waves that travel at the speed of sound through air), seismic waves (ground vibrations that travel through the earth), and ultrasound used for medical imaging (waves that travel through the body). Sound waves cannot travel through vacuum.

Hence option C is correct.

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After t = 4 seconds, ascertain the baseball's vx and vy horizontal and vertical speeds. Only the pressure applied on the object's vertical component—gravity—is recognized by us. This indicates that the perpendicular speed is unchanged.

A planet and maybe other bodies pull objects toward their centers through their gravitational force. The gravitational attraction of the sun keeps all of the planets in their orbit around it.

The size of an object and its distance from other things both have an impact on gravity. The mass of a thing is a gauge for its matter content. A heavier object will fall to the ground more quickly than a lighter one. Gravitational pull weakens when the separation between two objects grows.

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The resistance of the electric heating element can be determined using Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage across the element is given as 232 volts and the current flowing through it is 19 amperes. The correct option is 12 ohms.

Applying Ohm's law, we calculate the resistance as:

R = V / I

R = 232 volts / 19 amperes

R ≈ 12 ohms

Therefore, the resistance of the element is approximately 12 ohms. This means that for every ampere of current flowing through the element, there is a voltage drop of 12 volts. The resistance value of 12 ohms indicates that the element has a moderate level of electrical resistance, which allows it to generate heat efficiently when current flows through it. Thus, the correct option is 12 ohms.

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Answer:

1. sand and water

2. suspension

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The integral of 1/(1 + x²) is (1/2)ln|1 + x²| + C where C is the constant of integration.

Integration is a mathematical process of finding the antiderivative of a function. To integrate the given expression 1/(1 + x²), we will use the substitution method.

Let u = 1 + x², du/dx = 2x dx, then dx = du/2x and the integral becomes:

∫1/(1 + x²) dx = ∫1/u * (1/2x) du= (1/2)∫1/u du

The antiderivative of 1/u is ln|u| + C, where C is the constant of integration.

Therefore, the final solution of the integral is (1/2)ln|1 + x²| + C.

Let us work through the steps:

Step 1:Let u = 1 + x² and then differentiate both sides with respect to x to obtain du/dx. du/dx = 2x

Substitute 2x dx = du into the integral ∫1/(1 + x²) dx to get the integral in terms of u:∫1/u * (1/2x) du = (1/2) ∫1/u du

Step 2:Calculate the antiderivative of 1/u, which is ln|u|. Thus, the final solution is (1/2)ln|1 + x²| + C, where C is the constant of integration. The constant C will vary depending on the initial conditions of the problem.

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Answer: The mutual speed of the 95-kg halfback and the 85-kg cornerback immediately after the tackle is 9.6 m/s.

The mutual speed of the 95-kg halfback and the 85-kg cornerback immediately after the tackle can be calculated using the equation of conservation of momentum.

Momentum = Mass x Velocity.

Thus, the total momentum before the tackle = (95 kg x 4.1 m/s) + (85 kg x 5.5 m/s).

The total momentum after the tackle = (95 kg x v1) + (85 kg x v2).

Here, v1 and v2 are the velocities of the halfback and cornerback respectively. To solve for the mutual speed of the two players after the tackle, we can set the two equations of momentum equal to each other and solve for v1 and v2.

v1 + v2 = (4.1 m/s + 5.5 m/s)

v1 + v2 = 9.6 m/s

Therefore, the mutual speed of the 95-kg halfback and the 85-kg cornerback immediately after the tackle is 9.6 m/s.

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Net force on center particle = \(-3.26 * 10^{-3} N\), Net force on left particle = +1.28 x \(10^{-3} N\)

Coulomb's law states that the force between two point charges is given by: \(F = kq1q2/r^2\)

We can calculate the force between the center particle and the left particle, and the force between the center particle and the right particle:

F_left-center =\(kq\_leftq\_center/r\_left\ -center^2\)

\(F\_left-center = (9.0 * 10^9 N*m^2/C^2) * (-55 \mu C) * (+45 \mu C) / (0.72 m)^2\)

F_left-center = \(-1.28 * 10^{-3} N\)

F_center-right = \((9.0 * 10^9 N*m^2/C^2) * (+45 \mu C) * (-78 \mu C) / (0.72 m)^2\)

F_center-right = \(-1.98 * 10^{-3} N\)

To calculate the net force on the center particle:

Net force on center particle =\(F\_left-center + F\_center-right\)

Net force on center particle = \((-1.28 * 10^{-3} N) + (-1.98 * 10^{-3} N)\)

Net force on center particle = \(-3.26 * 10^{-3} N\)

To calculate the net force on the left particle:

Net force on left particle = - F_left-center

Net force on left particle = - \((-1.28 * 10^{-3 }N)\)

Net force on left particle = +1.28 x \(10^{-3} N\)

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