a 0.1 kg object oscillates as a simple harmonic motion along x axis with a frequency f=3.185 hz. At a position x1 , the object has a kinetic energy of o.7 j and a potential energy 0.3 J.The amplitude of oscillation A is:​

Answer:

Grams

Explanation:

Because that is the definition

Answer:

65 grams

Explanation:

The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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The length of the wire is 1.12m

What is Resistance ?

Resistance is opposition to the current flowing in the circuit.

Given ,

D= 0.6m

R= 0.6/2m

R= 28Ω

rho = 1Ωm

R= rho ×l/A

where R is resistance , l is length ,A is area and rho is resistivity

By applying value and Solving we get,

l= 1.12m

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The equation for the magnitude of the system's acceleration is a = g + (2M/m), where g is the acceleration due to gravity and M and m are the masses of the two objects.

Acceleration is the rate of change of velocity in an object over time. It is the rate of change of the speed of an object, and is defined as the change in velocity divided by the change in time. In other words, it is the rate at which an object's speed changes. Acceleration can be caused by a change in the speed of an object, or by a change in its direction. It is usually measured in meters per second squared (m/s2). Acceleration can be either positive or negative, depending on the direction of the change in velocity.

a = g + (2M/m)

a = 9.81 m/s^2 + (2(8 kg)/(3 kg)) = 13.87 m/s^2

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Answer:

Explanation:

4

The magnitude and direction of the resultant force on q3 is;

F_net = 32.2 N in the direction of q3

We are given;

q1 = +2 μC = 2 × 10^(-6) C

q2 = -3 μC = -3 × 10^(-6) C

q3 = +4 μC = 4 × 10^(-6) C

Now,from the given image, charges q1 and q2 attract charge q3.

Formula for force on a charge is;

F = (kq1 * q2)/r²

Where;

k = 9 × 10^(9) N.m²/C²

Thus;

F_y = kq1*q3/(r1)²

We have;

q1 = 2 × 10^(-6) C

r1 = 5 cm = 0.05 m

q3 = 4 × 10^(-6) C

Thus;

F_y = (9 × 10^(9) × 2 × 10^(-6) × 4 × 10^(-6))/(0.05^(2))

F_y = 28.8 N

Similarly;

F_x = kq2*q3/(r2)²

r2 = 5/(tan 30)

r2 = 8.66 cm = 0.0866 m

F_x = (9 × 10^(9) × -3 × 10^(-6) × 4 × 10^(-6))/(0.0866^(2))

F_x = -14.4 N

Thus, magnitude of resultant force is;

F_net = √((28.8)² + (-14.4)²)

F_net = √1,036.8

F_net = 32.2 N

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One end of an object has a definite positive charge, which is an instance of inverse-square or a positive or negative state that reacts to an electric field.

A positive or negative energy is what?

A fundamental characteristic of some of the constituent particles, of which all matter is made, is electric charge.

Electric charges come in two different flavours.

(+) positive charges

adverse charge (-)

A negative charge is present when there are more electrons than protons in an item.

When an atom has more protons than electrons, it has a positive charge.

Protons have a positive charge, whereas electrons have a negative charge.

The coulomb is the unit of charge (C).

Similar charges repel one another.

Similar charges are drawn to one another.

The positive charge produces an electric flux.

The negative charge is the point where electric flux ends.

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The value of the charge on each particle is \(1.05 x 10^-8 C\).

Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,

\(F = k(q1q2)/r^2\)

where F is the force of attraction or repulsion between the two charged particles,\(q1\) and \(q2\) are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of \(9.0 x 10^9 Nm^2/C^2.\)

In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,

\(0.5 N = k(q^2)/(0.1 m)^2\)

Solving for q, we get:

\(q = \sqrt{[(0.5 N)(0.1 m)^2/k]}\)

\(q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}\)

\(q = 1.05 x 10^-8 C\)

Therefore, the value of the charge on each particle is \(1.05 x 10^-8 C.\)

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The strength of gravity on the moon is approximately 1.6 J/kg.

The change in gravitational potential energy (GPE) is given by the equation:

ΔGPE = m * g * Δh

where ΔGPE is the change in gravitational potential energy, m is the mass of the object, g is the strength of gravity, and Δh is the change in height.

In this case, the astronaut's GPE increases by 880 J as he climbs up the ladder by 5 m. We can rewrite the equation as:

880 J = (110 kg) * g * (5 m)

To find the strength of gravity on the moon (g), we can rearrange the equation:

g = 880 J / (110 kg * 5 m)

g = 1.6 J/kg

Therefore, the strength of gravity on the moon is approximately 1.6 J/kg.

It's important to note that the value of gravity on the moon is significantly lower than that on Earth. The moon has about one-sixth the gravity of Earth, which means objects weigh less on the moon compared to Earth. This lower gravity is due to the moon's smaller mass and smaller radius compared to Earth.

As a result, astronauts experience a different gravitational environment on the moon, which affects their movements and the energy required to perform tasks such as climbing.

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Since focal length cannot be negative for a converging lens, we take the positive value: f ≈ 41 cm Option C

To determine the focal length of the lens, we can use the lens formula, which relates the object distance (u), image distance (v), and focal length (f) of a lens. The lens formula is given by:

1/f = 1/v - 1/u

In this case, the object distance (u) is 25 cm and the image distance (v) is 64 cm. We can substitute these values into the lens formula to solve for the focal length:

1/f = 1/v - 1/u

1/f = 1/64 cm - 1/25 cm

To simplify the equation, we can find a common denominator:

1/f = (25 - 64) / (64 * 25)

1/f = -39 / (64 * 25)

Now, we can invert both sides of the equation to solve for the focal length:

f = (64 * 25) / -39

f ≈ -41.03 cm

Since focal length cannot be negative for a converging lens, we take the positive value:

f ≈ 41 cm

Therefore, the correct answer is option C) 41 cm.

It's important to note that in the lens formula, distances are measured with respect to the lens, with positive values indicating distances on the opposite side of the incident light. The negative value obtained for the focal length indicates that the lens is a converging lens, as expected. Option C

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Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\(\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s\)

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

Answer:

65 miles per hour

Explanation:

Speed= Distance/Time

Following this formula, we just need to insert the values.

The distance is 125 miles and the time is 2 hours

125=2=65

65 miles per hour

The speed of the car was exactly 125 mi per 2 hours.

In a more familiar unit,  125mi/2hr = 62.5 mi/hr.

The displacement of a 1.5 kg mass is then determined using the formula x = F/k.  stretching a spring 2 cm from its equilibrium position need twice as much effort as stretching it a distance of x

W = 1/2kx2 = 1.96 Joules.

Actually, it requires more than twice as much labour since, as the spring extends, more power is needed to do so.

The shear strength and shear modulus of a compression spring formed of music wire with a 2mm diameter are 800 MPa and 80 GPa, respectively.

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Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

\(v_{f}=v_{o}+a*t\)

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

\(v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}\)

Answer:

b lie down and rest

Explanation:

The least effective way to treat heat exhaustion is applying sunscreen. The correct option is A.

Heat exhaustion is a situation characterized by excessive sweating and a rapid pulse as a consequence of the body's excessive heat.

It is one of three heat-related symptoms and signs, the mildest being heat cramps and the most severe being heatstroke.

The main symptom of heatstroke is a core body temperature of 104 F or or above, as measured with a rectal thermometer.

Changes in state of mind or behavior. Heatstroke can cause confusion, agitation, slurred speech, irritability, delirium, seizures, and coma.

Heat exhaustion or heatstroke can occur suddenly or gradually over several hours or days.

The best way to relieve from these symptoms are lying down and rest, drinking cold drinks, as well as taking a cool shower.

Thus, the correct option is A.

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Sedimentation ponds are useful for reducing the effects of soil erosion because : ( D ) They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

Sedimentation ponds are ponds that hold sediments and suspended solids discharged by runoff waters thereby preventing the soil sediments from reaching water bodies such as ponds, lakes and rivers.

The sediment ponds hold the water while allowing the debris and solids contained in the water to settle down.

Hence we can conclude that sedimentation ponds provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

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Answer:

K = 1800 kJ

Explanation:

Given that,

The speed of the object, v = 30 m/s

Mass of the object, m = 4000 kg

We need to find the kinetic energy of the object. The formula for the kinetic energy is given by :

\(K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ\)

So, the required kinetic energy is equal to 1800 kJ.

Answer:

meter

Explanation:

Answer: The International System of Units is a system of measurement based on 7 base units

Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.

Answer:

When light gets in contact with water, it emerges into the air, and then it speeds up, which makes it look shallower than it actually is.

Explanation:

Answer:

Refraction

Explanation:

Light rays reflected from the fish are refracted at the surface of the water, but the eyes and brain trace the light rays back into the water as thought they had not refracted, but traveled away from the fish in a straight line. This effect creates a "virtual" image of the fish that appears at a shallower depth.

Answer:

he fish would travel a horizontal distance of 1.78 meters during the 2 seconds of tail motion

Explanation:

The initial horizontal velocity of the bluefish is 0.45 m/s. When it begins tail motion, it experiences an additional force due to the thrust produced by the tail. The thrust produced by the tail is 0.65 N. We can use Newton's second law to find the acceleration produced by this force:

F = ma

0.65 N = 1.7 kg * a

a = 0.38 m/s^2

This acceleration will cause the velocity of the bluefish to increase over time. The distance the fish travels during this time can be calculated using the kinematic equation:

d = vit + 1/2 at^2

where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time. Since the fish is initially coasting horizontally, its initial vertical velocity is 0 m/s. Therefore, vi = 0.45 m/s. The time interval for which the fish is tail-motoring is not given, so let's assume it is 2 seconds:

d = (0.45 m/s)(2 s) + 1/2 (0.38 m/s^2)(2 s)^2

d = 1.78 meters

Therefore, the fish would travel a horizontal distance of 1.78 meters during the 2 seconds of tail motion.

Answer: net primary productivity

Explanation: i just took my test and this was correct!

Answer:

The time when the ball strikes the ground is closest to  \(t_t = 9.4 \ s\)

Explanation:

From the question we are told that

  The time of projection is t = 0.0 s

   The  distance of the point  from the ground  is  \(d = 90 \ m\)

    The  initial velocity of the ball is  \(v _i = 36 .2 \ m/s\)

generally the time required to reach maximum height is  

      \(t_r = \frac{g}{v}\)

Where is the acceleration due to gravity  with value  \(g = 9.8 \ m/s^2\)

Substituting values

        \(t_r = \frac{36.2}{9.8}\)

        \(t_r = 3.69 s\)

when returning the time and velocity at the roof level is  t =  3.69 s and  u = 36.2 m/s this due to the fact that  air resistance is negligible

   The final velocity at which it  hit the ground is

      \(v_f^2 = u^2 + 2ag\)

So  

    \(v_f = \sqrt{ u^2 + 2gs}\)

substituting values

    \(v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}\)

     \(v_f = 55.45 \ m/s\)

The time taken for the ball to move from the roof level to the ground is  

     \(t_g = \frac{v-u}{a}\)

substituting values

    \(t_g = \frac{55.45 -36.2}{9.8}\)

     \(t_g = 1.96 \ s\)

The total time for this travel is  

    \(t_t = t_g + 2 t_r\)

     \(t_t = 1.96 + 2(3.69)\)

      \(t_t = 9.4 \ s\)

The information is missing here but data that may support the existence of life in an exoplanet may include the discovery of atmospheric biological gases.

An exoplanet is any planet that surrounds another star beyond the sun, and therefore they are discovered by analyzing other galaxies in the Universe.

The existence of life in these exoplanets can be indirectly supported by the presence of gases that are only produced by organisms, which are detected by suitable telescopes that analyze emission spectra.

In conclusion, the information is missing here but data that may support the existence of life in an exoplanet include biological gases.

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Answer:

Distance

Explanation:

When using the length of the space between two points, the distance measures the length

Answer:

D.Distance

Explanation:

The speed of the block of wood of mass 1.5 kg after the bullet of mass 22 g emerges from the block of wood is 1.69 m / s

According to law of conservation of momentum,

\(m_{1}\) \(v_{1}\) + \(m_{2}\) \(v_{2}\) = \(m_{1}\) \(v_{1}\)' + \(m_{2}\) \(v_{2}\)'

Let the bullet be object 1 and the block of wood be object 2.

\(m_{1}\) = 22 g = 0.022 kg

\(m_{2}\) = 1.5 kg

\(v_{1}\) = 250 m / s

\(v_{1}\)' = 135 m / s

\(v_{2}\) = 0

After substituting these values,

( 0.022 * 250 ) + ( 1.5 * 0 ) = ( 0.022 * 135 ) + ( 1.5 * \(v_{2}\)' )

5.5 + 0 = 2.97 + 1.5 \(v_{2}\)'

\(v_{2}\)' = 2.53 / 1.5

\(v_{2}\)' = 1.69 m / s

Therefore, the block of wood would move at 1.69 m / s after the bullet emerges.

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The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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The compressive stress in the steel column is found to be approximately 397.6 MPa.

The formula for calculating the area of a circle can be used to determine the steel column's cross-sectional area (A),

A = π*(d/2)², diameter of the column is d,

A = π*(0.4/2)²

A = 0.1257m²

The compressive stress (σ) in the column can be calculated using the formula, σ = F/A, F is the load carried by the column is F.

σ = 50 MN/0.1257m²

σ = 397.6 MPa

Therefore, the compressive stress in the steel column is approximately 397.6 MPa.

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Answer:

please i dont understand this

Answer:As relative density is the ratio of similar quantities, it has no unit.So RD remains the same in both CGS and SI unit

Explanation: i did the test