A 2-kg mass hangs above a table on a 1-m string. The mass just clears the table. Use the idea of conservation of energy to describe how the pendulum can be used to move a 1-kg book across the table.

Answer:

I believe it's a Seismograph

Explanation:

Answer:

capacitive reactance equals inductive reactance

Solution :

Given :

Diameter, D = 30 m

∴ Radius, R = 15 m

Angular acceleration, α = 0.044 \($rad/s^2$\)

a). Velocity of the rider just as he completes a quarter of a turn is :

\($\omega_i = 0 \ rad/s$\)

\($\theta = \frac{\pi}{2}\ rad$\)

V = R ω

∴\($\omega^2_f=\omega^2_i + 2 \alpha \theta$\)

 \($\omega^2_f=0+ 2 \times 0.04 \times \frac{\pi}{2}$\)

\($\omega^2_f=0.1256$\)

\($\omega = \sqrt{0.1256}$\)

   \($=0.354 \ rad/s$\)

∴ \($V=R \omega_f$\)

     \($= 15 \times 0.354$\)

    = 5.32 m/s up

b). Tangential acceleration

   \($a_T= \alpha R$\)

        = 0.04 x 15

       \($= 0.600 \ m/s^2$\)  up

Radial acceleration,

\($a_r=\frac{V^2}{R}$\)

   \($=\frac{(5.32)^2}{15}$\)

   \($= 1.88 \ m/s^2$\) towards center.

c). Final angular velocity

   given : \($V_0 = 6 \ m/s$\)

              \($\omega_i = 0.354 \ rad/s$\)

              \($\alpha = 0.0400 \ rad/s^2$\)

\($\omega_f = \frac{6}{15} \ rad/s$\)

\($\omega^2_f=\omega^2_i + 2 \alpha \theta$\)

\($\left(\frac{6}{15}\right)^2=(0.354)^2 + 2 \times 0.04 \times \theta$\)

\($\theta = \frac{0.16-0.1256}{2 \times 0.04}$\)

  = 0.43 rad

or \($\theta = 0.43 \times \frac{180}{\pi}$\)

      \($24.6^\circ$\)

Answer:

Names of the Equations of Motion

1 The First Equation of Motion is, v=u+at is known as velocity – time relation. 2 The Second Equation of Motion is, s=ut+12at2 is known as position – time relation.

Answer:

conduction

Explanation:

Answer:

I think it is O only nonliving

Given :

Charge on human 1 , \(q_1=2\ C\) .

Charge on human 2 , \(q_2=-2\ C\) .

Force of attraction between them , F = 650 N .

To Find :

Force between them , F =- 650 N .( force between opposite charge is -ve )

Solution :

We know , electrostatics force between two charge is given by :

\(F=\dfrac{kq_1q_2}{r^2}\)

Here k is universal gravitational constant , \(k=6.67408 \times 10^{-11}\ m^3\ kg^{-1}\ s^{-2}\) .

Putting all given value in above equation we get :

\(-650=\dfrac{6.67408 \times 10^{-11}\times 2\times (-2)}{r^2}\\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times 2\times 2}{650}}\\\\r=6.40\times 10^{-7}\ m\)

Hence , this is the required solution .

If we rank these magnets from the strongest to the weakest magnetic field the correct order is 4, 3, 2, 1.

The magnetic field and radius are related in the context of a charged particle moving in a circular path under the influence of a magnetic field. When a charged particle moves in a circular path under the influence of a magnetic field, the force on the particle is directed toward the center of the circle. In this force, the radius can be expressed as r = mv / Bq.

This equation shows that the radius of the circular path is directly proportional to the velocity of the particle, and inversely proportional to the magnetic field strength and the charge of the particle.

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Answer:

W = F • ∆x

so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)

Answer:

192.6N

Explanation:

Let's consider the forces acting on the beam:

Weight of the beam (W): It acts vertically downward and has a magnitude of W = mass * gravitational acceleration = 39.4 kg * 9.8 m/s^2.

Force exerted by the link on the beam (F_link): It acts at an angle of 90 degrees with respect to the beam and has two components: the vertical component and the horizontal component.

Tension in the cable (T): It supports the far end of the beam and acts at an angle of 90 degrees with respect to the beam. Since the angle between the beam and the cable is 90 degrees, the tension in the cable only has a vertical component.

Let's break down the forces acting on the beam:

Vertical forces:

W (weight of the beam) - T (vertical component of tension) = 0

T = W

Horizontal forces:

F_link (horizontal component of the force exerted by the link) = ?

To find the magnitude of the horizontal component of the force exerted by the link on the beam (F_link), we need to consider the equilibrium of forces in the horizontal direction.

Since the beam is inclined at an angle of θ = 33.1 degrees with respect to the horizontal, the horizontal equilibrium equation can be written as:

F_link = W * sin(θ)

Let's substitute the given values:

W = 39.4 kg * 9.8 m/s^2

θ = 33.1 degrees

F_link ≈ (39.4 kg * 9.8 m/s^2) * sin(33.1 degrees)

Using a calculator, we find that the magnitude of the horizontal component of the force exerted by the link on the beam (F_link) is approximately 192.6 N.

Answer:

Explanation:physical

Answer:

mathematical

Explanation:

Answer: D) dedicate as many hours as needed to the work.

Explanation:

edge 2020

If the entrepreneur puts in dedication to duty, responsibility and long hours of work, the new business is more likely to survive.

A business refers to any venture that is set up with the intention of making profit. The proprietors of new businesses are called entrepreneurs. They are saddled with the responsibility of propagating the activities of the new business.

The entrepreneur must be ready to put in:

If the entrepreneur does all these, his/her new business is more likely to survive than ever.

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Because the mass and displacement are already given in Kg and m, respectively, in the first part of your question, there is no need to convert them. However, in the second part of your question, you must use the given equation to calculate the spring constant.

if the table data is given in grams and  cm you have to convert it using the following conversion,

1. To convert grams to kilograms, we divide the mass values by 1000.

2. To convert centimeters to meters, we divide the displacement values by 100.

But here in the given table it's already given the mass in kg and the displacement in meters (m). so no need to convert it.

Now comes the second part of your question,

To calculate the spring constants for the given data, we can use the equation:

k = -mg/Δx

where:

k is the spring constant (in N/m),

m is the mass (in kg), and

Δx is the displacement of the spring (in m).

Let's calculate the spring constants using the provided data:

Mass (kg): 0.05  0.1  0.2  0.3  0.4  0.5  0.6

Displacement of Spring (m): 0.012  0.027  0.065  0.1  0.135  0.17  0.199

Using the equation

k = -mg/Δx,

we can calculate the spring constant for each data point:

For the first data point (m = 0.05 kg, Δx = 0.012 m):

k = -0.05 kg * 9.8 m/s² / 0.012 m

k ≈ -40.833 N/m

Similarly, we can calculate the spring constants for the other data points:

For the mass of 0.05 kg, the spring constant is approximately -40.833 N/m.

For the mass of 0.1 kg, the spring constant is approximately -18.519 N/m.

For the mass of 0.2 kg, the spring constant is approximately -6.154 N/m.

For the mass of 0.3 kg, the spring constant is approximately -3.267 N/m.

For the mass of 0.4 kg, the spring constant is approximately -2.222 N/m.

For the mass of 0.5 kg, the spring constant is approximately -1.716 N/m.

For the mass of 0.6 kg, the spring constant is approximately -1.449 N/m.

Therefore, In the first part of the question, there is no need to convert the mass into kg and the displacement cm into m because it is already given in kg and m respectively, and in the second part question you have to calculate the spring constant using the given equation.

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Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 70 × 0.85

We have the final answer as

Hope this helps you

Answer:

We can use the formula:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred from the gold bar to the water is equal to the heat transferred from the water to the gold bar, since they reach thermal equilibrium. Therefore:

q_gold = q_water

We can solve for the temperature change of the gold bar:

q_gold = mcΔT_gold

q_water = mcΔT_water

Since the heat transferred is equal:

mcΔT_gold = mcΔT_water

Rearranging and solving for ΔT_gold:

ΔT_gold = ΔT_water(m_water/m_gold)

ΔT_water is the temperature change of the water, which is 17°C. m_water is 0.220 kg, and m_gold is 3 kg. c_gold is given as 129 J/kg K.

ΔT_gold = 17°C(0.220 kg/3 kg)(1/129 J/kg K) = 0.025°C

Therefore, the temperature change of the gold bar is 0.025°C when it is placed into 0.220 kg of water and thermal equilibrium is reached.

The assertion that ocean waves move on the water's surface is accurate. Therefore, choice A is right.

A wind wave, water wave, or tornado water wave is a surface wave that develops on a body of water's free surface as a result of the air flowing over the ocean's surface. The fetch is the distance at which two objects come into touch when facing the wind.

The wind is the most frequent source of waves. The friction between the wind and the surface of the water produces wind-driven waves, also known as surface waves. A wave crest is produced when wind continuously disturbs the water's surface in an ocean or lake.

The statements about ocean waves a true will be they travel on the surface of the water.

Thus, option A is correct.

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Answer:

We answer that before but I forgot

Answer:

the deeper into the ocean you go, the more pressure is exerted on you

Explanation:

Answer:

√2

Explanation:

From the question, we're given that the

Acceleration of the leaf is 1 m/s²

Change in displacement of the leaf is 1 m/s.

Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest

Now, to solve this, we don't the equation of motion to ur

S = ut + 1/2at², substituting the whole parameters, we then have

1 = 0 * t + 1/2 * 1 * t²

1 = 1/2 * t²

t²/2 = 1

t² = 2

t = √2 seconds

Therefore the time it takes the leaf to dislodge is 2 seconds

Energy effectiveness would be a term that refers to the proportion of input power over output. Power generation, as well as simply energy utilization, is the process of reducing the amount of energy used to produce goods and services.

Considering that,

A cookie contains 54.0 kcal of energy. An athlete utilizes the 54.0 kcal inside this cookie from input energy.

The following diagram illustrates the relationship among input as well as output energy:

Efficiency = output energy / input energy...(i)

Output energy = efficiency × input energy

By using equation (i)

⇒ output energy = 0.25 × 54 kcal = 13.5 kcal.

The lifting exercise has been performed n times for the output energy.

In terms of potential energy, such output energy could be written as follows:

Mass × gravity ×height.

So, energy per repetition = mgh = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J =  0.059 kcal.

So, Count of repetitions = sim of output energy / energy per repetition..(ii)

By using equation (ii)

Count of repetitions = 13.5 kcal / 0.059 kcal =229 repetitions.  

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Answer:

       y =\(\frac{1.22L}{D}\) \(\sqrt{\frac{h^2 m}{2eV} }\)

Explanation:

Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation

             λ= \(\frac{h}{p} = \frac{h}{mv}\)

Let's use conservation of energy for speed

starting point

             Em₀ = U = e V

final point

             Em_f = K = ½ m v²

             Em₀ = Em_f

             eV = ½ m v²

             v =\(\sqrt{\frac{2eV}{m} }\)

we substitute

             λ=  \(\sqrt{ \frac{h^2 m}{2eV}}\)

the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum

            a sin θ = m λ

first zero occurs at m = 1, also these experiments are performed at very small angles

            sin θ = θ

            θ = λ / a

This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain

           θ = 1.22 λ / D

where D is the diameter of the opening

we substitute

          θ = \(\frac{1.22}{D}\)   \sqrt{ \frac{h^2 m}{2eV}}  

this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians

            θ = y / L

when substituting

where y is the minimum distance that can be resolved for this acceleration voltage

            y =\(\frac{1.22L}{D}\) \(\sqrt{\frac{h^2 m}{2eV} }\)

To find the mass in kilograms, you need to know the object's weight in newtons and the acceleration due to gravity. The formula for finding mass is mass = weight / acceleration due to gravity. So if you have an object with a weight of 100 N and the acceleration due to gravity is 9.8 m/s^2, the mass would be 10.204 kg.

The mass of the block is 0.025 kg or 25 g, when the spring has k = 28 N/m, and compresses 0.11 m before bringing the block to rest.

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Phobos 1 was an uncrewed Soviet space probe of the Phobos Program launched from the Baikonur launch facility on 7 July 1988. Its intended mission was to explore Mars and its moon Phobos and Deimos.

Mars Pathfinder was the first-ever robotic rover to the surface of the red planet Mars. The Opportunity was the second of the two rovers launched in 2003. There are lots of asteroids in our solar system. Most of them are located in the main asteroid belt- a region between the orbits of Mars and Jupiter. The Trojan asteroids are affected by Jupiter's gravity.

The lowest gravitational potential energy of a system of particles is achieved when they form a sphere as opposed to a disk. This is why many smaller objects like asteroids and even Mars's two moons are shaped like lumpy boulders and so is our moon also spherical in shape.

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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