In part a, how would you have to adjust the time/div control to make a two-cycle sine wave?

When a beam of electrons passes through the same material, the electrons bounce off the gold, proving the particle nature of electrons.

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

Interference patterns which is a characteristic of waves, arise when X rays pass through a crystalline material like gold.

The X rays ionize the electrons and they jump off from the gold. This represents the particle nature of electrons.

Thus, when a beam of electrons passes through the same material, the electrons bounce off the gold, proving the particle nature of electrons.

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. Work done by a force depends upon force applied and displacement produced.

Question:

A block of mass 1kg is placed on an inclined plane at an angle of 30° to the horizontal. Calculate the frictional force, normal reaction, the coefficient of the static friction​

Answer:

Break the mg force into components in the plane of the inclined plank. Now force acting parallel is mgsin37°. Perpendicular force is mgcos37° which is equal to normal force.  

Now calculate max value of friction force which is 0.8*mg cos37°. Since its greater than pulling force mgsin37°, the body will be at rest and acceleration is zero.  

Force applied by surface is vector sum of normal force and friction force acting.

For the honeybee that circles a field looking for a flower upon which to land, we have:

a) The tangential velocity is 10.01 m/s.

b) The centripetal acceleration is 40.1 m/s².

c) The centripetal force is 1.60 N.

a) The tangential velocity can be calculated as follows:

\( v = \omega r = \frac{2\pi r}{T} \)  (1)

Where:

r: is the radius of the circle = 2.5 m

ω: is the angular velocity = 2π/T

T: is the period = 1.57 s

By entering the above values into equation (1), we have:

\( v = \frac{2\pi r}{T} = \frac{2\pi 2.5 m}{1.57 s} = 10.01 m/s \)  

Hence, the tangential velocity is 10.01 m/s.

b) The centripetal acceleration is related to the tangential velocity as follows:

\( a_{c} = \frac{v^{2}}{r} = \frac{(10.01 m/s)^{2}}{2.5 m} = 40.1 m/s^{2} \)

Therefore, the centripetal acceleration is 40.1 m/s².

c) The centripetal force is given by:

\( F = ma_{c} \)

Where:

m: is the honeybee's mass = 0.04 kg

\( F = ma_{c} = 0.04 kg*40.1 m/s^{2} = 1.60 N \)

Hence, the centripetal force is 1.60 N.

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Answer: 8.3 hours

Explanation:

time= distance divided by speed so:

150 divided by 18= 8.3 hours

Answer:

56

Explanation:

Its 56 just because

Answer:

wave w is a sound wave

wave x and y are light waves

Answer:

D

Explanation:

The magnitude of the EMF induced at the ends of the skipping rope is 3.4 × 10^-4 V. The energy stored in the magnetic field of an inductor with 8.0 mH of inductance when a steady current of 4.0 A flows through it is 64 mJ.

(a) To calculate the magnitude of the EMF induced at the ends of the skipping rope, we can use the formula ε = BLVsinθ, where ε is the induced EMF, B is the magnetic field, L is the length of the conductor, V is the velocity of the conductor, and θ is the angle between the velocity of the conductor and the magnetic field.

In this case:

B = 5.0 × 10^-5 T (given)

V = 8.2 m/s (given)

L = 12 m (given)

θ = 90° (as it is perpendicular to the magnetic field)

First, we calculate the value of B:

B = BVsinθ / L

B = (5.0 × 10^-5 T) × (8.2 m/s) × sin 90° / 12 m

B = 3.42 × 10^-6 T

Substituting the value of B in the first equation, we get:

ε = BLVsinθ

ε = (3.42 × 10^-6 T) × (12.0 m) × (8.2 m/s) × sin 90°

ε = 3.4 × 10^-4 V

Therefore, the magnitude of the EMF induced at the ends of the skipping rope is 3.4 × 10^-4 V.

(b) To calculate the energy stored in the magnetic field of an inductor, we can use the formula W = (LI^2) / 2, where W is the energy stored, L is the inductance, and I is the current flowing through the inductor.

In this case:

L = 8.0 mH

I = 4.0 A

Using the formula, we find:

W = (LI^2) / 2

W = (8.0 mH) × (4.0 A)^2 / 2

W = 64 mJ

Therefore, the energy stored in the magnetic field of an inductor with 8.0 mH of inductance when a steady current of 4.0 A flows through it is 64 mJ.

In the given problems, we calculated the magnitude of the EMF induced at the ends of the skipping rope using the magnetic field, length, velocity, and angle. We also determined the energy stored in the magnetic field of an inductor using the inductance and current.

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Answer:

a)  E_{z} = 0, b) E_{z} = 0, c)    z = 1.73 [1 + \(\sqrt{1 - \frac{4R^2}{3} }\)], d)   \(E_{max}\)Emax = 9.7  10¹⁰ N / C

Explanation:

For this exercise we use the expression

          E = k ∫ dq / r²

By applying this expression to our problem of a ring of radius R, with a perpendicular axis in the z direction, we can calculate the electric field for a small charge element

          dE = k dq / r²

In the attachment we can see a diagram of the electric field, it is observed that the fields perpendicular to the z axis cancel and the field remains in the direction of the axis

           d\(E_{z}\)= dE cos φ

we substitute

         E_{z} = k∫  \(\frac{dq}{r^2}\)  cos φ

let's write the expressions

          r² = R² + z²

          cos φ = z / r

we substitute in the integral, where we see that the load differential does not depend on the distance and the value of the total load is + Q

          E_{z} = k  \(\frac{1}{ (R^2 +z^2) } \ \frac{z}{ (R^2 + z^2)^{1/2} }\) ∫  dq

          E_{z} = k Q  \(\frac{z}{ (R2+z^2)^{3/2} }\)  

This is the expression for the electric field in the axis perpendicular to the ring, we analyze this expression to answer the questions

a) the magnitude of the field at z = 0

         E_{z} = 0

b) the magnitude of the field for z = inf

        when z »R the expression remains

          E_{z} = k \(\frac{z}{z^{3} }\) Q

          E_{z} = k Q \(\frac{1}{z^2}\)

therefore when the value of z = int the field goes to E_{z} = 0

c) In value of z for which the field is maximum.

    We have an extreme point when the first derivative is equal to zero

     \(\frac{dE_z}{dz } = k Q [ (R^2 +z^2)^{3/2} - z \ 3 \frac{z}{ (R^2 +z^2)^{1/2} } = 0\)

we solve

    (R² + z²)^{ 3/2} = 3 z² /(R² +z²) ^{1/2}

    (R² + z²)² = 3z²

    r² + z² = √3    z

    z² –1.732 z + R² = 0

we solve the quadratic equation

       z = [1.732 ± \(\sqrt{3 - 4R^2}\)]/ 2 = [1.73 ± 1.73 \(\sqrt{ 1 - \frac{4 R^2}{3} }\)   ] / 2

       z = 0.865 [1 ± \(\sqrt{1 - \frac{4R^2}{3} }\)]

Therefore there are two points where the field has an extreme point one, one is a maximum and the other a minimum, as we have already determined a minimum at z = 0 the maximum point must be

  z = 0.865 [1 + \(\sqrt{1 - \frac{4R^2}{3} }\)]

d) the value of Emax

         z₁ = 0.865 [1+\(\sqrt{1 - \frac{4 \ 0.04^2}{3} }\)) = 1.73 [1 + √0.99786]

          z₁ = 1.729 m

         z₂ = 0.865 [1 - √0.99786 ]

         z₂ = 0.0011 m≈ 0

for which the field has a maximum value substituting in equation 1

            \(E_{max} = 9 10^9 \ 9 \ \frac{1.729}{(0.04^2 + 1.729^{3/2})}\)

            \(E_{max}\) = 81 10⁹   \(\frac{1.729}{1.4408}\)

            \(E_{max}\)Emax = 9.7  10¹⁰ N / C

The kinetic energy is high because, the mass of the object is large and thus has a great momentum and its velocity low because it has a large mass.

We know that kinetic energy of an object K = 1/2mv² where m = mass of the object and v = velocity of the object and its momentum, p = mv

For a given value of kinetic energy, the mass of the object m = 2K/v². This implies that m ∝ 1/v².

So, as the mass of the object increases, its velocity decreases.

So. an object with large mass has low velocity while an object with small mass has high velocity.

Also, the kinetic energy, K = 1/2mv². So, K ∝ mv². So, if m is large and v is small, the product mv² is large and thus K = 1/2mv² is high.

So, the kinetic energy of heavy vehicles is high because, the mass of the object is large and thus has a great momentum and its velocity low because it has a large mass.

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The required total momentum of the system at this instant is calculated to be 9.16 kg m/s.

Given that,

Mass of the first cart, m₁ = 2.5 kg

Mass of the second cart, m₂ = 1.3 kg

Velocity of the first cart before collision, u₁ = 4.6 m/s

Velocity of the second cart before collision, u₂ = -1.8 m/s

The total momentum of the system before collision is calculated as follows,

P t = P₁ + P₂

P t = m₁ u₁ + m₂ u₂ = 2.5 × 4.6 + (1.3 × -1.8) = 11.5 - 2.34 = 9.16 kg m/s

Thus, total momentum of the system of the two carts at this instant is 9.16 kg m/s.

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The acceleration of the masses is 6.67 m/s², and the tension in the rope is 33.33 N.

To find the acceleration, we can use the formula for net force acting on the system, which is the difference between gravitational forces acting on the masses:
F_net = m1 * g - m2 * g,
where g is the acceleration due to gravity (9.81 m/s²).
Then, we can divide the net force by the total mass (m1 + m2) to find the acceleration: a = F_net / (m1 + m2).
In this case, a = (5 * 9.81 - 10 * 9.81) / (5 + 10) = 6.67 m/s².
To find the tension in the rope, we can use the formula T = m1 * (g - a), where T is the tension.
Plugging in the values, we get T = 5 * (9.81 - 6.67) = 33.33 N.
Thus, the acceleration of the masses is 6.67 m/s², and the tension in the rope is 33.33 N.

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Answer:

An open circuit is one where the continuity has been broken by an interruption in the path for current to flow. A closed circuit is one that is complete, with good continuity throughout

The amount of compression the femur can withstand before breaking can be calculated using the following formula:

Compression = Young's Modulus x Length / \((3 x 10^8)\)

Using the given values, the amount of compression is calculated to be 8.97 mm.

To further explore this topic, you may want to learn more about Young's Modulus, which is a measure of the stiffness of a material. You can also learn about the different ways that compression can affect bones, such as by causing them to become more brittle, or increasing the risk of fracture.

Additionally, you can learn about the different types of fracture, such as compression fractures and stress fractures, and how they are treated. Finally, you can explore the effects of different types of loading on bones, such as the impact of compressive forces and the effects of repeated loading.

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The kinetic energy at the top of the rock trajectory is 43.57 Joules

The kinetic energy formula and the procedure we will use is:

K.E. = 1/2 m (v*cosθ)²

Where:

Information of the problem:

Applying kinetic energy formula we get:

K.E. = 1/2 m (v*cosθ)²

K.E. = 1/2 * 0.145 kg (32 m/s*cos40°)²

K.E. = 1/2 * 0.145 kg *(24.513 m/s)²

K.E. = 1/2 * 0.145 kg *600.91 m²/s²

K.E. =  87.132 kg m²/s² / 2

K.E. =  43.57 Joules

It is the energy possessed by a body due to its relative motion. It is usually expressed in Joules (J).

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The power loss in a kilometer of 10-gauge copper wire, when it carries 0.95 × 102 A, is 30,800 watts. To calculate the power loss in a kilometer of 10-gauge copper wire, we need to use the formula for power loss, which is P = I^2R, where I is the current and R is the resistance.

We first need to calculate the resistance of the wire using the formula R = (ρL)/A, where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area.

The cross-sectional area of 10-gauge wire is 5.26 mm^2. The length of the wire is 1000 meters. Substituting the values in the formula, we get:

R = (1.72 x 10^-8 x 1000) / 5.26 x 10^-6 = 3.27 Ω

Now, we can calculate the power loss using the formula:

P = (0.95 x 10^2)^2 x 3.27 = 3.08 x 10^4 W

Therefore, the power loss in a kilometer of 10-gauge copper wire, when it carries 0.95 × 102 A, is 30,800 watts.

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Answer:

Explanation:

c = 180

i have same problem and it work

The angle between the directions of a and b is 43.95° (to two decimal places).To determine the angle between the directions of a and b, the dot product of the two vectors a and b must be found.

The formula for the dot product of two vectors a and b is given as follows;

a·b = |a| |b| cosθ Where,|a| is the magnitude of vector a|b| is the magnitude of vector bθ is the angle between vectors a and b Using the given values in the question, we can find the angle between the directions of a and b;

a·b = |a| |b| cosθcosθ

= (a·b) / (|a| |b|)cosθ

= (14.0) / (17.0)(13.0)cosθ

= 0.72θ

= cos⁻¹(0.72)θ = 43.95°

Therefore, the angle between the directions of a and b is 43.95° (to two decimal places).

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The angle between the directions of vectors a and b is approximately 86.8 degrees.

To find the angle between the directions of vectors a and b, we can use the dot product formula:

a · b = |a| |b| cos(θ),

where a · b is the dot product of vectors a and b, |a| and |b| are the magnitudes of vectors a and b, and θ is the angle between the two vectors.

Given:

|a| = 17.0 units,

|b| = 13.0 units,

a · b = 14.0.

Rearranging the formula, we have:

cos(θ) = (a · b) / (|a| |b|).

Substituting the given values:

cos(θ) = 14.0 / (17.0 * 13.0).

Calculating the value:

cos(θ) ≈ 0.06243.

To find the angle θ, we can take the inverse cosine (arccos) of the calculated value:

θ ≈ arccos(0.06243).

Using a calculator or trigonometric tables, we find:

θ ≈ 86.8 degrees (rounded to one decimal place).

Therefore, the angle between the directions of vectors a and b is approximately 86.8 degrees.

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Answer:

fffffff

Explanation:

Answer:

Accuracy

Explanation:

Accuracy means making measurements that are close to the value precision means making measurement that are close in value to eachother but not necessarily close to the true value.

I hope this helps! If not sorry.

Our "cosmic address" from small to large are earth, solar system, milky way galaxy, local group, local supercluster, universe (option A)

The "smallest" is earth, it is the planet where we live and the third of the eight planets in the solar system.

The solar system is a collection of celestial bodies consisting of a star called the sun and all the objects that are bound by its gravitational force. And the milky way galaxy is the galaxy that contains our solar system. It is a barred spiral galaxy 100,000-120,000 light years in diameter containing 100-400 billion stars.

The local group is the group of galaxies that includes the milky way among others. It consists of more than 54 galaxies, including dwarf galaxies.

The local supercluster is one of the largest known cosmic structures in the universe. They are large, threadlike formations, with a typical length of 50 to 80 megaparsecs h-1, which form the boundaries between the great voids in the universe.

The universe is isotropic, the distance to the edge of the observable universe is approximately the same in all directions. That is, the observable universe is a spherical (spherical) volume centered on the observer, regardless of the shape of the universe as a whole.

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Answer:

Our "cosmic address" from small to large are earth, solar system, milky way galaxy, local group, local supercluster, universe (option A)

Answer:

Because even if iron is magnetic its not a magnet it self unless rubbed a magnetic object

Answer:

Reaction is 20N.

Explanation:

In general, Newton's third law of motion explains that when a force is applied on an object, a reaction of equal magnitude applies in the reverse direction of the force.

i.e F = -R

This implies that, reaction is a force in the opposite direction to that of the force applied.

Force can also be related to the weight of an object as;

F = W = mg

Thus to prevent the 20N stone from falling, a force of 20N is applied in the opposite direction. So the reaction, R is 20N.

Drag is the force that acts against the motion of an object through a fluid. It is determined by several factors such as the frontal area, body shape, surface texture, and depth.

However, one of these factors does not affect drag, and that is depth. Depth refers to the distance of an object from the surface of a fluid. It does not play a significant role in determining drag because the force of drag is primarily caused by the resistance of the fluid molecules to the object's motion.

Therefore, a change in depth does not significantly impact the resistance of the fluid and, as a result, does not affect the drag.

Drag is determined by all of the following EXCEPT:

d. depth.

Drag is the force that opposes an object's motion through a fluid, such as air or water. It is influenced by factors such as frontal area (a), body shape (b), and surface texture (c).

Frontal area affects the amount of fluid displaced by the object, body shape determines how easily the fluid flows around the object, and surface texture influences the amount of turbulence created as the fluid moves across the object's surface.

However, depth (d) does not play a direct role in determining drag, as it refers to the vertical distance of a submerged object, which is unrelated to the resistance it encounters in moving through a fluid.

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The jet aircraft will fly for 200 minutes to travel a distance of 2000 miles with a velocity of 600 miles/hour.

The velocity of a moving object can be described as a vector measurement of the direction and rate of motion.

Velocity can be defined as a vector quantity with magnitude and direction.

v = d / t

where d is the distance,  v is the velocity, and t is the time.

Given the velocity of the jet aircraft, v = 600 miles/hour

The distance traveled by the jet aircraft, d = 2000 miles

The time is taken by jet aircraft = 2000/600 = 3.33 hr

The time of flying in minutes = 3.33 × 60 = 200 min

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Answer:

Explanation:

The time taken can be found by using the formula

\( t = \frac{d}{v} \\ \)

d is the distance

v is the velocity

From the question we have

\(t = \frac{3.5}{4.8} \\ = 0.72916...\)

We have the final answer as

Hope this helps you

The rectangular loop of copper wire cable transports electrons.

Using Wire, your team can interact and exchange information quickly, securely, and always in context. This includes messages, files, conference calls, and private conversations. Your team can communicate whether they are in the workplace or on the go because to Wire's compatibility with any hardware and operating system.

You can use Wire to make secure voice, video, and messaging calls. It employs wireless data, which can reduce the cost of phone calls & SMS messages (through a cell plan or wi-fi). For those of us who wish to contact or text pals without using up a costly phone plan, this is fantastic.

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