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

Answers

Answer 1

To create a two-cycle sine wave on an oscilloscope, you need to adjust the time/div (time per division) control accordingly.

The time/div control determines the horizontal scaling of the waveform displayed on the screen.

A sine wave completes one cycle when it goes from its starting point, through its peak, back to its starting point, and then through its trough, finally returning to the starting point. In other words, it completes one full oscillation.

To create a two-cycle sine wave, you want the waveform to complete two full oscillations within the visible horizontal width of the oscilloscope screen. Therefore, you need to adjust the time/div control so that it represents the time it takes for two cycles to occur within one division on the screen.

The specific adjustment required will depend on the frequency of the sine wave you are working with. Let's assume you know the frequency of the sine wave and want to adjust the time/div control accordingly. Here's a general method to achieve this:

Determine the period of the sine wave: The period is the time it takes for one complete cycle of the waveform. It is the reciprocal of the frequency. If you know the frequency of the sine wave, you can calculate the period using the formula: period = 1 / frequency.

Determine the time it takes for two cycles: Multiply the period by 2 to get the time it takes for two cycles to occur.

Adjust the time/div control: Look for the time/div knob or button on your oscilloscope. Turn or press it to adjust the time per division. The available options may be labeled on the knob/button or displayed on the screen. Choose a setting that represents the time it takes for two cycles to occur within one division.

Fine-tune if necessary: If the two-cycle waveform is not precisely fitting within one division, you may need to adjust the time/div control further to achieve the desired display.

Remember, the specific steps and controls can vary depending on the oscilloscope model you are using. Consult the oscilloscope's user manual or refer to the manufacturer's instructions for precise details on adjusting the time/div control.

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Related Questions

Which of the following correctly lists our "cosmic address" from small to large?

a. Earth, solar system, Milky Way Galaxy, Local Group, Local Supercluster, universe

b. Earth, solar system, Local Group, Local Supercluster, Milky Way Galaxy, universe

c. Earth, Milky Way Galaxy, solar system, Local Group, Local Supercluster, universe

Answers

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)

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​

Answers

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.

Drag is determined by all of the following EXCEPT:
a. frontal area.
b. body shape.
c. surface texture.
d. depth.

Answers

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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Why heavy vehicles usually have low velocity but possess high kinetic energy

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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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two masses are tied to the same rope which hangs over a massless and frictionless pulley. m1 is 5kg and m2 is 10kg. what is the acceleration of the masses and the tension in the rope?

Answers

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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It takes 6s for a stone to fall from the top of a building to the ground calculate i.the height of the building ii.the speed of the stone just before it strikes the ground (g=10ms^-2

Answers

Answer:

fffffff

Explanation:

Problem 10:00-gauge copper wire has a diameter of 9.266 mm and a resistivity of 1.72 x 10-8 2.m. Calculate the power loss in a kilometer of such wire, in watts, when it carries 0.95 × 102 A. P=

Answers

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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A 0.145 kg rock is thrown with a speed of 32.0 m/s at an angle of 400. what is its kinetic energy at the top of its trajectory?

Answers

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:

v = speedm = massθ = angle Joules = Kg m²*s²

Information of the problem:

v = 32 m/sm = 0.145 Kgθ = 40°

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

What is kinetic energy?

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

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A 0.145 kg rock is thrown with a speed of 32.0 m/s at an angle of 400. what is its kinetic energy at

There is a paperweight that is known to be 72 ml, and has a mass of 184.32g,
what is the block made of?

Answers

Answer:

56

Explanation:

Its 56 just because

vector has a magnitude of 17.0 units, vector has a magnitude of 13.0 units, and ab has a value of 14.0. what is the angle between the directions of a and b?

Answers

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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fill in the blanks (about force)
depends on _____ and _____

Answers

Depends on who and where I’m just answering

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

Los Angeles is about 2000 miles from Chicago. A jet aircraft can fly about 600 miles/hours. How long will it take to fly from Chicago to Los Angeles? How long in minutes

Answers

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

What is the velocity?

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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How long does it take to run 150 meters at the speed of 18 km

Answers

Answer: 8.3 hours

Explanation:

time= distance divided by speed so:

150 divided by 18= 8.3 hours

Interference patterns, a characteristic of waves, arise when X rays pass through a crystalline material like gold. What occurs when a beam of electrons passes through the same material

Answers

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

What are 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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in a copper wire, electrons move to the right. the direction of the magnetic field is toward you. in which direction is the magnetic force on the copper wire?

Answers

The rectangular loop of copper wire cable transports electrons.

What is the purpose of the wire app?

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.

Who is calling from Wire?

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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A measurement has high
when it is very close to the
true value?

Answers

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.

A cat is running at 4.65 m/s and has 91.7 J of kinetic energy. What is the cats mass

Answers

To find the mass of the cat, you can use the formula for kinetic energy, which is:

KE = (1/2) * m * v^2

where KE is the kinetic energy, m is the mass, and v is the velocity.

You can rearrange this formula to solve for mass:

m = (2 * KE) / v^2

Plugging in the values given, you get:

m = (2 * 91.7 J) / (4.65 m/s)^2

Solving this equation gives a mass of approximately 2.07 kilograms.

A 0.04 kg honeybee circles a field looking for a flower upon which to land. The radius of the circle she travels is 2.5 m and it takes her 1.57 s to complete the circle.

Find the following:
a) Tangential Velocity
b) Centripetal Acceleration
c) Centripetal Force.​

Answers

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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Open and Closed circuit similarities

Answers

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

A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the
ground. How long will it take, in seconds, for the ball to hit the ground?

Answers

Answer:

0.73 s

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

0.73 s

Hope this helps you

Write any two conditions at which a body become weightless.​

Answers

When a body falls freely under the influence of the force of gravity only. When the acceleration of a falling body is equal to the acceleration due to gravity.

Two students standing 12.0 m apart rotate a conducting skipping rope at a tangential speed of 8.2 m/s. The vertical component of the Earth’s magnetic field is 5.0 × 10-5 T. Find the magnitude of the EMF induced at the ends of the skipping rope.
(c) How much energy is stored in the magnetic field of an inductor of 8 mH when a steady current of 4.0 A flows in the inductor?

Answers

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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Bone ha a Young’ modulu of about
1. 8 × 1010 Pa. Under compreion, it can
withtand a tre of about 1. 65 × 108 Pa before breaking. Aume that a femur (thigh bone) i 0. 49 m
long, and calculate the amount of compreion
thi bone can withtand before breaking. Anwer in unit of mm

Answers

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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Which of these energy transformations is not part of how a nuclear power plant
provides electrical energy to a home?
changing electrical energy to chemical energy
changing thermal energy to mechanical energy
changing nuclear energy to thermal energy
changing mechanical energy to electrical energy

Answers

I think the second one

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R. The central perpendicular axis through the ring is a z-axis, with the origin at the center of the ring.


(a) What is the magnitude of the electric field due to the rod at z = 0? N/C

(b) What is the magnitude of the electric field due to the rod at z = infinity? N/C

(c) In terms of R, at what positive value of z is that magnitude maximum? R

(d) If R = 4.00 cm and Q = 9.00 �C, what is the maximum magnitude?N/C

Answers

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

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius

A 2.5-kg cart is rolling along a frictionless, horizontal track towards a 1.3-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is 4.6 m/s, and the second cart's velocity is -1.8 m/s. (a) What is the total momentum of the system of the two carts at this instant

Answers

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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An iron nail does not pick up paper clips as a magnet would.

Answers

Answer:

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

An 800-kg car traveling in a straight line at a speed of 30 m/s applies its brakes. During braking, there is a net force of 2000 N in the horizontal direction on the car. What distance will the car move from the time it applies its brakes until it has stopped?
a.12
b.75
c.180
d.300
e.60000

Answers

Answer:

Explanation:

c = 180

i have same problem and it work

The answer for your question is C=180

Which statement is best supported by the table?
Wave Wand Y are light waves.
Wave X is a sound wave.
Wave W is a light wave.
Waves X and Y are sound waves.
Waves Y is a sound wave.
Waves X and W are light waves.
o Wave W is a sound wave.
Waves X and Y are light waves.

Which statement is best supported by the table?Wave Wand Y are light waves.Wave X is a sound wave.Wave

Answers

Answer:

wave w is a sound wave

wave x and y are light waves

Answer:

D

Explanation:

Cuanta fuerza se necesita lara evitar que una piedra de 20N caiga?

Answers

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.

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