59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?​

Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree

During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.

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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is  4.51 sec.

The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.

Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,

force = ma = μmg, putting the value,

a = - 1.175 m/sec²

now using equation of motion

v = u + at

t = 4.51 sec

So, time taken by the puck to come to rest is 4.51 sec.

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Answer:

B

Explanation:

The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.

Answer:

Explanation:

Impossible to say without knowing how high point D is.

If we ignore friction, the energy converted from potential energy will exist as kinetic energy.

let d be the height in meters of point D "in the air"

mg(19 - d) = ½mv²

m is common so divides out

g(19 - d) = ½v²

v = √(2g(19 - d))

no sense in making g any more precise than the height. g = 9.8

v = √(2(9.8)(19 - d))

v = √(19.6(19 - d))

Answer:

200 m/s

Explanation:

as momentum is a product of mass and speed, and mass is not changing, four times the speed will result in four times the momentum.

p = mv

4p = m(4v)

At 200m/s velocity of the train, the momentum of the train will be four times its original momentum. This is because the velocity is directly proportional to the momentum of an object.

Momentum is the tendency of an object to remain in motion. It is the product of the mass of the object and the velocity of the object. It is denoted by "p". Momentum is a vector quantity because it has both the magnitude and direction. The unit of Momentum is g.m/s.

The momentum can be calculated with the formula:

p = m × v

where, p = momentum of the object,

m = mass of the object,

v = velocity of the object.

The momentum is directly proportional to the mass and velocity of the object. Hence, to increase the momentum of the train by four times the mass or velocity of the train needs to be increased by four times.

The increase in the velocity of the train by four times will increase the momentum of the train.

Original Velocity of the train = 50 m/s

Increased Velocity of the train = 50 × 4 = 200 m/s

Increased new Momentum of the train = mass × new velocity

Increased Momentum = 20 metric ton × 200m/s

Increased Momentum = 4000 metric ton. m/s

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The wavelength will remain unchanged.

Explanation:

The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is

[tex]v = \lambda\nu[/tex] (1)

so if we double both the velocity and the frequency, the equation above becomes

[tex]2v = \lambda(2\nu)[/tex] (2)

Solving for the wavelength from Eqn(2), we get

[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

Answer:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

Explanation:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

[tex] \boxed{ \sf \: better \: luck \: at \: your \: class}[/tex]

A uniform solid sphere revolves around a vertical support on a frictionless bearing.

A solid disk is covered with a light string that circles the sphere's equator, over which is suspended a mass. The mass M of the hanging mass, disk, and sphere is the same.

The radius R of the sphere and disk is the same. If the string does not slip on the sphere or disk, determine the acceleration of the hanging mass. If necessary, explain your response in terms of M, R, and g.

A uniform solid sphere revolves around a vertical support on a frictionless bearing, as seen in the illustration below. Around the sphere's equator, a thin cable travels over a solid surface.

Therefore, A uniform solid sphere revolves around a vertical support on a frictionless bearing.

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Answer:

I think that those are all true but I'm not completely sure.

The force will be reduced to 1/4 of the original

Explanation:

According to Coulomb's law, the force between two charges [tex]q_a\:\text{and}\:q_b,[/tex] separated by a distance r is given by

[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]

where k is the Coulomb constant.

Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as

[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]

[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]

Therefore, the force will be reduced to a quarter of its original value.

Hope you could understand.

If you have any query, feel free to ask.

The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1.

     a) The stopping distance is: x = 270 m

     b) The initial velocity is: v₀ = 17.3 m / s

     c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

Given parameters

To find

Question 1.

    a) Minimum braking distance.

    b) If the distance is x = 40.0 m, what speed should the vehicles have?

    c)  Conclusive importance of physics in daily life.

Question 2.

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

         v² = v₀² - a2 x

Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.

Newton's second law states that the net force is proportional to the mass and the acceleration of the body.

          F = ma

Where F is force, m is mass and acceleration.

In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body

        fr = m a

        a =[tex]\frac{fr}{m}[/tex]  

        a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]  

        a = 3.75 m / s²

This acceleration is in the opposite direction to the speed.

Let's find the distance needed to stop, the final speed is zero.

          0 = v₀² - 2 ax

           x = [tex]\frac{v_o^2 }{ 2a}[/tex]  

Let's calculate.

          x = [tex]\frac{45^2 }{2 3.75}[/tex]  

          x = 270 m

This is the minimum distance that the two vehicles must separate to avoid a collision.

b) We look for speed.

        v₀ = [tex]\sqrt{2ax}[/tex]  

        v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]  

        v₀ = 17.3 m / s

c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.

           0 = v₀² - 2ax

Let's substitute.

             a = [tex]\frac{v^2}{2d}[/tex]  

They ask the distance traveled if this car traveled from an initial speed 2v.

             0 = v² - 2 a x

             x = [tex]\frac{v^2}{2a}[/tex]  

We substitute

            x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]  

            x = 4 d

In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1

       a) The stopping distance is: x = 270 m

       b) The initial velocity is: v₀ = 17.3 m / s

        c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

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Answer:

The distance of man from the mirror is 3 m

Refer to the attachment

[tex]\Large\textsf{Hope \: It \: Helped}[/tex]

The distance of the man from the mirror is 3m

From the characteristics of image formed in a plane mirror,

⇒ From the last point,

⇒ From the question,

Hence, The distance of the man from the mirror is 3m

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Answer:

When buying a margarine, people should look for trans fat and saturated fat content. If it is over 2%, then it is not healthy. If it is lower that means it is healthy and fresh!

Explanation:

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

The wavelengths of the constituent travelling waves is calculated as follows;

[tex]L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}[/tex]

for first mode: n = 1

[tex]\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm[/tex]

for second mode: n = 2

[tex]\lambda = \frac{2L}{2} = L = 100 \ cm[/tex]

For the third mode: n = 3

[tex]\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm[/tex]

For fourth mode: n = 4

[tex]\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm[/tex]

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

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Answer:

Light to

Electrical to

mechanical and sound

Answer: D

Explanation:

they are all odorless, colorless, monatomic gases with very low chemical reactivity.

Brainliest?

The elements in group 18 that they rarely react with any other elements. The correct option is D.

The periodic table, also known as the periodic table of the elements, is an arrangement of chemical elements in rows and columns.

It is widely used in chemistry, physics, and other sciences, and is widely regarded as a chemistry icon.

The periodic table is worked upon by scientists to quickly refer to information about an element, such as its atomic mass and chemical symbol.

The arrangement of the periodic table also allows scientists to detect trends in element properties such as electronegativity, ionization energy, and atomic radius.

The noble gases are found at the far right of the periodic table and were previously known as "inert gases" due to their filled valence shells, which make them extremely nonreactive. In comparison to other element groups, noble gases were discovered relatively late.

Thus, the correct option is D.

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Answer:

W = 5W

Explanation:

1W = 1J/s

W = Watts

J = Joules

s = seconds

W = J/s

W = 300J/60s

W = 5W

Answer:

Explanation:

homework

or just play some 2012 math game

Answer:

I think calculating distance (if thats an option)

Explanation:

Really hope this helped!

Remain the same.

Explanation:

The kinetic energy of the ejected electrons, and thus the resulting photocurrent, does not depend on the intensity of the incident radiation. Instead, it depends on the frequency. So since the stopping potential is used to reduce the photocurrent to zero and the photocurrent does not depend on the intensity, the stopping potential remains the same.

Answer:

Explanation:

I will us θ as the angle above horizontal as I don't know how to make your chosen symbol for angle.

           KEy = PE

½m(vsinθ)² = mgh

                h = (vsinθ)²/2g

There will be exchange of heat in the system, the tub filled with ice will gain heat, while the bottle of liquid will lose heat.

The flow is made possible by means of convection which implies heat flow from region of higher concentration to region of low concentration

by actual movement of material particles

There are basically three modes of heat transfer

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Explanation:

if you go to connexus and are doing the science

Thermal Energy Unit Test 10 qestions 8th grd unit 5 test lesson 12 here are the correct answers

1. A. metal is heated from room temperature to 200'C.

2. B. The particles will have more space between them as steam, but they will be moving at the same speed in both states.

3. C. The potential energy decreases due to the tighter arrangement of the particles.

4. B. an increase in heat and an increase in kinetic energy until a phase change occurs.

5. C. thermometer

6. B. 4.

7. D. convection

8. C. Air moves from the areas of higher temperature to areas of lower temperature.

9. C. The temperature of the ice increases, while the temperature of the water decreases

10. A. heat transfer by radiation.

i hope this helps and please let me know if it does! = ) <3

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

Answer:

Explanation:

For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.

For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.

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Answer:

true answer this question

Answer:

11.573

Explanation:

f = m*a

where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration

so, we solve for a

a = f/m

a = 22/1.901

a = 11.573

Answer:

more likely it is to have gravity

Answer:

do it got a picture

Explanation:

Answer:

a

Explanation:

the answer is a because it is feasible