name a substance which will undergo change E​

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

The force that will move the box at constant velocity must be a little more than 64 N.

The coefficient of sliding friction is obtained from the formula;

μ= F/R

Where;

F = frictional force

μ = coefficient of sliding friction

R = Normal reaction

It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.

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Answer:

b. diverging, converging

Answer:diverging, converging

Explanation:

Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.

Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry

Answer:

vddvdvvdvffdfvd

Explanation:

Answer:

It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.

Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.

This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.

The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.

It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].

The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.

Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.

Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

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Answer:

bowling ball, basket ball, tennis ball

Explanation:

The statement shows a case of rotational motion, in which the disc decelerates at constant rate.

i) The angular acceleration of the disc ([tex]\alpha[/tex]), in revolutions per square second, is found by the following kinematic formula:

[tex]\alpha = \frac{\omega_{f}-\omega_{o}}{t}[/tex] (1)

Where:

If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]t = 15\,s[/tex], then the angular acceleration of the disc is:

[tex]\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}[/tex]

[tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex]

The angular acceleration of the disc is [tex]\frac{1}{36}[/tex] radians per square second.

ii) The number of rotations that the disk makes before it stops ([tex]\Delta \theta[/tex]), in revolutions, is determined by the following formula:

[tex]\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex] (2)

If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex], then the number of rotations done by the disc is:

[tex]\Delta \theta = 3.125\,rev[/tex]

The disc makes 3.125 revolutions before it stops.

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Answer:

it right congratulations

Answer:

A

Explanation:

Simply option A is correct

It's a bit of a common sense question

At the highest point it's velocity will be automatically 0 since it is in rest and so is the acceleration

Answer:

Mass of a object 75 Kilograms

Explanation:

Net force acting on an object, [tex]Fnet = 225N[/tex]

Acceleration produced, [tex]a = 3.0m/s^2[/tex]

According to Newton's second law :

F = m a

[tex]M =\frac{F}{a}[/tex]

[tex]m =\frac{225N}{300m/s^2}[/tex]

[tex]m= 75 Kg[/tex]

So, the mass of an object is 75kg.

Hence, this is the required solution.

Answer:

75 kg.

Explanation:

Net Force

[tex]\sf \longmapsto F_{net} = 225N[/tex]

Acceleration produced

[tex]\sf \longmapsto \: a = 3 .0 m / {s}^{2}[/tex]

According to Newton's 2nd Law –

F = m•a

[tex]\sf \longmapsto \: M = \frac{F}{A} [/tex]

[tex]\sf \longmapsto \: M = \frac{225N}{300m/s ^{2} } [/tex]

[tex]\sf \longmapsto \: M = \: 75 \: kg[/tex]

Therefore, the mass of an object is 75 kg.

Answer:

b

Explanation:

hope helpful

have a great day

Answer:

for part A the answer is D and for part B the answer is F

Answer:

include the statements pls so i can choose wich one it is and tell you

Explanation:

Answer:

the ball will go up 3s and down 3s

v=gt

where t=3s and g=9.8m/s^2

distance=v0(t)+(1/2)gt^2

where initial velocity (v0)=0

Explanation:

The speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

As given in the problem we have to find the speed required to throw a ball straight up and it returns 6 seconds later,

S = ut + 1/2*a*t²

0 = u×6 + 0.5×(-9.81)×6²

0 = 6u - 176.8

6u = 176.8

u = 176.8/6

u = 29.43 meters / seconds

Thus, the speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

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Answer:

Electrostatic phenomenon is a from the forces that electric charges exert on each other.

Answer:

0.4 g/cm

Explanation:

density (g cm ³) = mass (g)

÷

volume (cm³)

Answer:

C. 4 g/cm

Explanation:

Use the formula:

density = mass ÷ volume

Sub in the values:

density = 36 ÷ 9 = 4 g/cm

Answer = 4 g/cm

Answer:

As thermal energy increases,there is more particle movement

The definition of moment of inertia and kinetic energy we can find the results for the questions are:  

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

a) The moment of inertia is a scalar quantity that has the same function as mass, it is given by

             I = m r²

They indicate that the length of the bar is y₁ = 5.70 m, if the two rigid bars form the square that supports the masses, the distance from the center to the end where each mass is:  

            r = y₁ / 2

             r = 5.70 / 2 = 2.85 m

The moment of inertia of the system is.

         I = 4 m r²

Let's calculate.

        I =4 m 2.85²

        I = 32.49m kg m²

If we assume that the mass of each point particle is:

        m = 1 kg

         I = 32.49 kg m²

b) They ask for the rotational kinetic energy, if the angular velocity is.

           w = 5.60 rad / s

The kinetic energy for rotational motion is.

         K = ½ I w²

Let's calculate.

          K = ½ 32.49 5.60²

          K = 509.4 J

In conclusion with the definition of moment of inertia and kinetic energy we can find the results for the questions are:

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

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Answer:

C, B, and D are ones i think i know that can occur during a physical change

Explanation:

At the maximum height the particles position has been shown as [tex]\frac{R}{3}[/tex].

The given parameters;

The velocity of the particle is given as;

[tex]v^2 = \frac{GM}{2R}[/tex]

The total mechanical energy of the particle is given as;

[tex]-\frac{GMm}{R} + \frac{1}{2} mv^2 = - \frac{GMm}{R} + \frac{1}{2} mv^2[/tex]

At maximum height the final velocity of the particle is zero.

[tex]-\frac{GMm}{R} + \frac{1}{2} mv^2 = - \frac{GMm}{R +h}[/tex]

[tex]-\frac{GMm}{R} + \frac{1}{2} m\times \frac{GM}{2R} = - \frac{GMm}{R +h}\\\\-\frac{GMm}{R} + \frac{GMm}{4R} = - \frac{GMm}{R +h}\\\\- \frac{3GMm}{4R} = - \frac{GMm}{R +h}\\\\- \frac{3}{4R} = - \frac{1}{R+ h} \\\\3(R+ h) = 4R\\\\3R + 3h = 4R\\\\3h = 4R - 3R\\\\3h = R\\\\h = \frac{R}{3}[/tex]

Thus, at the maximum height the particles position has been shown as [tex]\frac{R}{3}[/tex].

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Answer:

Gravitional pull or gravity

Explanation:

Its that simple

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d

         sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           [tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]

        θ₂ = 0.181 rad

we use the equation of refraction.

         [tex]\theta_r[/tex]  = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )

λ₁ = 400 nm  

       θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  [tex]\frac{1 sin 0.181}{1.33}[/tex]

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

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Answer:

Explanation:

Your numbers seem wonky, so I'll just assume that the initial displacement is a distance A (Amplitude) from the equilibrium position. Spring constant = k

Initial potential energy is

PE = ½kA²

As potential energy and kinetic energy are constantly exchanging in SHM,

the position x where half of the original spring potential exists is found where

½kx² = ½(½kA²)

    x² = ½A²

    x = (√0.5)A

    x ≈ 0.707A

just plug in your actual starting position A

With A = 5.2 cm

x = 3.67695... 3.7 cm

Answer:

s times t

Explanation:

The equation for distance is D = st

Answer:

D is the correct answer (D=S*T)

This is an algebraic law; if you are to move the denominator or numerator to the other side of the equation, you must multiply it by the other side.

S=D/T is the same as D=S*T.

so distance=speed times time.

so if you are going 50 miles per hour, you multiply that times time, say one hour. you get your D distance which is 50 miles.

Changes that would results in an increase in the kinetic energy of the cannonball in a closed system, when it is fired from the cannon is B:increase the mass of the cannon or decrease the mass of the cannonball.

Therefore, option B is correct.

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Explanation:

M.A = load / Effort

efficiency = M.A/V.R X 100

75 = M.A / 4 X 100

75 = 25 X M.A

M.A = 75/25 = 3

M.A = load / effort

3 = 100/E

E = 100/3 = 33.333

A machine whose efficiency is 75% is used to lift a load of 100m. the effort needed to put into the machine is 33.33 meters if the velocity ratio is 4.

Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.

Mechanical advantage = output force(load) /input force (effort)

Simple machines like static pulleys, as well as a movable pulleys, provide a great amount of mechanical advantage.

A machine whose efficiency is 75% is used to lift a load of 100m.calc if it has a velocity ratio of 4.

As we know that the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

mechanical advantage  = velocity ratio   ×Efficiency

       =4 × 0.75

       = 3

The mechanical advantage is 3 , which means the ratio of load to effort is 3

load / effort = 3

100 m/effort =3

effort = 33.33 m

Thus, the effort put into the machine comes out to be 33.33 m.

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Answer: See explanation

Explanation:

Please see attached picture. Due to the fact one of the words in the equation F=ma contains a word not allowed.

By Newton's second law, the net force on the crate is

∑ F = 300 N - f = (45 kg) (4.44 m/s²)

where f is the magnitude of friction. Solve for f :

300 N - f = (45 kg) (4.44 m/s²)

f = 300 N - (45 kg) (4.44 m/s²)

f = 100.2 N ≈ 100 N (D)

Answer:

configuration of string:

Node - Antinode - Node    or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string