A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simultaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?

Answer: C. A hot non-polar substance

Explanation: I am guessing you have options for this, because I've already had this question at school. So C.

The volume of the parallelepiped is 83 cubic units.

The volume of a parallelepiped with sides give as a = ( 7,2 ,4 ) , b = ( 4,7 ,6 ) and c = ( 3,4 ,7 ).

The volume of a parallelepiped with adjacent sides a, b, and c is given by the scalar triple product (a × b) · c.

First, need to calculate the cross product of vectors a and b

a × b =

\(\left[\begin{array}{ccc}i &j&k\\7&2&4\\4&7&6\end{array}\right]\)

= (2 × 6 - 4 × 7) i - (7 × 6 - 4 × 4) j + (7 × 7 - 2 × 4) k

= -8 i - 26 j + 45 k

Now, calculate the scalar triple product

(a × b) · c = (-8)(3) + (-26)(4) + (45)(7) = 83

Therefore, the volume of the parallelepiped is 83 cubic units.

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Answer:

Check Newton's Rings:

d = height of air film

s = distance from center to ring being considered

R = radius of circle considered

The approximate formula is:

d = s^2 / (2 R)   or s = (2 R d)^1/2

If we just use 4000 mi for R and 1/4 mi for d the height

we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi

The reactants in the given chemical equation are methane (CH₄) and oxygen (O₂), while the products are carbon dioxide (CO₂) and water (H₂O).

During a chemical reaction, the atoms of the reactants reorganize to form new compounds in the products. A chemical change is this atomic rearrangement.

A chemical change is the transformation of one or more substances into new ones with distinct chemical and physical properties. New substances are created when the original substances' atoms are rearranged and new chemical bonds are formed during a chemical change.

Typically, a transfer of energy in the form of heat, light, or sound occurs in conjunction with this kind of change. Reactions like combustion, oxidation, decomposition, and synthesis are all examples of chemical changes. Physical changes, on the other hand, involve alterations in a substance's physical properties without altering its chemical composition.

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Answer:

Vf =3.87[m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

\(v_{f}^{2}=v_{o}^{2} +2*a*x\)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (from the rest)

a = acceleration = 3 [m/s²]

x = distance = 2.5 [m]

Now replacing:

\(v_{f}^{2}=0+2*3*2.5\\v_{f}^{2} =15\\v_{f}=\sqrt{15} \\v_{f}=3.87[m/s]\)

Answer:

The speed of car A before collision is 3.5 km/h.

Explanation:

Mass of car A = 690 kg eastwards

Mass of car B = 520 kg at 74 km/h west wards

Distance, s = 6 m

coefficient of friction = 0.75

Let the speed after collision is v.

Use third equation of motion

\(v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h\)

Let the initial speed of car A is v'.

Use conservation of momentum

690 x v' - 520 x 74 = (690 + 520) x 33.8

690 v' + 38480 = 40898

v' = 3.5 km/h

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

              \(a_{t}\) = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

                    \(a_{c}\) = v2/R

In the general case, both the module and the address change

             a = Ra (  a_{t}^2 +   a_{c}^2)  

The velocity of an object changes if it undergoes uniformly acceleration

motion by considering if it is:

This is the rate of change of an object's position with time. If the object has

positive constant acceleration, the slope goes upward while if it is a

negative constant acceleration, the slope goes downward.

The direction doesn't change as a result of the uniform speed and

direction that is involved.

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The length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

Given data: r = 0.2 mm = 0.2 x 10^-3m Resistivity = 3 x 10^-6 ohm R = 15.552 ohm

Formula Used: Resistivity (ρ) = (RA)/L

Where, R is resistance, A is the area of cross-section, L is the length of the wire.

Resistance (R) = ρ (L/A)

Multiplying A on both sides, we get

Resistance (R) x A = ρ L ... equation (1)

Area of the cross-section of a wire of radius (r) is given by, A = πr^2

where, π is a constant whose value is 3.14

Substituting the given values, we get

A = πr^2= π (0.2 x 10^-3m)^2= 1.2566 x 10^-7 m^2

Substituting the values of R, A and ρ in equation (1), we get

Length of wire (L) = (Resistance x Area) / Resistivity= (15.552 ohm x 1.2566 x 10^-7 m^2) / (3 x 10^-6 ohm)= 6.5268 m

Therefore, the length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

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Sound waves are a type of mechanical wave that propagate through a medium, typically air but also other materials such as water or solids.

Frequency: the frequency of a sound wave refers to the number of cycles or vibrations it completes per second and is measured in Hertz (Hz).

Amplitude: the amplitude of a sound wave refers to the maximum displacement or intensity of the wave from its equilibrium position. It represents the loudness or volume of the sound, with larger amplitudes corresponding to louder sounds and smaller amplitudes corresponding to softer sounds.

Wavelength: the wavelength of a sound wave is the distance between two consecutive points in the wave that are in phase, such as from one peak to the next or one trough to the next. It is inversely related to the frequency of the wave.

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Answer:

I KNOW THE ANSWER IT WILL COST 30$

Explanation:

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

Answer:

The angular acceleration is 4.5 rad/s^2.

Explanation:

Acceleration, a = 26.5 m/s2

length, L = 5.89 m

The angular acceleration is

\(\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2\)

The equivalent resistance of the circuit shown  is 23 ohms.

Option A is correct.

Resistance is  described as the opposition that a substance offers to the flow of electric current.

In a series circuit, all components are connected end-to-end to form a single path for current flow.

In a parallel circuit, all components are connected across each other with exactly two electrically common nodes with the same volt.

We then 1/R = 1/100 + 1/100 + 1 /(50+ 50) +  1 /(50+ 50)

I/R = 0.04

R = 25 ohms.

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The negative influences that must be avoided in order to achieve your goal are excuses, procrastination and distractions.

A negative influence is someone or something that induces negative or bad action in someone. This bad influence can create bad energy within and corrupt good mind.

Thus, the negative influences that must be avoided in order to achieve your goal are excuses, procrastination and distractions.

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The maximum height the crate reaches is 0.856 m.

To determine the maximum height the crate reaches, we need to use the conservation of mechanical energy:

Initial potential energy = final kinetic energy + final potential energy + work done by friction

Let's calculate each term:

Initial potential energy:

PE₁ = mgh₁ = 60.0 kg * 9.81 m/s² * 1.46 m = 868.98 J

Final kinetic energy:

KE₂ = (1/2)mv², where v is the final velocity of the crate after passing over the rough patch.

To find v, we can use the work-energy principle:

Work done by friction = change in kinetic energy

F * d = (1/2)mv² - 0

where F is the force of kinetic friction and d is the distance over which the force acts.

F = μ_k * N, where N is the normal force acting on the crate.

Since the crate is on an incline, we need to find the components of the gravitational force and the normal force acting on it:

mgsinθ = 60.0 kg * 9.81 m/s² * sin(25.0°) = 243.2 N

mgcosθ = 60.0 kg * 9.81 m/s² * cos(25.0°) = 550.2 N

N = mgcosθ = 550.2 N

F = μ_k * N = 0.34 * 550.2 N = 187.07 N

Now we can solve for v:

187.07 N * 1.60 m = (1/2) * 60.0 kg * v² - 0

v = √(2 * 187.07 N * 1.60 m / 60.0 kg) = 2.47 m/s

Final kinetic energy:

KE₂ = (1/2)mv² = (1/2) * 60.0 kg * (2.47 m/s)² = 183.08 J

Final potential energy:

PE₂ = mgh₂

At the maximum height, the final velocity of the crate is zero, so all the final energy is potential energy. Therefore:

PE₂ = PE₁ - KE₂ - work done by friction

mgh₂ = 868.98 J - 183.08 J - 187.07 N * 1.60 m

Solving for h₂, we get:

h₂ = (868.98 J - 183.08 J - 187.07 N * 1.60 m) / (60.0 kg * 9.81 m/s²) = 0.856 m

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Answer:

a

The  winning boat is  boat A

b

\(v_{avg} =0 m/s \)

Explanation:

From the question we are told that

  The  width of the lake is  \(w =  68 \  km  \)

   The  speed of boat A to and fro  is  \(v_a  =  68km/h\)

   The speed of boat B going is  \(v_b  =  34 km/h\)

    The speed of boat B coming back is  \(v_B =  102 \ km/h\)

Generally the time taken by boat A is mathematically represented as

    \(t_A  =   \frac{w}{ v_a}  + \frac{w}{ v_a}\)

      \(t_A  =  \frac{68}{68}  +  \frac{68}{68}\)

     \(t_A  =  2 \  hours\)

Generally the time taken by boat B is mathematically represented as

      \(t_B  =   \frac{w}{ v_b}  + \frac{w}{ v_B}\)

      \(t_B  =   \frac{68}{ 34}  + \frac{68}{ 102}\)

       \(t_B  =  2.67 \ hours \)

The  winning boat is  boat A

The  average velocity is mathematically represented as  

      \(v_{avg} =  \frac{v_a  -  v_a}{ d}\)

Here d is the total displacement of the winning boat which is  0 m

So

        \(v_{avg} =  \frac{68  -  68}{0}\)

         \(v_{avg} =0 m/s \)

Answer:

The velocity of an object must include these two things displacement and time.

Explanation:

Hope this helps you. Have a nice day^_^

Please mark as brainliest. It helps a lot. :)

Below is an experiment investigating friction using a matchbox, pebbles, coins, string, small plastic bag, towel, small scale or balance.

Procedure:

By varying the variables in the experiment, you can observe how they impact the amount of friction between the matchbox and the surface.

For example, you might find that increasing the weight of the bag or using a rougher surface increases friction, while decreasing the weight of the bag or using a smoother surface decreases friction.

Friction is a force that opposes motion between two surfaces that are in contact. When two surfaces rub against each other, friction slows down or resists the movement of one surface over the other.

Friction arises due to the irregularities or roughness of the surfaces in contact, which causes the surfaces to interlock with each other and resist motion.

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Answer:

B

Explanation:

no reason for this answer

Explanation:

Yes, there are differences in the shapes of position-time graphs for uniform acceleration and uniform deceleration in different directions. Let's consider each case separately:\(\hrulefill\)

(1) - Uniform acceleration towards the positive direction:

In this case, the object is moving in the positive direction with a constant acceleration. The displacement-time graph will typically be a curve that starts from an initial position and shows a steady increase in displacement over time. The shape of the graph will depend on the specific acceleration value.

(2) - Uniform acceleration towards the negative direction:

In this case, the object is moving in the negative direction with a constant acceleration. The displacement-time graph will also be a curve, but it will show a steady decrease in displacement over time.

(3) - Uniform deceleration towards the positive direction:

In this case, the object is initially moving in the positive direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a positive slope and gradually levels off.

(4) - Uniform deceleration towards the negative direction:

In this case, the object is initially moving in the negative direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a negative slope and gradually levels off.

Answer:

To find the resultant force and its direction, we can use vector addition.

First, let's break down force B and force C into their horizontal and vertical components:

Horizontal component of force B:

Bx = 20N * cos(140°)

Vertical component of force B:

By = 20N * sin(140°)

Horizontal component of force C:

Cx = 16N * cos(290°)

Vertical component of force C:

Cy = 16N * sin(290°)

Now, let's add up the horizontal and vertical components of all the forces:

Horizontal component of resultant force:

Rx = Ax + Bx + Cx

Vertical component of resultant force:

Ry = Ay + By + Cy

To find the magnitude of the resultant force (R), we use the Pythagorean theorem:

R = sqrt(Rx^2 + Ry^2)

To find the direction (θ) of the resultant force, we can use the inverse tangent function:

θ = atan(Ry / Rx)

Plugging in the given values:

Ax = 12N (horizontal component of force A)

Ay = 0N (vertical component of force A)

Bx = 20N * cos(140°)

By = 20N * sin(140°)

Cx = 16N * cos(290°)

Cy = 16N * sin(290°)

Now let's calculate the values:

Bx = 20N * cos(140°) ≈ -11.55 N

By = 20N * sin(140°) ≈ 9.56 N

Cx = 16N * cos(290°) ≈ 13.82 N

Cy = 16N * sin(290°) ≈ -5.45 N

Rx = 12N + (-11.55N) + 13.82N ≈ 14.27 N

Ry = 0N + 9.56N + (-5.45N) ≈ 4.11 N

R = sqrt(14.27^2 + 4.11^2) ≈ 14.98 N

θ = atan(4.11 / 14.27) ≈ -15.58°

The magnitude of the resultant force is approximately 14.98 N, and the direction is approximately -15.58° (or approximately -45° to force A).

Note: The negative sign indicates that the resultant force is in the opposite direction to force A.

If a frictionless plane is 10.0 m long and inclined at 36.0°.

We can use conservation of energy to find the distance that the first sled travels up the incline. The potential energy of the sled at the bottom of the incline is zero, and its kinetic energy is:

KE = (1/2)mv^2

where m is the mass of the sled and v is its speed. At the point where the sled stops, all of its kinetic energy has been converted into potential energy, so we can write:

mgh = (1/2)mv^2

where h is the height that the sled has traveled up the incline. Solving for h, we get:

h = (v^2)/(2g)

where g is the acceleration due to gravity. Using the given values, we have:

h = (6.00 m/s)^2 / (2 * 9.81 m/s^2) = 1.83 m

So the first sled travels a distance of 10.0 m - 1.83 m = 8.17 m up the incline.

b. To find the initial speed of the second sled, we can use conservation of energy again. At the top of the incline, the sled has potential energy:

PE = mgh

where h is the height of the incline. As the sled slides down the incline, its potential energy is converted into kinetic energy:

KE = (1/2)mv^2

where v is the speed of the sled at the bottom of the incline. We can equate these two expressions and solve for v:

mgh = (1/2)mv^2

v = sqrt(2gh)

Using the given values, we have:

v = sqrt(2 * 9.81 m/s^2 * 10.0 m * sin(36.0°)) = 12.2 m/s

So the second sled must be released from the top of the incline with an initial speed of 12.2 m/s.

The position of the sled as a function of time is given by:

y = -0.5gt^2 + Vi*t + h

where y is the vertical position of the sled, t is the time, Vi is the initial speed of the sled, and h is the height of the incline. At the bottom of the incline, y = 0, so we can solve for the time it takes for the second sled to reach the bottom:

0 = -0.5gt^2 + Vi*t + h

t = (Vi ± sqrt(Vi^2 - 2gh)) / g

Since we want both sleds to reach the bottom at the same time, we set the time for the first sled to slide down the incline equal to this expression for t and solve for Vi:

t = sqrt(2h/g) = sqrt(2 * 1.83 m / 9.81 m/s^2) = 0.619 s

0 = -0.5gt^2 + Vi*t + h

Vi = (h - 0.5gt^2) / t

Vi = (1.83 m - 0.5 * 9.81 m/s^2 * (0.619 s)^2) / 0.619 s

Vi = 5.68 m/s

So the initial speed of the second sled is about  5.68 m/s

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Answer:

See a tractor is more slow but has a greater force and a car is fast but has a slower force

Explanation:

so your answer is Tractors have more force then cars

The rank of the forces in order of increasing magnitude is Fg < FB < FA.

option D is the correct answer.

The net force on an elevator moving upwards is determined by the force of gravity acting downwards and the normal force of the elevator acting upwards.

That is, the two forces acting on a person when he is moving in an elevator are:

When the two forces are of equal magnitude, the elevator will be static or moving with constant velocity.

When the magnitude of the two force are unequal, then the elevator will be accelerating upward or downward.

Since the elevator is moving upwards, it implies that the normal force is greater than the force of gravity acting downwards.

The box at the bottom will feel much heavier due to the weight of box and gravity acting downwards.

FA = FB + Fg

Thus, the force exerted on box A is the greatest, followed by the force on box B and then, the smallest is force of gravity.

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Answer:

a first-hand or contemporary account of an event or topic.

Explanation:

Answer:100

Explanation:

Answer:

All planets including the outer larger planets were formed at the same time somewhere around 4.5 Billion years ago.

Explanation:

the young sun drove away most of the gas from the inner solar system, leaving behind the rocky cores also known as the terrestrial planets.

Answer:

a) a = 1 [m/s²]

b) a = 0.25 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

a)

Between 0 and 2 seconds

where:

Vf = final velocity = 2 [m/s]

Vo = initial velocity = 0

a = acceleration [m/s²]

t = time = 2 [s]

Note: the positive sign of acceleration in the previous equation allows us to appreciate that the velocity is increasing

2 = 0 + a*2

a = 2/2

a = 1 [m/s²]

b)

Between 2 and 4 seconds

where:

Vf = final velocity = 2.5 [m/s]

Vo = initial velocity = 2

a = acceleration [m/s²]

t = time = (4 - 2) = 2 [s]

2.5 = 2 + a*2

2.5 - 2 = 2*a

0.5 = 2*a

a = 0.25 [m/s²]