The gravity of Earth is attracting a person towards the center with 500N of gravitational force. The person is exerting a reactionary force on the Earth with how much force?

Answer:

Explanation:

1. Place a cup of hot tea on a flat surface.

2. Place a thermometer in the tea and record the temperature.

3. Place a fan in front of the cup of tea and turn it on.

4. Place the thermometer in the tea again and record the temperature.

5. Compare the two temperatures and observe the difference.

6. The difference in temperature is an indication of the thermal energy that has been dissipated from the cup of hot tea.

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

\(F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz\)

The frequency of the simple pendulum is given as;

\(F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m\)

Thus, the length of the simple pendulum is 0.25 m.

Answer:

Look below.

Volume expansion of a gas              Gas thermometer

Volume expansion of a liquid           Laboratory or clinical thermometer

Volume expansion of a solid     Bi-metallic strip thermometer

Pressure change of a fixed mass of gas    Constant – volume gas  thermometer

Answer:

A. pi1 = 690 kg m/s

B. Δp1 = - 690 kg m/s

C. The two changes in momenta are equal in magnitude and opposite in direction.

D. v2i = -6.16 m/s

Explanation:

Part A.

The momentum can be calculated as mass times velocity, so the initial momentum of the 92-kg player is equal to

pi = mv

pi = (92 kg)(7.5 m/s)

pi = 690 kg m/s

So, the initial momentum of the 92-kg player was 690 kg m/s

Part B.

After the collision, each player has zero velocity, so the final velocity will be 0 m/s. Then, the change in momentum is equal to

Part C.

In a closed system the momentum is conserved, so to conserve the momentum the change in momentum of the 92-kg player should be equal in magnitude but opposite in direction. Then, when we add them, the sum should be equal to 0. So, the answer is:

The two changes in momenta are equal in magnitude and opposite in direction.

Part D.

Taking into account part C, we can write the following equation

Since the final velocity of the 112-kg player is also 0 m/s, we can solve for v2i as

Replacing the values, we get

Therefore, the answers are:

A. pi1 = 690 kg m/s

B. Δp1 = - 690 kg m/s

C. The two changes in momenta are equal in magnitude and opposite in direction.

D. v2i = -6.16 m/s

Answer:

F=ma

Explanation:

Force is equal to mass multiplied by area

Force= mass × area

In turn the formula can be twisted so

Force/mass= area

Force/area= mass

Answer:

It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.

Answer:

7 protons, 8 neutrons, 7 electrons

Explanation:

Protons, electrons and neutrons are three sub-atomic particles present in an atom. Protons and neutrons are present in the nucleus while electrons are present outside the nucleus.

The number of protons, which is also the ATOMIC NUMBER, is equal to the number of electrons in a neutral atom (no overall charge). In this case, the number of protons is 7, however, since the charge of this Nitrogen atom is 0, it means that the number of electrons is also 7.

The mass number, denoted by A, is the sum of the number of protons and neutrons in an atom i.e.

A = number of proton + number of neutron.

In this case of Nitrogen isotope where the mass number is 15, the number of neutron will be mass number - number of protons i.e. 15 - 7 = 8.

Therefore, the number of protons, neutrons and electrons are 7, 8 and 7 respectively.

The voltage across the wire is 0.635 V.

The resistance of the wire can be calculated using the formula:

R = (ρL)/A

where ρ is the resistivity of the wire, L is its length, A is its cross-sectional area, and R is its resistance.

The cross-sectional area of the wire can be calculated using the formula for the area of a circle:

A = πr^2

where r is the radius of the wire.

Substituting the given values, we get:

A = π(0.50 x 10^-3 m)^2 = 7.85 x 10^-7 m^2

R = (100 x 10^-8 O * m)(1.0 m) / (7.85 x 10^-7 m^2) = 1.27 Ω

The voltage across the wire can be calculated using Ohm's law:

V = IR

where I is the current through the wire.

Substituting the given values, we get:

V = (0.50 A)(1.27 Ω) = 0.635 V

Therefore, the voltage across the wire is 0.635 V.

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Answer:

◆ See the attachment photo.

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Answer:

thermal energy

Explanation:

This is the definition of the thermal energy of a substance.  It's the sum of the KE and PE of all its particles.

Answer

1) AlH3= trigonal planar

2) CH2F2= tetrahedral

3) PH3= trigonal pyramidal

4) O3= bent

Explanation:

I took the test

Answer:

- They need oxygen to burn fuel

- Aerodynamics

- Extreme temperatures

- Radiation

- Pressure issues

Explanation:

A airplane is a heavier-than-air aircraft kept aloft by the upward thrust exerted by the passing air on its fixed wings and driven by propellers, jet propulsion, etc.

Aeroplanes cannot travel in space for several reasons:

They need oxygen to burn fuel - Aeroplane engines rely on the oxygen in the atmosphere to burn fuel and generate thrust. In space, there is no atmosphere so there is no oxygen for the engines to work.

Aerodynamics - Aeroplane wings generate lift by interacting with the air. In space, there is no air so wings would be unable to generate any lift. Aeroplanes rely on aerodynamics to fly which does not work in space.

Extreme temperatures - In space, temperatures can range from -150 degrees Celsius to 150 degrees Celsius. Aeroplanes are designed to operate within a much narrower temperature range. The extreme cold and heat of space could damage aeroplane components.

Radiation - In space, there are high levels of radiation from the Sun and cosmic rays. Aeroplane bodies are not designed to shield against this type of radiation and it could damage electronics and affect aeroplane systems.

Pressure issues - Aeroplanes are designed to withstand air pressures at altitudes up to around 12 kilometers. In low-Earth orbit and beyond, the air pressure is essentially zero. This extreme change in pressure could cause structural damage to the aeroplane.

In summary, while aeroplanes are designed to fly through the Earth's atmosphere, they lack the key features needed to operate in the extreme environment of outer space like spaceships. Aeroplanes require things like oxygen, aerodynamics and being able to withstand changes in pressure - all of which do not exist or work the same way in space.

Explanation:

The wing is pushed up by the air under it. Large planes can only fly as high as about 7.5 miles. The air is too thin above that height. It would not hold the plane up.

The maximum height of the child is 12.14 m.

We define gravity as: a body is drawn toward the center of the earth or any other physical body with mass by the force known as gravity.

Given parameters:

Mass of the child; m = 21 kg.

Gravitational potential energy at maximum height; E =  2.5 x 10³ J

Let her maximum height is h. Then Gravitational potential energy at maximum height; E = mgh = 21 × 9.8 × h joule.

So, 21 × 9.8 × h =  2.5 x 10³

⇒ h = 12.14 m.

The maximum height of the child is 12.14 m.

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Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

\(T=\dfrac{38}{8}\\\\T=4.75\ s\)

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

\(T=2\pi \sqrt{\dfrac{m}{k}}\)

k = spring stiffness constant

So,

\(k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m\)

When the cord is in air,

mg=kx

x = the extension in the cord

\(x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m\)

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.

The given parameters;

The period of oscillation of the bungee jumper is calculated as follows;

\(T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s\)

The spring stiffness constant is calculated as follows;

\(T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m\)

The extension of the cord is calculated as follows;

\(F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m\)

The unstretched length of the bungee cord is calculated as;

\(\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m\)

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Answer:

a)  The electric field at point P has no component in the x and z directions.

b) dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)Ey = = 2kλd / y₀( d² + 4y₀²))^1/2

d) Ey = 411.84 N/C

Explanation:

a)

from the uploaded image;

the electric field will only be in y-direction.

Therefore the electric field at P have no component in the  x and z directions.

b)

dEy = dEsin θ

dE = kdq / r²

from figure

sinθ = y₀ / √( x² + y₀²)

r = √( x² + y₀²)

dq = λdx

dEy = kdq sinθ / r²

dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)

the net electric field at p is ,

Ey = ∫^d/2_-d/2  kλdxy₀ / (√( x² + y₀²))^3/2

= kλy₀ ∫^d/2_-d/2  dx / (√( x² + y₀²))^3/2

= 2kλd / y₀( d² + 4y₀²))^1/2

d)

let y₀ = 15 × 10⁻²m,  λ = 3.5 nC/m , d = 1.5m

Ey = 2kλd / y₀( d² + 4y₀²))^1/2

Ey = 2(9×10⁹ N.m²/C²)(3.5×10⁻⁹m)(1.5m) / 0.15×(1.5)² + 4(0.15)²))^1/2

Ey = 411.84 N/C

The distance traveled by the person to come to a complete stop in 38 ms at a constant acceleration of 60 g is approximately 273.42 meters.

The distance that a person covers to come to a complete stop in 38 ms at a constant acceleration of 60 g can be calculated using the kinematic equation.

The formula is given by, d = (v^2 - u^2) / 2a, where d is the distance traveled, v is the final velocity, u is the initial velocity, and a is the acceleration given in g units.

To solve the problem, we need to first convert the acceleration given in g units to meters per second squared (m/s²). We know that 1 g is equivalent to 9.8 m/s².

Hence, 60 g is equivalent to 60 × 9.8 m/s² = 588 m/s².

Substituting the values in the above formula, we get,d = (0 - u^2) / 2a= u^2 / 2a, since the final velocity is 0 when the person comes to a complete stop= u^2 / 2 × 588= u^2 / 1176 m

The time taken, t = 38 ms = 0.038 s.

Now, we know that acceleration, a = (v - u) / t.

We can rearrange the above equation to find the final velocity, v. We get,v = u + at

Substituting the values, we get,588 = u + (588 × 0.038)u = 588 - (588 × 0.038)u = 567.816 m/s

Using the value of u, we can now find the distance traveled using the kinematic equation as, d = u^2 / 1176= (567.816)^2 / 1176≈ 273.42 m.

Therefore, the distance traveled by the person to come to a complete stop in 38 ms at a constant acceleration of 60 g is approximately 273.42 meters.

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\({\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\\)

\(\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }\)

\(\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }\)

\(\\\)

\({\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\\)

\(\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }\)

\(\\\)

\({\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\\)

☯ As we know that,

\(\\\)

\(\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}\)

\(\\\)

\(\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }\)

\(\\\)

\(\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}\)

The potential energy of the system have been observed as 0.7 J.

We know that the potential energy is the energy that is possessed by an object that is found at a given height. The implication of this is that the potential energy depends on the height of the object.

If we decrease the height of the object then the potential energy of the body would also change. We have been told that the mass of the balls are 30 g each hence a total mass of 120 g or 0.12 Kg.

The change in the potential energy is;

0.12 * 9.8 * (0.9 - 0.3)

= 0.7 J

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The acceleration of the ship is –0.0012 m/s²

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

Velocity = displacement / time

Velocity = 2200 / 425

Velocity = 5.2 m/s

The acceleration of the ship can be obtained as follow:

a = (v – u) / t

a = (5.2 – 5.7) / 425

a = –0.5 / 425

a = –0.0012 m/s²

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Answer:

J = FΔt

J = mΔv

J = Δp

Explanation:

Answer:

two identical

Explanation:

Answer:

7.43m/s

Explanation:

Given parameters:

Initial velocity = 1.7m/s

Height  = 2.5m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we apply one of the motion equation. The most fitting one given the parameters is;

           V²  = U² + 2gh

where V is the final velocity

           U is the initial velocity

          g is the acceleration due to gravity = 9.8m/s

           h is the height

Input the parameters and solve;

    V² = 2.5² + 2(9.8)(2.5)

    V = 7.43m/s

Answer:

U = 41.65 J

Explanation:

We have just to calculate it with the following expression of gravitational potential energy:

U = m*g*h

Where:

m: mass

g: gravity acceleration;

h: height.

So:

U = m * 9.8 * 0.85

U = 8.33 * m

Then, it is just putting the mass value of the dog in that equation to obtain the value of potential energy acquired by the dog when jumping.

The mass of this kind of dog is between 3.6kg - 5kg, so, the maximum value of the potential energy can be like this:

U = 5 * 9.8 * 0.85

U = 41.65 J

The work done by the force on the block is approximately 311.2 Joules.

To calculate the work done by the constant force of magnitude F = 45 N over a distance of 8 m at an angle of 30 degrees to the horizontal, we need to find the component of the force that acts parallel to the displacement.

The horizontal component of the force can be calculated using trigonometry:

F_horizontal = F * cos(angle)

            = 45 N * cos(30 degrees)

            = 45 N * (√3 / 2)

            ≈ 38.9 N

Now, we can calculate the work done by the force using the equation:

Work = Force * Distance * cos(theta)

where theta is the angle between the force and the displacement.

Work = F_horizontal * Distance * cos(0)

     = 38.9 N * 8 m * cos(0)

     = 38.9 N * 8 m

     = 311.2 Joules

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Answer:

D) 1003 N​

Explanation:

Given the following data;

Mass of man = 85 kg

Acceleration of elevator = 2 m/s²

Acceleration due to gravity, g = 9.8 m/s²

To find the force exerted by the man on the floor;

Force = mg + ma

The aim of this study is to develop a Sun-Earth-Moon model that will serve to describe "the real and apparent movements of the Sun-Earth-Moon system and the results of these movements".

A fundamental topic in astronomy, that is difficult To properly understand the why requires three-dimensional thinking and imagination. In this context;

• Students have difficulty understanding basic concepts of astronomy; H.

such as creating seasons, phases of the moon, lunar and solar eclipses that require three-dimensional thinking must be properly understood. So to teach difficult abstract concepts,

• To solve the problem of the lack of materials and tools related to matter.

• To enhance learning retention by engaging students' multiple senses,

• Capturing concepts visually and verbally increases skills. Remember the image of this concept when you meet its verbal equivalent, or reversed. Therefore, to help students remember what they are learning,

• Have students see tools, objects, and events that they cannot see perceive with their five senses,

• Show students tools, objects, and events that they cannot reach

or bring to class,

• Attract attention and arouse interest in a topic, needed a model to grow.

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Potential energy (PE) = m g h

Kinetic energy (KE ) =1/2 m v^2

Where:

m = mass = 0.06 kg

g = acceleration of gravity = 9.8 m/s^2

h = heigth = 1237 meters

v= speed

PE = 0.06 * 9.8 * 1237 = 727.356 J

PE = KE

727.356J = 1/2* 0.06 * v^2

Solve for v

727.356 J / ( 1/2 * 0.06 ) = v^2

24,245.2 = v^2

√24,245.2 = v

v = 156 m/s

The time elapsed between the first splash and the second splash is approximately 0.69 seconds.

To calculate this, we consider the motion of two rocks thrown simultaneously from a bridge. Heather throws a rock straight down with a speed of 14 m/s, while Jerry throws a rock straight up with the same speed.

We use the equation for displacement in uniformly accelerated motion: s = ut + (1/2)at^2.

For Heather's rock, which is thrown downwards, the initial velocity (u) is positive and the acceleration (a) due to gravity is negative (-9.8 m/s^2). The displacement (s) is the height of the bridge (46 m).

Solving the equation, we find two possible values for the time (t): t ≈ -4.91 s and t ≈ 1.91 s.

Since time cannot be negative in this context, we discard the negative value and consider t ≈ 1.91 s as the time it takes for Heather's rock to hit the water.

For Jerry's rock, thrown upwards, we use the same equation with the same initial velocity and acceleration. The displacement is also the height of the bridge, but negative.

Solving the equation, we find t ≈ -5.68 s and t ≈ 1.22 s. Again, we discard the negative value and consider t ≈ 1.22 s as the time it takes for Jerry's rock to reach its maximum height before falling back down.

To find the time difference between the first and second splash, we subtract t ≈ 1.91 s (Heather's rock) from t ≈ 1.22 s (Jerry's rock). This gives us a time difference of approximately 0.69 seconds.

Therefore, the time elapsed between the first splash and the second splash is approximately 0.69 seconds.

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To better understand crash dynamics we have to look at "the laws of motion."

The laws of motion

The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.

It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion:Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.

Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.

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The hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

This question involves vector addition, the resolution of vectors, the use of bearings, and trigonometry in the calculation of the hiker's movement.

This may appear to be a difficult problem, but with some visual aid and the proper use of mathematical formulas, the issue can be addressed correctly.

Resolution of VectorThe resolution of a vector is the process of dividing it into two or more components.

The angle between the resultant and the given vector is equal to the inverse tangent of the two rectangular components.

Angles will always be expressed in degrees in the solution.

The sine, cosine, and tangent functions in trigonometry are denoted by sin, cos, and tan.

The tangent function can be calculated using the sine and cosine functions as tan x = sin x/cos x. Also, in right-angled triangles, Pythagoras’ theorem is used to find the hypotenuse or one of the legs.

Distance Travelled North/SouthThe hiker traveled North for the first part of the hike and South for the second.

The angles that the hiker traveled in the first part and second parts are 53 degrees and 17 degrees, respectively.

The angle between the two is (180 - 53 - 17) = 110 degrees.

The angle between the resultant and the Northern direction is 110 - 53 = 57 degrees.

Using sine and cosine, we can calculate the north/south distance traveled to be 5 sin 57 = 4.02 km, and the east/west distance to be 5 cos 57 = 2.93 km.

Distance Travelled East/WestThe hiker walked East for the second part of the hike.

To calculate the distance travelled East/West, we must first calculate the component of the first part that was East/West.

The angle between the vector and the Eastern direction is 90 - 53 = 37 degrees.

Using sine and cosine, we can calculate that the distance travelled East/West for the first part of the hike is 5 cos 37 = 3.88 km.

To determine the net distance travelled East/West, we must combine this component with the distance travelled East/West in the second part of the hike.

The angle between the second vector and the Eastern direction is 17 degrees.

Using sine and cosine, we can calculate the distance traveled East/West to be 3 sin 17 = 0.86 km.

The net distance traveled East/West is 3.88 + 0.86 = 4.74 km.

Therefore, the hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

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