An apple with mass M is hanging at rest from the lower end of a light vertical rope. A dart of mass M/4 is shot vertically upward, strikes the bottom of the apple, and remains embedded in it.

Distance = (speed) x (time)

Distance = (375 km/hr) x (2 hr)

Distance = 750 km

(That's about 233 miles per hour.  A hot-air balloon moving at that speed is in big serious trouble.  I hope there's nobody trapped in the basket.)

Jack speed as he hits the water is 14.13m/s

What is speed?
The speed of an object, also known as v in 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, making it a scalar quantity. The instantaneous speed is the upper limit of the average speed as the duration of the time interval approaches zero. The average speed of an item in a period of time is equal to the distance travelled by the object divided by the duration of the period. Velocity and speed are not the same thing.

The only force on Jack is gravity, which points straight down, and has an acceleration of 9.91m/s².  Whether he's falling off a bridge or simply standing still on the surface of the earth, the acceleration due to gravity is the same.

If you're looking for the speed of Jack as he hits the water, you just need to solve for v in the equation\($v^2 = v_0^2 + 2ad$\).  
Here,
v is Jack's speed as he hits the water,
\($v_0$\) is Jack's initial speed (which is zero since he's not moving when he jumps),
a is Jack's acceleration (which is 9.91m/s²), and
d is Jack's displacement (which is the height of the bridge, which you said is 10m).  
This gives you an answer of \($v = \sqrt{2ad} = \sqrt{2(9.91)(10)} = 14.13 \ \mathrm{m/s}$\)

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Answer:

P = 0.98 W

Explanation:

Given that,

Mass, m = 1 kg

Height, h = 0.9 m

(a) We need to find power must the battery supply to the vehicle for a run time of 9 seconds. Power is given by work done divided by time. It can be given by :

\(P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{1\times 9.8\times 0.9}{9}\\\\P=0.98\ W\)

So, the required power is 0.98 W.

Answer: Gravitational Forces

Explanation:

Answer:

e

Explanation:

the field points to the right

If loop is kept horizontally  and direction of current is clockwise , then according to Fleming’s Left Hand Rule the direction of magnetic field will be towards right .

Fleming’s Left Hand Rule states that if we arrange our thumb, forefinger and middle finger of the left-hand perpendicular to each other, then the thumb points towards the direction of the force experienced by the conductor, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the electric current.

If loop is kept horizontally  and direction of current is clockwise , then according to Fleming’s Left Hand Rule the direction of magnetic field will be towards right .

correct answer e) The field points to the right

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Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

Answer:

Change its velocity by 10 m/s in 1s

Explanation:

Based on the diagram, the correct statement is: Government and consumers make economic decisions in a mixed economy, while consumers make economic decisions in a market economy.

In a mixed economy, economic decisions are made by both the government and consumers.

The government plays a significant role in regulating and influencing economic activities through policies, regulations, and interventions.

In market economy, economic decisions are primarily made by consumers. The market forces of supply and demand dictate the allocation of resources, production levels, and pricing.

The freedom to buy and sell whatever they choose is what ultimately determines how commodities and services are produced and distributed.

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Hi there!

Recall the equation for work given a potential difference and charge:
\(\large\boxed{W = qV}\)

W = Work (J)
q = charge (C)
V = Potential difference (V)

The work is equivalent to the change in KINETIC ENERGY, and only depends on the particle's charge and the potential difference that it experiences. Mass has no impact on this.

Thus:

\(W' = (2Q)(V) = 2W\)

This is equivalent to:

\(\large\boxed{2W = \text{ d) } 2K}}\)

After time t, the position function of the ball is determined as \(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

The given parameters;

The position function of the ball after time t, is calculated as follows;

\(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.

Thus, after time t, the position function of the ball is determined as \(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

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Answer:

Substitute the numbers for YOUR problem this is NOT the answer.

Explanation:

False. E=hf, where h is Planck's constant, c is the speed of light, f is the frequency, and is the wavelength; and E=hc/, where E is directly proportional to frequency and inversely proportional to wavelength.

With respect to the wavelength of the radiation, photon energy is inversely proportional.

Two formulas can be used to determine a photon's energy: E = h f is a formula that can be used if the photon's frequency is known. This equation, sometimes known as Planck's equation, was created by Max Planck.

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The short summary of the concept of dark matter and dark energy is given here.

An estimated 85% of the universe's mass is assumed to be made up of dark matter, a hypothetical type of stuff.  Because it does not appear to interact with the electromagnetic field—that is, it does not absorb, reflect, or emit electromagnetic radiation—dark matter is referred to as being "dark," making it challenging to detect.

Numerous astrophysical observations support the existence of dark matter, including gravitational effects that cannot be described by the gravity theories currently in use without the presence of more matter than can be observed. Because of this, the majority of scientists believe that dark matter is prevalent in the universe and has significantly influenced both its structure and evolution.

Dark energy is an undiscovered type of energy that has the largest effects on the cosmos according to physical cosmology and astronomy. Supernova observations provided the first direct proof for its existence by demonstrating that the cosmos is expanding faster than ever rather than at a fixed rate. It is necessary to understand the universe's origins and initial elements in order to comprehend its evolution.

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Answer:

The  drag coefficient is  \(D_z = 1.30512\)  

Explanation:

From the question we are told that

     The density of air is  \(\rho_a = 1.21 \ kg/m^3\)

     The diameter of bottom part is  \(d = 0.15 \ m\)

The  power trend-line  equation is mathematically represented as

      \(F_{\alpha } = 0.9226 * v^{0.5737}\)

let assume that the velocity is  20 m/s

Then

      \(F_{\alpha } = 0.9226 * 20^{0.5737}\)

       \(F_{\alpha } = 5.1453 \ N\)

The drag coefficient is mathematically represented as

      \(D_z = \frac{2 F_{\alpha } }{A \rho v^2 }\)

Where  

     \(F_{\alpha }\) is the drag force

      \(\rho\) is the density of the fluid

       \(v\) is the flow velocity

       A is the area which mathematically evaluated as

       \(A = \pi r^2 = \pi \frac{d^2}{4}\)

substituting values

     \(A = 3.142 * \frac{(0.15)^2}{4}\)

     \(A = 0.0176 \ m^2\)

Then

   \(D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }\)

   \(D_z = 1.30512\)  

Speed is defined as a change in position over time. Position's derivative is velocity, and velocity's derivative is acceleration.

                             S = d/t

                               v = v 0 + a t

                          v = s/t

                      a = (v - v0) / t = Δv / t

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C. 10,200 N is how much energy was lost due to air resistance while she was bouncing

The energy lost due to air resistance while the bungee jumper was bouncing can be calculated by finding the total mechanical energy of the system at the beginning of the jump and comparing it to the total mechanical energy at the end of the jump.

At the beginning of the jump, the total mechanical energy is given by:

Ei = mgh

where m is the mass of the bungee jumper, g is the acceleration due to gravity, and h is the height of the bridge. Therefore, at the beginning of the jump:

50 x 30 x 10 - 1/2 x 30^2 x 10

= 15000 - 4500

= 10,200 N

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Answer:

The pressure at Mombasa is 101.396 KPascal.

Explanation:

Given: Reading on barometer = 760 mm

                                                  = 0.76 m

           Density of mercury = 1.36 x \(10^{4}\) kg/\(m^{3}\)

In this case, the pressure can be expressed as:

P = σhg

Where: σ is the density of mercury, h is height of mercury in the barometer, and g is the acceleration due of gravity.

But, g = 9.81 m/\(s^{2}\)

So that,

P = 1.36 x \(10^{4}\) x 0.76 x 9.81

  = 101396.16 Pascal

P = 101.395 KPascal

The pressure at Mombasa is 101.396 KPascal.

Answer:Magnetic fields from two sources add up as vectors at each point, so the strength of the field is not necessarily the sum of the strengths1. Magnetic fields are vectors, which means they have direction as well as size. Therefore, the sum of two magnetic fields is not simply the sum of their magnitudes2.

Explanation:

Answer:

a) 103.176 m / min

b) 1751.28 meters

Explanation:

Given:-

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

                         \(E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t\)

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

                         ( 0 ≤ t ≤ 15 ) mins                  

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

Find:-

a) Find the value of \(\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\) using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?  

Solution:-

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

                          \(f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx\)

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

                         \(E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\\)

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

                         \(E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\\)

- Complete the square in the denominator:

                          \(E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\\)

- Use the following substitution:

                          \(u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du\)

- Substitute the relations for (u) and (dt) in the above E_avg expression.

                          \(E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du\)

- Use the following standard integral:

                          \(arctan(u) = \int {\frac{1}{u^2 + 1} } \, du\)

- Evaluate:

                         \(E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |\)

- Apply back substitution for ( u ):

                        \(E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\\)

- Plug in the limits and find Emma's average velocity:

                        \(E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}\)

Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of  0≤t≤15 is given by the relation:

                    \(S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\\)

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of  0 ≤ t ≤ 15 is given by the relation:

                    \(S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\\)

 Apply integration by parts:

  \(S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\\)

 Re-apply integration by parts 2 more times:

 \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\\)             \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\\)    

\(S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\\)

- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L )  from S ( E ):

                    \(S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\\)

- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

              \(\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671\)

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value:

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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A hollow glass sphere has a density of 1.3g/cm at 20 C. Glycerine has a density of 1.26 g/cm at 20 C.

To determine the temperature at which the hollow glass sphere begins to float in glycerine, we need to calculate the density of glycerine at various temperatures and compare it to the density of the glass sphere.

The density of glycerine changes with temperature, so we need to use a density-temperature chart or equation to determine the density of glycerine at different temperatures.

Assuming the hollow glass sphere has a uniform wall thickness, we can calculate its volume by subtracting the volume of the hollow interior from the volume of the whole sphere

Volume of sphere = (4/3)π\(r^{3}\)

Volume of hollow interior = (4/3)π\((r-t)^{3}\)

Volume of glass wall = (4/3)π(\(r^{3}\) - \((r-t)^{3}\)), where t is the thickness of the glass wall.

From the density and volume of the glass sphere, we can determine its mass

Mass of glass sphere = Density of glass sphere x Volume of glass sphere

Next, we can use Archimedes' principle to determine the volume of glycerine displaced by the glass sphere when it is submerged in the glycerine

Volume of glycerine displaced = Mass of glass sphere / Density of glycerine at the given temperature

When the glass sphere floats, the volume of glycerine displaced will be equal to the volume of the glass sphere. Thus, we can set the two volumes equal to each other and solve for the temperature at which the density of glycerine matches the density of the glass sphere

Volume of glass sphere = Volume of glycerine displaced

(4/3)π\(r^{3}\) - (4/3)π\((r-t)^{3}\) = Mass of glass sphere / Density of glycerine at the given temperature

Hence, for the temperature requires knowing the radius and thickness of the glass sphere and the mass of the sphere.

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Work done by \(\overrightarrow w\)= 255 kg x 9.8 m/s² x 7 m = 17,890 J, while work done by \(\overrightarrow T\) 1 = 0 = perpendicular to displacement of piano.

Work and displacement are two terms related to physics. Work is a measure of energy and is defined as the amount of energy transferred to an object when a force is applied to it over a certain distance. In other words, work is the result of a force acting on an object over a distance. Displacement is a measure of the change in position of an object over a given distance, which is the difference between the initial and final positions of the object.

A) The work done by \(\overrightarrow w\) is 7m × 255kg × 9.81m/s² = 17,570 joules.

B) The work done by \(\overrightarrow T\) 1 is 7m × 255kg × 9.81m/s² × cos(30°) = 13,177 joules.

C) The work done by \(\overrightarrow T\) is 7m × 255kg × 9.81m/s² × cos(60°) = 4,393 joules.

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Complete question:

The speed of the microwaves is 3 × 10^10 meters per second. This result indicates that microwaves, like all electromagnetic waves, travel at the speed of light in a vacuum.

The speed of a wave can be calculated using the formula: speed = frequency × wavelength. In this case, the frequency of the microwaves is given as 10,000 million Hz, which is equivalent to 10,000 × 10^6 Hz or 10^10 Hz. The wavelength is given as 0.03 m.

Plugging these values into the formula, we have:

Speed = (10^10 Hz) × (0.03 m)

Simplifying the calculation, we find:

Speed = 3 × 10^10 m/s

Therefore, the speed of the microwaves is 3 × 10^10 meters per second. This result indicates that microwaves, like all electromagnetic waves, travel at the speed of light in a vacuum. The speed of light is approximately 3 × 10^8 meters per second, so microwaves have a slightly higher speed due to their longer wavelength. It's important to note that the speed of light is a fundamental constant of nature and does not depend on the properties of the specific electromagnetic wave being considered.

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Answer:

Since velocity is considered a vector quantity and vectors show you in what direction it goes the object wont be moving forward but backwards because its a negative and negatives go to the left and that would be considered backwards.

Explanation:

Hope this helps

Answer:

Answer is D: D.Fp and Fg

Explanation:

I did the quiz some time ago and rember it hope it helps.

Answer:

i just finish its d

Explanation: