An airplane starts from rest and accelerates at a constant 3.00 m/s^2 It’s final displacement is 600m. What is the time value?

Answer:

The relationship is \(E=\frac{ kQ}{d^2}\)

Explanation:

The electric field strength is denoted by the symbol E,

the test charge is denoted be q and the source charge be Q

distance is denoted by d

Then the equation can be rewritten in symbolic form as

Electric field strength is = Force/charge

\(E=\frac{F}{q}------1\)

we know that the formula for force is given as

\(F= \frac{kQq}{d^2} ------2\)

where \(k= 9*10^9 N.m^2/C^2\)

and d is the separation distance between charges

We can insert the expression for Force in equation one

we have

\(E= \frac{kQ\sout{q}/d^2}{\xout{q}}--------3\)

We can strike out both qs in the numerator and denominator we have

\(E=\frac{ kQ}{d^2}\)

Answer:

:)

Explanation:

:(

Answer:

resultant force = 14 N  ( East direction)

Explanation:

A =   √( (4*7)² + (4*7)² )

A = 39.6 N

B = 4 * 7

B = 28 N

C = 2 * 7

C = 14 N

∑ y forces = Ay - B = (4*7) - 28 = 0

∑ x forces = Ax - C = (4*7) - 14 = 14 N

so the resultant force = 14 N  ( East direction)

Answer:

F • t= m•∆v

Explanation:

the impulse experienced by the object equals the change in momentum of the object. in equation form: F • t= m•∆v

Answer:

0.784 A

Explanation:

From the question,

Note that the current in the circuit is the same as the current flowing through the inductor since they are both connected in series.

I = VL/XL....................... Equation 1

Where I = current flowing through the circuit, VL = Voltage drop across the inductor,  XL = reactance of the inductor.

XL = 2πfL................. Equation 2

Given: f = 28.9 Hz, L = 1.07 H, π = 3.143

XL = 2(3.143)(28.9)(1.07)

XL = 194.38 Ω.

VL = V-Vf

VL = 170-17.6

VL = 152.4 V

Substitute these values into equation 1

I = 152.4/194.38

I = 0.784 A

The current in the circuit when combination is connected  should be 0.784 A.

SInce

we know that

I = VL/XL....................... Equation 1

Here,

I = current flowing through the circuit,

VL = Voltage drop across the inductor,  

XL = reactance of the inductor.

And,

XL = 2πfL................. Equation 2

Here

f = 28.9 Hz, L = 1.07 H, π = 3.143

So,

XL = 2(3.143)(28.9)(1.07)

XL = 194.38 Ω.

Now

VL = V-Vf

VL = 170-17.6

VL = 152.4 V

Now

I = 152.4/194.38

I = 0.784 A

hence, The current in the circuit when combination is connected  should be 0.784 A.

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Answer:

Explanation:

PE = mgh

h = PE/mg

h = 2000/(65.0(9.81))

h = 3.1365168979...

h = 3.14 m

a VERY strong man if he can do that!

The weight of the fourth casting is 48 kg.

Given - The weight of three castings -

45.302kg

44.982kg

45.716kg

To find - The average weight if the 4th casting is taken the average weight becomes 46kg what is the weight of the 4th casting​

Solution -

Let's calculate the average weight of the first three castings using the given weights:

Average weight = (Weight of first casting + Weight of second casting + Weight of third casting) / 3

Average weight = (45.302 kg + 44.982 kg + 45.716 kg) / 3

Average weight = 136.000 kg / 3

Average weight = 45.3333... kg (rounded to four decimal places)

Now, let's calculate the weight of the fourth casting by using the average weight and the desired overall average weight:

Total weight of all four castings = Average weight * 4

Total weight of all four castings = 46 kg * 4

Total weight of all four castings = 184 kg

To find the weight of the fourth casting, we subtract the total weight of the first three castings from the total weight of all four castings:

Weight of the fourth casting = Total weight of all four castings - Total weight of first three castings

Weight of the fourth casting = 184 kg - 136 kg

Weight of the fourth casting = 48 kg

Therefore, the weight of the fourth casting is 48 kg

12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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Answer:

There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we will look at time, speed, and velocity to expand our understanding of motion.

A description of how fast or slow an object moves is its speed. Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction. Because speed is a rate, it depends on the time interval of motion. You can calculate the elapsed time or the change in time, Δt, of motion as the difference between the ending time and the beginning time

The SI unit of time is the second (s), and the SI unit of speed is meters per second (m/s), but sometimes kilometers per hour (km/h), miles per hour (mph) or other units of speed are used.

When you describe an object's speed, you often describe the average over a time period. Average speed, vavg, is the distance traveled divided by the time during which the motion occurs.

vavg=

distance

time

You can, of course, rearrange the equation to solve for either distance or time

time =

distance

vavg

.

distance = vavg × time

Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is

vavg =

distance

time

=

150 km

3.2 h

47 km/h.

A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time, however, is its instantaneous speed. A car's speedometer describes its instantaneous speed.

Explanation:

Answer:

 h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Explanation:

Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’

let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy

starting point. Starting point

         Em₀ = U = m g h

final point. Just before the crash

         Em_f = K = ½ m v²

as there is no friction the mechanical energy is conserved

         Em₀ = Em_f

         mg h = ½ m v²

         v = √2gh

this speed is the same for the two bodies.

Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment

initial instant. Just when the superball starts contacting the ground

      p₀ = M v

this velocity is negative because it points down

final instant. Just as the superball comes up from the floor

      p_f = M v '

the other body does not move

      p₀ = p_f

     - m v = M v '

       v ’= -v

Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.

Let's use the subscript 1 for the marble and the subscript 2 for the superball

Third part. The superball and the marble collide

the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved

initial instant. Moment of shock

        p₀ = M \(v_{1'}\)+ m v_2

final instant. When the marble shoots out.

        P_f = Mv_{1f'}+ m v_{2f}

        p₀ = p_

        M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}

        M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})

in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved

       K₀ = K_f

       ½ M v_{1'}² + m v₂² = M v_{1f'}²  + ½ m v_{2f}²

        M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)

Let's set the relation  (a + b) (a-b) = a² - b²

      M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

let's write our two equations

           M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f})                 (1)

           M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

if we divide these two expressions

           (v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )

we substitute this result in equation 1 and solve

          v_{1f'}= (v₂ + v_{2f}) - v_{1'}

          M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})

           -M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}

          -M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}

          v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}

          v_{2f} = - \(( \frac{ M-m}{M +m } )\)) v₂ + 2 \((\frac{M}{M+m})\) v_{1'}

now we can substitute the velocity values ​​found in the first two parts

          \(v_{2f}\) = - ( \frac{ M-m}{M +m  } ) √2gh + 2(\frac{M}{M+m}) √2gh

we simplify

          v_{2f} = [( \frac{ M-m}{M +m  } ) + 2 (\frac{M}{M+m})] \(\sqrt{2gh}\)

let's call the quantity in brackets that only depends on the masses

          A = ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]

           v_{2f}= A \sqrt{2gh}

in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball

finally with the conservation of energy we find the height that the marble reaches

Starting point

          Emo = K = ½ mv_{2f}²

Final point

          Emf = U = m g h'

          Em₀ = Em_f

          ½ m v_{2f}² = m g h ’

          h ’= ½ v_{2f}² / g

         h ’= ½ (A \sqrt{2gh})² / g

         h ’= A² h

         h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

The object's speed shortly before it lands on the earth is 40 m/s.

The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.

The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;

The kinematic equation is as follows;

v = in which

v = 0+10 (4)

The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,

the object's final velocity before impact is 40 m/s.

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Answer:

201.8 days

Explanation:

The activity of a radioactive sample as a function of times is :

\($R=R_0e^{\frac{0.693t}{T_{1/2}}}$\)

Here, \($R_0$\) = the initial activity

          \($T_{1/2}$\) = half life

          t = elapsed time

Now rearranging the equation for time, t, we get:

\($\frac{R}{R_0}=e^{-\frac{0.693t}{T_{1/2}}}$\)

\($\ln\left(\frac{R}{R_0}\right)=-\frac{0.693t}{T_{1/2}}$\)

\($t=\frac{-\ln\left(\frac{R}{R_0}\right)T_{1/2}}{0.693}$\)

\($t=\frac{-\ln\left(\frac{25}{350}\right)\times 53}{0.693}$\)

 = 201.8 days

Therefore, the required time is 201.8 days

Answer:

Ok. Thanks.

I'll try it out.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

This observation that electromagnetic radiation behaves like particles is known as the photoelectric effect.

It provides evidence that electromagnetic radiation exhibits particle-like properties. When EMR with sufficient energy (above a certain threshold) interacts with a metal surface, it can cause the ejection of electrons from the metal.

This behavior indicates that EMR behaves as discrete packets of energy called photons, which transfer their energy to the electrons and cause their release. The photoelectric effect supports the particle nature of EMR and is a fundamental concept in the field of quantum mechanics.

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Answer:

watch mark rober on yt he explains how to win

Explanation:

look up "Mark rober egg drop"

Answer:

does anyone have the lab report???

Explanation:

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

0.4

Explanation:

\( \frac{1}{2} mv ^{2} \)

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

Answer:

d = 711.42 m

Explanation:

Given that,

The average velocity of a runner, v = 4.26 m/s

Time, t = 167 s

We need to find the runner's displacement.

The average velocity of an object is equal to the displacement per unit time. It can be given by :

\(v=\dfrac{d}{t}\\\\d=vt\\\\d=4.26\ m/s\times 167\ s\\\\d=711.42\ m\)

So, the runner's displacement is 711.42 m.

Answer:

The correct answer to the question will be compass i.e compass is not a part of a DC motor. Various parts of a DC motor are coil of wire, armature, field magnets, brushes and commutator. The coil of wire plays a vital role in DC motor.

Explanation:

Answer:

It has a mass of 40 kg.

Explanation:

Because Force = mass x Acceleration or F = m a, we could say that the mass is force/acceleration which in your case is 2,400/60 which equals 40 kg.

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .

For calculating the Acceleration , we will use the Formula :-

\( \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} \)

Where ,

Therefore , by Substituting the given values in the above Formula :-

\( \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } \)

\( \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} \)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}\)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}\)

\( \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} \)

Hence :-

\( \underline {\rule {212pt} {4pt}} \)

Answer:

-2

Explanation:

Answer:

the average acceleration due to gravity on earth is 9.8m/s2 and on moon it is 1/6 of the earths acceleration due to gravity

Answer:

earth attracts moon by means of acceleration due to gravity

Explanation:

The correct answer is "c. horizontal component of the normal force." When a railroad train rounds a banked track, the centripetal force needed comes from friction, but from the horizontal component of the normal force.

It is the component of the normal force that acts perpendicular to the direction of motion. The normal force is the force exerted by a surface on an object that is in contact with it. It acts perpendicular to the surface and helps to support the weight of the object. In the case of a train rounding a banked track, the normal force acts perpendicular to the surface of the track and helps to keep the train from sliding off the track.

The horizontal component of the normal force is the component of the normal force that acts in the direction of motion. It acts parallel to the surface of the track and helps to provide the centripetal force needed to keep the train moving in a circular path.

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The wavelength, A, of a photon that has an energy of E= 4.73 × 10⁻¹⁹ J is 4.17 × 10⁻⁷ m.

Photon energy can be calculated using the equation E = hc/λ, where λ is the wavelength of the photon.

We know that the photon energy, E, is given as E = 4.73 × 10⁻¹⁹ J.

Additionally, we know that Planck's constant, h, is equal to 6.626 × 10⁻³⁴ J·s and the speed of light, c, is equal to 2.99 × 10⁸ m/s.

Substituting the known values into the equation for photon energy, we get: E = hc/λ4.73 × 10⁻¹⁹ J = (6.626 × 10⁻³⁴ J·s)(2.99 × 10⁸ m/s)/λ

Simplifying the expression above, we get:λ = hc/Eλ = (6.626 × 10⁻³⁴ J·s)(2.99 × 10⁸ m/s)/(4.73 × 10⁻¹⁹ J)λ = 4.17 × 10⁻⁷ m

Therefore, the wavelength, A, of a photon that has an energy of E= 4.73 × 10⁻¹.

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Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = \( \frac{1}{2} \times m \times {v}^{2} \)

Plugging the values,

\( = \frac{1}{2} \times 2 \times {(6)}^{2} \)

Reduce the numbers with Greatest Common Factor 2

\( = {(6)}^{2} \)

Calculate

\( = 36 \: joule\)

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= \(\frac{1}{2}\) × m × v²

By substituting the values, we get

= \(\frac{1}{2}\) × 2 × (6)²

=  \(\frac{1}{2}\) × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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The seven risk factors for heart attack are Age, Tobacco usage, high blood pressure, high cholesterol level, obesity, diabetes and metabolic syndrome.

Some other risk factors for heart attack are:

To prevent heart attack, one must,

Therefore, the seven risk factors for heart attack are Age, Tobacco usage, high blood pressure, high cholesterol level, obesity, diabetes and metabolic syndrome.

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Answer:

h = 1.07 m

Explanation:

It is given that,

The average sped of the kingfisher, v = 4.6 m/s

Time, t = 1.4 s

Final speed, v = 0 (as it hits the water)

We need to find the height from which the bird dove. Let the height is h. According to the conservation of energy.

\(mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(4.6)^2}{2\times 9.8}\\\\h=1.07\ m\)

So, the bird dove from a height of 1.07 m.