How much force is needed to accelerate a 1400 kg car from rest to 14 meters/sec in 7 seconds?

So in the position of 1 and 5 height is maximum from the ground and here velocity is also minimum so potential energy is maximum and in the position of no 3 velocity maximum so kinetic energy will be maximum.

We know total energy that is kinetic energy + potential energy is always constant so where kinetic energy is maximum there potential energy will be minimum and where potential energy is maximum there kinetic energy will be minimum.

we know potential energy=m g h

where h=height

So in the position of 1and 5 height is maximum from the ground and here velocity is also minimum so potential energy is maximum and in the position of no 3 velocity maximum so kinetic energy will be maximum.

Kinetic energy, which may be seen in the motion of an object, particle, or group of particles, is the electricity of motion. Any moving item uses kinetic energy, such as a person who walks, a baseball being thrown, a piece of food falling from a table, or a charged particles in an electric field.

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Answer:

number 3

Explanation:

the effort is almost equal to the load

The ball was in motion for 16.7 seconds before coming to rest.

Let us assume that the ball was in motion for a time equal to 't' seconds before coming to rest.

Given in the question

Initial velocity = u = 50 m/s

Final Velocity = v = 0 m/s ( As the ball stops moving after some time, its final velocity hence becomes zero. )

Acceleration = a = -3 m/s²

To find

Time = t seconds

Using the first equation of motion,

v = u + at

Put in the value in the equation, we get

0 = 50 + (-3)(t)

3t = 50

t = 50/3

t = 16.66 seconds

t = 16.7 seconds ( rounding off to one decimal place )

Therefore, the ball was in motion for 16.7 seconds before coming to rest.

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A shaded pole is a short-circuited winding, which consists of a single turn of copper wire and acts on only a portion of the stator windings.

Most single-phase motors just require the two start leads to be switched in order to turn the rotor in the opposite direction.

An AC motor that starts using a shaded stator pole. A partial short-circuited winding that affects only some of the stator windings, made up of a single turn of copper wire.

The motor is momentarily changed into a two-phase motor during the starting time to get around this problem and make it self-start. A single-phase motor’s stator is equipped with an additional winding, referred to as the starting winding, for this function (or auxiliary).

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The first acceleration is 20276.6 m/s² and the second one is 4427.2 m/s².

                         [tex]\begin{aligned}\\\\\omega &= 2\pi.f\\\\&= 2\pi.\big(\frac{4300\,\text{rpm}}{60\,s})\\\\&= 450.29\, rad/s\end{aligned}[/tex]

                         [tex]\begin{aligned}\\\\a &= r\omega^2 \\\\&= 0.1\, m\,\times\,(450.29\,rad/s)^2\\\\&= 20276.1\, m/s^2 \end{aligned}[/tex]

                            [tex]\begin{aligned}\\\\v &= \sqrt{2.\times 9.8\,m/s^2\,\times 1.0\,m }\\\\&= 4.427\,m/s\\\\a &=\frac{\Delta v}{\Delta t}\\\\&= \frac{0-4.427\,m/s}{0.001\,s}\\\\&= -4427\, m/s^2\cdots(\text{- indicates a deceleration})\end{aligned}[/tex]

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the smallest deceleration needed, in m/s^2, to avoid rear-ending the car

will we of a = -12.5m/s^2.

A reduction in velocity, or momentum, is referred to as a deceleration. The slowing of a rise is referred to as a deceleration. For instance, the wind has a certain amount of momentum. We may say that the wind is decelerating when its momentum slows down, i.e., when it blows less forcefully.

To avoid rear-ending the car

Reduce final velocity of the car to less or equals to the speed of bad driver's car before car travels 20 m

Because if final velocity of car is less than or equal to the speed of bad driver's car then car can't able to cross bad driver's car.

Final velocity of the car (Vf) = 20 m/s

Using equation of motion

[tex]v^{2} _{f} = v^{2} _{i} + 2as\\ a = v^{2} _{f} - v^{2} _{i} / 2s\\= 20 ^{2} - 30 ^{2}/ 2 *20\\= -12.5m/s^{2}[/tex]

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The exoplanet where the human expedition landed has a surface gravity acceleration of -0.20 meters per square second.

The relationship between a moving body's distance, velocity, acceleration, and time is known as the equation of motion.

Here, is the beginning body, is the final body, is the acceleration of the body, and is the distance travelled. This is the third equation of motion for distance.

The greatest distance leapt by one of the explorers is 19.5 metres, as stated in the problem.

The starting velocity is 2.80 m/s.

It is necessary to determine the gravitational acceleration on the exoplanet's surface.

The explorers' velocity is currently zero. As a result, according to the third equation of motion, the explorers are moving downhill as shown by the negative sign.

Thus, the gravitational acceleration on the surface of the exoplanet is -0.20 meter per squared second.

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The work done by tension force in pendulum is zero.

We need to know about pendulum experiment to solve this problem. The pendulum is a simple experiment to calculate gravitational acceleration.

The work done is proportional to force and displacement. It can be determined as

W = F . s

where W is work done, F is force and s is displacement.

From the question above, the given parameters are

mass is m  

string of length is l

angle is θ

F = T

s = 0

In pendulum motion, there is no displacement so the work done by the tension force is zero.

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Answer:

Q=Mcø

528=m×4200×50

m=528÷(4200×50)

m=0.002514kg

2.514g

Compare the energy of a 1-megaton hydrogen bomb to the energy released by a major earthquake. Earthquake/Ebomb = 5.

Compare the energy of a 1-megaton hydrogen bomb to the energy released by a major earthquake. Earthquake/Ebomb = 5.

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Ammonia is a weak base and Potassium hydroxide is a strong base, then then the solution of potassium hydroxide will always have a higher pH than a solution of ammonia, therefore the correct answer is option C.

A chemical compound is a combination of two or more  either similar or dissimilar chemical elements.

As given in the problem, ammonia is a weak base. Potassium hydroxide is a strong base then we have to find out which statement is correct from the given list of statements,

Since ammonia is a weak base and potassium hydroxide is a strong base, their solutions will always have a higher pH than ammonia solutions. Option C is the right response.

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Answer:

penalty

Explanation:

Penalties can gain your team yards if assessed against the opposition...their are multiple penalties which are assessed 10 yards

Answer:

77ruririfusjd u 6r7r u y9r8

A star is moving directly toward earth at a large fraction of the speed of light. The light we observe from this star compared to the light received from the same star if it were at rest relative to earth will be blueshifted.

The amount of light emitted by a star coming towards earth would have a shorter wavelength as compared to other sources of light. This refers to the doppler shift principle. According to the doppler shift principle, if an observer is moving in relation to a source, then the change in frequency in the light waves would be relative to that observer.

In this case, we will observe that the star approaching earth will shift its spectral lines toward the blue end of the light series.

Since blue light has a shorter wavelength, therefore, the large fraction of the speed of light coming out from a star would correspond to the blue end of the light spectrum hence called blueshift.

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gravitational potential energy was gained when it was winched up 15 m is  73500 J.

GPE is turned

into kinetic energy as the pumpkin drops

Then,

pumpkin speed is 17.14 m/s.

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The prevention factors required to prevent mother-to-child transmission are:  

Mother-to-child transmission also referred to as Perinatal is described as  when HIV is transmitted from the mother to the child during pregnancy, childbirth, or breastfeeding. For ladies who are pregnant, HIV counseling and testing is the primary and most basic step in lessening mother-to-child transmission.  Prenatal Treatment , when taken all through pregnancy, conveyance, and breastfeeding, can decrease the chance of HIV transmission to a newborn child to less than 5%. Ladies who test HIV-positive amid their pre-birth visits at EGPAF-supported locales are quickly connected to long-lasting Craftsmanship and counseled on the benefits of enrollment and maintenance on treatment.  Safe childbirth is an imperative thought for all hopeful moms and especially for HIV-positive ladies. Post-natal care is eesential for both the mother and the baby especially since it includes child wellness services such as nutrition, immunization,  and growth monitoring.

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The distance the soccer ball cover, until it stops, is 62.5m

The distance the ball rolls up to before coming to rest may be calculated using Newton's third equation of motion:

v² = u² + 2as .......(i)

Where v ⇒ Final velocity

u ⇒ Initial velocity

a ⇒ Acceleration

s ⇒ Displacement

Now as per the question,

v = 0 m/s

u = 25 m/s

a = -5m/s²

Putting these values in equation (i)

0 = (25)² - 2× 5 × s

0 = 625 - 10s

10s = 625

s = 625/10

s = 62.5m

Therefore, the distance the ball cover, until it stops, is 62.5m.

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A girl delivering newspapers covers her route by traveling so resultant displacement is 5 blocks.

Her net x (east-west) displacement is –3.00+0+6.00= +3.00 blocks, while her net y (north-south) displacement is 0+4.00+0=+4.00 blocks.

The magnitude of the resultant displacement is

R= [tex]\sqrt{x^{2} +y^{2} } =\sqrt{3^{2} + 4^{2} }[/tex] = 5 blocks

and the angle made by the resultant

= [tex]\theta[/tex] = [tex]tan^{-2} =[/tex] 4/3

= 53.1°

The resultant displacement is 5 blocks at 53.1°

The term “displacement” refers to a shift in an object’s position. It is a vector quantity with a magnitude and direction. The symbol for it is an arrow pointing from the initial location to the ending place.

For instance, if an object shifts from location A to position B, its position changes.

An object’s displacement is equal to its velocity times its acceleration times its time squared plus one-half (t2).

A scalar quantity known as distance measures “how much ground an object has traversed” while moving.

The term “how far out of place an object is” is referred to as displacement, a vector quantity.

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If the starship drops the engine force output to 3000 N of force he ship's acceleration going to be at acceleration, a = 4.28 m/s².

In the given equation we can see the different force working on engine so,

To find the acceleration first we need the mass of engine

Using formula,

F = m x v

10,000 - 3000 N = m x 10 m/s

7000N = m x 10 m/s

m = 700 kg

Now as we have mass of the engine,

By the formula,

a = F / m

F = 3000 N

m = 700 kg

So, putting values we have :

a = 3000 / 700

a = 4.28 m/s²

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Weight of the vehicle on Earth is 6736.72 kg.

What is the weight of an object?

The weight of an object is the force exerted on it owing to gravity in science and engineering. Weight is defined as a vector quantity in certain standard textbooks, as the gravitational force exerted on the item. Others describe weight as a scalar quantity, the magnitude of gravity.

Weight on Mars = (Weight on earth/10) x 3.711

Given, weight on mars = 2500 kg

Therefore, weight on Earth = 6736.72 kg.

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The expression for horizontal magnitude acceleration of van is a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m

Given that,

The magnitude of the force applied by Hans on van = [tex]F_{H}[/tex]

The magnitude of the force applied by Frans on van = [tex]F_{F}[/tex]

Let us consider the horizontal magnitude of acceleration as a and the mass of van as m.

The horizontal forces acting on the van are [tex]F_{H}[/tex] and [tex]F_{F}[/tex].

∑[tex]F_{Hor}[/tex] = [tex]F_{H}[/tex] + [tex]F_{F}[/tex]

m a =  [tex]F_{H}[/tex] + [tex]F_{F}[/tex]

a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m

The vertical forces here are the force due to weight and the normal force due to its weight on ground.

Therefore, the expression for horizontal magnitude acceleration of van is a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m

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Surface Charge density on each plate  sigma =+- Q/a = +4.9×10²⁴/4.6×10⁻²C/m² and Electric field is found to be 0.12×10³⁸N/C.

Given, here the cross sectional area of the conducting plate (A)

A = 460 cm² = 460 x 10-4 m² = 4.6 x10-²m² and distance between plate, d = 4.0 cm = 4x10² m²

Charge on the plate, Q = ne ,where n= 1×10¹²

Q = ± 10¹² x4.9x 10¹²c = ± 4.9x 10²⁴ c

Now, we have to find charge density on each plate  sigma =+- Q/a = +4.9×10²⁴/4.6×10⁻² C/m²

σ= 1.09× 10²⁶ C/m²

Now we have to find electric field between the plates.

E= σ /E°= 1.09× 10²⁶/ 8.85×10⁻¹²

E= 0.12×1038N/C

Charge density is a measure of the amount of electric charge present in a space per unit volume in one, two, or three dimensions. The quantity of electric charge per surface area or volume, respectively, is known as the linear surface or volume charge density.

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The tension in the string is 19.8 N.

Find the tension in the string:

The net force on the block will be:

T - Fr = Fnet = ma

Where T is the tension and Fr is the frictional force

T - ukR = ma

Where R is the upwards normal force equal to the block's weight. R = mg

therefore,

T - uk mg = ma

T - 0.2(5 x 9.8) = 5(2)

=> T = (0.2 x 5 x 9.8) + 5(2) = 19.8 N

This is the tension in the string.

For the free body diagram, draw a block and show a rope connected horizontally to it. The tension in the string will be along the rope outwards from the block. Now, the frictional force will be between the ground and the block and it will be in the opposite direction to the direction of tension of the rope. The force due to gravity will act vertically downwards while the normal force R will act vertically upwards.

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The temperature from t=0 to 10min with a step size of 2 min is as follows:

t (in min)   0      2       4           6         8         10

T ( in °C )  70  68.1  66.27  64.514  62.82  61.2

We know that,

The solution of first order differential equation is

dT / dt = - k ( T - [tex]T_{a}[/tex] )

where,

T = Initial temperature

[tex]T_{a}[/tex] = Ambient temperature

Given that,

k = 0.019 / min

[tex]T_{a}[/tex] = 20 °C

Using Euler method

[tex]T_{i} ^{'}[/tex] = - 0.019 ( [tex]T_{i}[/tex] - 20 )

[tex]T_{i+1}[/tex] = [tex]T_{i}[/tex] + ( [tex]T_{i+1}[/tex] * h )

With t = 0 to 10, h = 2, the iterative solution is as follows

At i=1, [tex]t_{i}[/tex] = 0, [tex]T_{i}[/tex] = 70

[tex]T_{i} ^{'}[/tex] = - 0.019 ( 70 - 20 ) = - 0.95

[tex]T_{i+1}[/tex] = 70 - ( 0.95 * 2 ) = 68.1

At i=2, [tex]t_{i}[/tex] = 2, [tex]T_{i}[/tex] = 68.1

[tex]T_{i} ^{'}[/tex] = - 0.019 ( 68.1 - 20 ) = - 0.914

[tex]T_{i+1}[/tex] = 68.1 - ( 0.914 * 2 ) = 66.27

At i=3, [tex]t_{i}[/tex] = 4, [tex]T_{i}[/tex] = 66.27

[tex]T_{i} ^{'}[/tex] =  - 0.019 ( 66.27 - 20 ) = - 0.88

[tex]T_{i+1}[/tex] = 66.27 - ( 0.88 * 2 ) = 64.514

At i=4, [tex]t_{i}[/tex] = 6, [tex]T_{i}[/tex] = 64.514

[tex]T_{i} ^{'}[/tex] =  - 0.019 ( 64.514 - 20 ) = - 0.846

[tex]T_{i+1}[/tex] = 64.514 - ( 0.846 * 2 ) = 62.82

At i=5, [tex]t_{i}[/tex] = 8, [tex]T_{i}[/tex] = 62.82

[tex]T_{i} ^{'}[/tex] =  - 0.019 ( 62.82 - 20 ) = - 0.8136

[tex]T_{i+1}[/tex] = 62.82 - ( 0.8136 * 2 ) = 61.2

At i=6, [tex]t_{i}[/tex] = 10, [tex]T_{i}[/tex] = 61.2

Euler's method is used for solving ordinary differential equation with a given initial value. It is a first order numerical procedure.

The given question is incomplete. The complete question is:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature), dT/dt = -k(T-Ta) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute), and Ta = the ambient temperature (°C). Suppose that a cup of coffee originally has a temperature of 70 °C. Use Euler's method to compute the temperature from t = 0 to 10 min using a step size of 2 min if Ta = 20 °C and k = 0.019/min.

Therefore, the temperature from t =0 to 10 min with a step size of 2 min is as follows:

t (in min)   0      2       4           6         8         10

T ( in °C )  70  68.1  66.27  64.514  62.82  61.2

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the frequency of each wavelength of electromagnetic radiation is

A)frequency of wavelength 632.8nm(red light from helium–neon laser)

is 4.738×10¹⁴ s⁻¹

B)frequency of wavelength 503 nm (maximum solar radiation)

is5.960×10¹⁴ s⁻¹

C)frequency of wavelength 0.052nm (medical x-rays)

is5.765×10¹⁸s⁻¹

A)

632.8 nm (wavelength of red light from helium–neon laser)

1 nm = 10⁻⁹ m

632.8 nm = 632.8×10⁻⁹ m = 6.328×10⁻⁷ m

The speed of light is c = 2.998×10⁸ m/s

λ = 6.328×10⁻⁷ m

the frequency can be calculated as

ƒ = c / λ

ƒ = (2.998×10⁸m/s) / (6.328×10⁻⁷ m)

ƒ = 4.738×10¹⁴ s⁻¹

frequency of wavelength 632.8nm is 4.738×10¹⁴ s⁻¹

B)

503 nm (wavelength of maximum solar radiation)

1 nm = 10⁻⁹ m

503 nm = 503×10⁻⁹m = 5.03×10⁻⁷ m

c = 2.998×10⁸ m/s

λ = 5.03×10⁻⁷ m

the frequency can be calculated as;

ƒ = c / λ

ƒ = (2.998×10⁸ m/s) / (5.03×10⁻⁷m)

ƒ = 5.960×10¹⁴ s⁻¹

frequency of wavelength 503 nm is5.960×10¹⁴ s⁻¹

C)

0.052 nm (wavelength contained in medical x-rays)

1 nm = 10⁻⁹ m

0.0520 nm = 0.0520×10⁻⁹ m = 5.20×10⁻¹¹m

Where the speed of light is

c = 2.998×10⁸ m/s

λ = 5.20×10⁻¹¹ m

the frequency can be calculated as

ƒ = c / λ

ƒ = (2.998×10⁸ m/s) / (5.20×10⁻¹¹ m)

ƒ = 5.765×10¹⁸s⁻¹

frequency of wavelength 0.052nm is5.765×10¹⁸s⁻¹

A)frequency of wavelength 632.8nm(red light from helium–neon laser)

is 4.738×10¹⁴ s⁻¹

B)frequency of wavelength 503 nm (maximum solar radiation)

is5.960×10¹⁴ s⁻¹

C)frequency of wavelength 0.052nm (medical x-rays)

is5.765×10¹⁸s⁻¹

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OD. 2kg of liquid oxygen at -225 *C has the least thermal energy.

The thermal energy for a phase of matter is given by the formula

Q = mCΔT where

m = mass

C = Specific heat constant

T = Temperature

For Oxygen, the Specific heat constant = 0.92 J/gK

Let's calculate the thermal energy for each case, assuming that the initial temperature is 0 Kelvin

OA.

m= 5000 g

ΔT = 303 K

So Q =5000×0.92×303 = 1393800 J = 1393.8 KJ

OB.

m= 5000 g

ΔT = 48 K

So Q =5000×0.92×48 = 220800 J = 220.8 KJ

OC.

m= 2000 g

ΔT = 293 K

So Q =2000×0.92×293 = 539120 J = 539.12 KJ

OD.

m= 2000 g

ΔT = 48 K

So Q =2000×0.92×48 = 88320 J = 88.32 KJ

Therefore, OD 2kg of liquid oxygen at -225 *C has the least thermal energy which is 88.32 KJ.

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Answer:

2 kg of liquid oxygen at -225°C

Explanation:

B has the same temperature, yet it's more mass. So, to use this as an example--B has the most thermal energy. If the mass decreased, then the thermal energy would decrease. More mass=more thermal energy. Hence, why the mass decreasing to 2 kg would equal the least amount of thermal energy.

The substante that is insoluble and floats is the wood chip and the one that is insoluble is a sugar cube.

Solubility is a property that a substance has to dissolve in another that is called a solvent, this can also depend on the conditions of the temperature to which the solution can be exposed and also the pressure. The solubility will depend on the properties of each of the components, it is not always the same in all types of solutions.

When no more of the solute can dissolve in a solution, the solution is said to be saturated.

Therefore, we can confirm that the substante that is insoluble and floats is the wood chip and the one that is insoluble is a sugar cube.

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