write an expression for the normal force, fn, that the block experiences in terms of the elevator's acceleration, the block's mass, and the acceleration of gravity.

Answer:

i. Cyclist A travelled 15 km 1 h cyclist A had started his journey from point X

  Cyclist B travelled 18 km 1 h cyclist A had started his journey from point X

ii. cyclist B overtake cyclist A 6 km from the same starting point X.

Explanation:

From the question,

- Cyclist A and B cycled at an average speed of 15 km/h and 20 km/h respectively.

- Cyclist B started his journey 6 mins after cyclist A had started.

Let the cyclist A time be t.

Then, we can write that

For Cyclist A

Speed = 15 km/h

Time = t mins

For Cyclist B

Speed = 20 km/h

Time = (t - 6) mins

i) To determine the distances travelled by cyclist A and B 1h cyclist A had started his journey,

For Cyclist A

Speed = 15km/h

Time = 1h = 60 mins

From the formula

Speed = Distance / Time

Then,

Distance = Speed × Time

Putting the values into the equation,

Distance = 15km/h × 1h

Distance = 15 km

∴ Cyclist A travelled 15 km 1 h cyclist A had started his journey from point X

For cyclist B

Speed = 20km/h

Time = 1h - 6mins = 60mins - 6mins = 54mins = 54/60 hour = 0.9 h

Also, from

Distance = Speed × Time

Putting the values into the equation

Distance = 20km/h × 0.9h

Distance = 18 km

∴ Cyclist B travelled 18 km 1 h cyclist A had started his journey from point X

ii) To determine the distance cyclist B overtake cyclist A, that is, when the distance covered by cyclist A equals that covered by cyclist B.

First, we will determine the time at which the distances covered by both cyclists were equal.

From

For Cyclist A

Speed = 15 km/h

Time = t hour  

Distance = Speed × Time

Distance = 15t km

For Cyclist B

Speed = 20 km/h

Time = (60t - 6) mins = (t - 0.1) hour

Distance = 20 × (t - 0.1) = (20t - 2) km

Equate the distances

15t = 20t - 2

15t - 20t = -2

-5t = -2

5t = 2

t = 2/5

t = 0.4 hour

Hence, cyclist B overtake cyclist A 0.4 hour after cyclist A had started.

For the distance cyclist B overtake cyclist A,

From

Distance = (20t - 2) km

Distance = (20×0.4 - 2) km

Distance = (8 - 2) km

Distance = 6 km

Hence, cyclist B overtake cyclist A 6km from the same starting point X.

The energy an object has due to its motion is called kinetic energy.

The work done on the block by the force of friction as the block moves from a to b is equal to -8 J.

Given:

Distance, d = 10 m

The kinetic energy of block (a) = 15 J

The kinetic energy of block (b) = 7J

The work done is given by:

Work done by friction = change in kinetic energy

The change in kinetic energy is:

KE = 7 - 15

KE = -8 J

Hence, the work done on the block by the force of friction as the block moves from a to b is equal to -8 J.

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#Complet question is:
A block slides on a rough horizontal surface from point A to point B. A force with magnitude F= 9N acts on the block between A and B, as shown. F 40 A B Points A and B are 10 m apart. If the kinetic energies of the block at A and B are 15 J and 7 J respectively, how much work is done on the block by the force of friction as the block moves from A to B?

Answer:

See below

Explanation:

For the, just remember the equation

F = m a

1000 N = 1980 kg  * a

1000/1980 = a = .505 m/s^2

Answer:

True

Explanation: Magic

Answer:

yurrrrr 221n54

Explanation:

yurrrrrrrrrrrrrrrrrrrrrr

Answer:

b

Explanation:

EDGE 2021

The indirect method is used to determine total power in a parallel circuit through the following:

This involves the circuit which has branches and a portion of the total current flows through it as a result.

In determining the total power in a parallel circuit, the indirect method is used only if the parameters mentioned above are available.

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Answer:4.39

Explanation:For acellus students the answer is 4.39

Law of attraction and Law of radius does not come in Kelper's law.

In astronomy, Kepler's legal guidelines of planetary movement, published via Johannes Kepler between 1609 and 1619, describe the orbits of planets across the solar. The laws changed the heliocentric theory of Nicolaus Copernicus, changing its circular orbits and epicycles with elliptical trajectories, and explaining how planetary velocities vary.

The elliptical orbits of planets were indicated through calculations of the orbit of Mars. From this, Kepler inferred that different our bodies within the solar device, including those farther far from the sun, additionally have elliptical orbits, the 2nd one law allows to set up that when a planet is toward the solar, it travels faster. The 3rd law expresses that the farther a planet is from the sun, the slower its orbital pace, and vice versa.

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Ploughing field with harrow pulled by a tractor being a modern technology​ is true.

This involves the use of scientific knowledge to tackle different life problems and has different sectors.

Ploughing field with harrow pulled by a tractor involves machines and not manual labour which makes it a modern technology.

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Answer:

They are all stratovolcanoes.

Volcanic fissures are linear vents on the surface of a volcano through which lava erupts.

Shield volcanoes are long, flat volcanic islands that form as a result of effusive eruptions and have varying eruption styles.

Dome volcanoes, as their nomenclature implies, look like a dome or a semi-circle and, like shield volcanoes, are formed as a result of effusive eruptions. They are also classified by their plinian eruption style and highly silicic and therefore viscous magma.

Cinder cones are steep, conical hills that are made up of pyroclastic fragments that accumulate around a vent with 30–40° slopes and are monogeneric, meaning that they erupt only once.

Stratovolcanoes are made up of multiple layers of lava flow, hence their famously perfect cone shape as they are continuously rebuilt, and have 50–70° slopes. They also have a plinian eruptive style and parasitic cones.

Lastly, calderas are craters that were former stratovolcanoes with depressions caused by collapsed magma chambers, the result of a final (and most probably large-scale) eruption, and have a diameter of at least two kilometers.

Answer: They are all stratovolcanoes.

Explanation:

An object's speed increases from 0 to 2 m/s, due to an amount of work w1, and then increases from 2 m/s to 4 m/s due to an amount of work w2.  The true statement would be option D. W1 < W2.

To determine the relationship between the amounts of work (w1 and w2) required to increase the speed of an object, we need to consider the work-energy principle.

The work done on an object is equal to the change in its kinetic energy. Mathematically, we have:

Work = Change in Kinetic Energy

Let's analyze the given scenario:

The speed increases from 0 to 2 m/s due to an amount of work w1.

The speed then increases from 2 m/s to 4 m/s due to an amount of work w2.

We know that the change in kinetic energy is related to the work done:

Change in Kinetic Energy = Work

For the first increase in speed (0 to 2 m/s), the change in kinetic energy is given by:

ΔKE1 = (1/2) * m * \((2^2 - 0^2)\) = 2m

For the second increase in speed (2 m/s to 4 m/s), the change in kinetic energy is given by:

ΔKE2 = (1/2) * m * \((4^2 - 2^2)\) = 10

Now, let's compare w1 and w2:

If w1 = w2, it would mean that the change in kinetic energy for both cases is the same (ΔKE1 = ΔKE2 = 2m). However, we found that ΔKE2 = 10m.

Therefore, the correct answer is:

D. W1 < W2

The amount of work required to increase the speed from 2 m/s to 4 m/s (w2) is greater than the amount of work required to increase the speed from 0 to 2 m/s (w1).

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The object's velocity accelerating from rest from the applied force and time is closest to 4 m/s.

The velocity of the object can be determied by applying Newton's second law of motion as shown below;

F = ma

where;

The acceleration of the object is calculated as follows;

a = F/m

at time, t 4.0 s, the corresponding force = 2 N

a = 2/2

a = 1 m/s²

Velocity of the object is calculated as;

v = u + at

v = 0 + 1 x 4

v = 4 m/s

Thus, the object's velocity accelerating from rest from the applied force and time is closest to 4 m/s.

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Answer:

Explanation:

chemical energy  :)   yay...  we can live  :D  

According to the question **Effect of the Sun shrinking on Earth's orbit.**

The length of a year would **decrease to 1/100 as long** if the Sun shrunk from its current diameter to 1/10 its current diameter while maintaining the same mass. This decrease in the Sun's diameter would result in a decrease in the gravitational pull experienced by the Earth, leading to a reduction in the orbital period.

According to Kepler's third law of planetary motion, the square of a planet's orbital period is proportional to the cube of its average distance from the Sun. As the Sun's diameter decreases, the average distance between the Sun and the Earth would remain relatively unchanged. Therefore, with a smaller diameter, the gravitational force exerted by the Sun on the Earth would be weaker, causing the Earth to orbit at a faster rate.

Hence, the length of a year would decrease significantly, becoming approximately 1/100 as long compared to its original duration.

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Apply Newton's second law

\(\\ \tt\hookrightarrow F=ma\)

\(\\ \tt\hookrightarrow F=1167(4)=4668N\)

Now

\(\\ \tt\hookrightarrow W=Fd\)

\(\\ \tt\hookrightarrow d=\dfrac{W}{F}\)

\(\\ \tt\hookrightarrow d=\dfrac{105}{4668}\)

\(\\ \tt\hookrightarrow d=0.022m\)

Now

According to third equation of kinematics

\(\\ \tt\hookrightarrow v^2=u^2+2as\)

\(\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)\)

\(\\ \tt\hookrightarrow v^2=100+8(0.022)\)

\(\\ \tt\hookrightarrow v^2=100+0.176\)

\(\\ \tt\hookrightarrow v^2=100.176\)

\(\\ \tt\hookrightarrow v=10.001m/s\)

The launch proyectiles of kinematics  allows to find the maximum initial vertical velocity of the body so that it just reaches the ceiling

             v_{oy} = 2.56 m / s

Given parameters

To find

Projectile launching is an application of kinematics where on the x axis there is no acceleration or and on the y axis the acceleration is the acceleration of gravity (g = 9.8 m / s ^ 2)

In this case, the maximum vertical velocity that the body can have occurs when the velocity on the ceiling is zero.

            v_y² = v_{oy}² - 2 g y

where v and v_{oy} are the initial velocity at the ceiling e initial, respectively, g the acceleration of gravity e and the height

           0 = v_{oy}² - 2 g y

           v_{oy} = \(\sqrt{2gy}\)

           v_{oy} = \(\sqrt{2 \ 9.8 \ 3}\)

           v_{oy} = 2.56 m / s

In conclusion with the kinematic of launch projectiles we can find the maximum initial vertical velocity of the body so that it just reaches the ceiling

             v_{oy} = 2.56 m / s

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Answer:

because sound needs medium to transmit.

The sound is produced by vibration and if there is no medium there is no vibration and no vibration means no sound.

The wavelength of light that is deep in the infrared region is around 10 micrometers (μm).

The wavelength of light is the distance between two consecutive crests or troughs in a wave. It is represented by the Greek letter lambda (λ) and is measured in meters (m), centimeters (cm), or micrometers (μm). Light waves are a type of electromagnetic radiation that is characterized by their wavelength and frequency. The electromagnetic spectrum is the range of all possible wavelengths of electromagnetic radiation. This spectrum is divided into different regions based on the wavelength of the radiation, with each region having its unique characteristics. The infrared region of the electromagnetic spectrum is the region between microwaves and visible light, with wavelengths ranging from 0.75 μm to 1000 μm.

Infrared radiation is used in a variety of applications, such as remote temperature sensing and communication. The deep infrared region of the electromagnetic spectrum is the region with the longest wavelengths in the infrared region. The deep infrared region has wavelengths of between 10 μm and 1000 μm, with a peak wavelength of around 10 μm. This wavelength is also known as the thermal radiation or thermal infrared region because objects at room temperature emit radiation at this wavelength. This region is used for a variety of applications, such as thermal imaging and night vision.

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Answer:

\(2380\ \text{m}\)

\(7.93\times 10^{-6}\ \text{s}\)

Explanation:

t = Temps mis par la foudre pour parcourir la distance = 7 secondes

v = Vitesse du son dans l'air = 340 m / s

La distance est donnée par

\(s=vt\\\Rightarrow s=340\times 7\\\Rightarrow s=2380\ \text{m}\)

L'éclair est tombé à une distance de \(2380\ \text{m}\)

c = Speed of light = \(3\times 10^8\ \text{m/s}\)

Le temps est donné par

\(t=\dfrac{s}{c}\\\Rightarrow t=\dfrac{2380}{3\times 10^8}\\\Rightarrow t=7.93\times 10^{-6}\ \text{s}\)

Le temps mis par la lumière pour parcourir la distance est de \(7.93\times 10^{-6}\ \text{s}\).

Answer:

gamma rays

Explanation:

hope it helps

A spring scale shows a net force of 0. 8 n acting on a 1. 5-kg mass, here  the acceleration of the object if the net force is decreased to 0.2N.  The acceleration of the object if the net force is decreased is          0.13 m/s².

Given data to find the acceleration,

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

The second law says that the acceleration of an object is related to the net force applied in the object divided by the mass of the object.

This can be written as:

F = m*a

Where F is force, m is mass of the object and a is the acceleration of the object.

Here, the mass used in state 1 and 2 remains the same, at 1.5 kg

In the state 1 ,

ΣF=0.8 N

m=1.5 kg

The acceleration, a:

a =∑F/m

  = 0.8 /1.5

  = 0.53 m/s².

In the state 2,

ΣF=0.2 N

m=1.5 kg

The acceleration, a:

a =∑F/m

  = 0.2 /1.5

  = 0.13 m/s².

The acceleration of the object if the net force is decreased to 0. 2 n is 0.13m/s².

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Hi there!

The equation for the electric field of a point charge is given by:

\(E = \frac{kq}{r^2}\)

E = Electric Field (N/C)
k = Coulomb's Constant

r = distance (m)

We can rearrange to solve for 'q':'

\(q^2 = \frac{Er^2}{k}\\\\q = \sqrt{\frac{Er^2}{k}}\)

Plug in the values and solve:

\(q = \sqrt{\frac{(1236)(4^2)}{(8.99*10^9)}} = \boxed{0.00148 C}\)

The  converging lens with the object placed beyond the focal point of the lens will create an image larger than the object. This is because the converging lens bends the light rays towards each other, causing them to converge at a point beyond the focal point, which results in a magnified image.

To explain further, when light rays pass through a converging lens, they converge at a point called the focal point. If the object is placed at the focal point, the light rays will converge to a single point, but no image will be formed. However, if the object is placed beyond the focal point, the light rays will converge and form a real image that is larger than the object.

On the other hand, a diverging lens spreads out the light rays, causing them to diverge away from each other. Therefore, neither of the arrangements that involve a diverging lens will create an image larger than the object.

In summary, a converging lens with the object placed beyond the focal point is the arrangement that will create an image larger than the object.

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The language of the given pushdown automaton (PDA) can be described as follows:

The PDA has a stack alphabet consisting of symbols 'a', 'b', 'z', '6', '1', '۸', '2', '9', '0', 'x', 'y'. 'z' represents the stack end symbol.

The transitions of the pushdown automaton (PDA) are as follows:

Thus, the language accepted by this PDA is characterized by a sequence of 'a's followed by a sequence of 'b's, where the number of 'b's is three times the number of 'a's, and each 'b' is followed by a corresponding sequence of '90', '91', '92', 'a', 'b', 'b', 'b', '1', 'b', 'b', '1', and ending with '1'.

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Considering the stress, the minimum width of the aluminum plate required to withstand a force of 49,700 N with no permanent deformation is 795.2 cm.

Stress, which is defined as force per unit area, may be used to indicate the aluminum's yield strength. As a result, we can determine the stress that corresponds to the aluminum's yield strength using the formula below:

Stress = yield strength equals 125 MPa or 125 N/cm2.

The aluminum plate's stress cannot be greater than its yield strength to prevent irreversible deformation of the material.

The force applied to the plate's surface may be related to its area using the stress formula:

Stress = force/area

area = 49,700 N / 125 N/cm² = 397.6 cm²

The area of the plate is the product of its width and thickness. Let's assume that the plate has a rectangular shape and solve for the minimum width required to achieve the required area:

area = width x thickness

width = area / thickness

         = 397.6 cm² / 0.5 cm

         = 795.2 cm

Therefore, the minimum width of the aluminum plate required to withstand a force of 49,700 N with no permanent deformation is 795.2 cm.

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A.The incline angle needs to be around 78.52° for the sphere to accelerate linearly at 0.70g.

B. The right answer is c. More than 0.70g of acceleration would be experienced.

(a) To find the incline angle for the uniform solid sphere with linear acceleration 0.70g, we can use the equation for the acceleration of a rolling sphere down an incline:

a = (5/7) * g * sin(theta)

Where a is the linear acceleration, g is the gravitational acceleration (9.81 m/s²), and theta is the incline angle. We are given a = 0.70g, so we can solve for theta:

0.70g = (5/7) * g * sin(theta)

Divide both sides by g:

0.70 = (5/7) * sin(theta)

Now, divide both sides by (5/7):

(0.70 * 7)/5 = sin(theta)

0.98 = sin(theta)

To find the angle, take the inverse sine of 0.98:

theta = sin^(-1)(0.98) ≈ 78.52°

So, the incline angle must be approximately 78.52° for the sphere to have a linear acceleration of 0.70g.

(b) If a frictionless block were to slide down the incline at that angle, its acceleration magnitude would be more than 0.70g. This is because the sphere's acceleration is reduced due to its rotation. A frictionless block doesn't have any rotation, so its acceleration down the incline is given by:

a_block = g * sin(theta)

Since sin(theta) is greater than the (5/7) factor in the sphere's acceleration equation, the acceleration of the frictionless block will be more than 0.70g. Therefore, the correct answer is c. The acceleration magnitude would be more than 0.70g.

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A uniform solid sphere rolls down an incline, the incline angle if the linear acceleration of the center of the sphere is to have magnitude 0.70g can be determined using-

The linear acceleration of the center of a uniform solid sphere rolling down an inclined plane is determined by the inclination angle of the plane.

It is determined by the formula a= g sin θ, where g is the acceleration due to gravity and θ is the angle of inclination.

For the linear acceleration of the center of the sphere to have a magnitude of 0.70g, we substitute 0.70g for a in the formula.

\(θ = sin⁻¹ (0.70) = 44.92°\)

Therefore, the inclination angle must be approximately 44.92 degrees for the linear acceleration of the center of the sphere to have a magnitude of 0.70g.

(b) If a frictionless block were to slide down the incline at that incline angle, would its acceleration magnitude be more less than or equal to 0.70g?

The acceleration of a frictionless block sliding down an inclined plane is determined by the formula a= g sin θ, where g is the acceleration due to gravity and θ is the angle of inclination.

Since the inclination angle of the plane is the same as in part (a), we can substitute 44.92° for θ in the formula

\(a= g sin θ = 9.81 × sin 44.92° = 6.78 m/s²\)

Since the magnitude of the acceleration of the block is less than 0.70g, the acceleration magnitude would be less than 0.70g. Therefore, option (b) is the correct answer.

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Answer:

Answer:

Carbon Dioxide