A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to the coil. Suppose that the magnetic field is varied so that B increases from 0.10 T to 0.35 T in 2 ms. Find the induced emf in the coil.

The charge q(t) on the capacitor would be, \(q(t)= 0.25c(1-e^{-t/0.04s})\)

And the current I(t) in the circuit will be,\(I(t)= 0.25mA(1-e^{-t/0.04s})\)

To find the charge q(t) on the capacitor at any time t, we can use the equation for the charge on a capacitor in an RC circuit:

\(q(t)= Q_{max} (1-e^{-t/RC})\)

where Qmax is the maximum charge on the capacitor, R is the resistance, C is the capacitance, and t is time.

To find Qmax, we can use the equation for the maximum charge on a capacitor in an RC circuit:

Qmax = E × C

where E is the electromotive force.

So, we have:

Qmax = 100 V × 10⁻⁴F = 0.01 C

Using the given values of R and C, we have:

R*C = 400 ohms × 10⁻⁴F = 0.04 s

Substituting these values into the equation for q(t), we get:

\(q(t)= Q_0.01(1-e^{-t/0.04C})\)

To find the current I(t), we can use Ohm's law and the equation for the charge on a capacitor:

I(t) = (1/R) × d(q(t))/dt

where d(q(t))/dt is the derivative of q(t) with respect to time.

Taking the derivative of q(t), we get:

\(dq(t)/dt= 0.01C (1-e^{-t/0.04C})\)

Substituting this into the equation for I(t), we get:

\(I(t)= (1/400ohm)(0.01/0.04s) (1-e^{-t/0.04C})\)I

Simplifying, we get:

\(I(t)= 0.25mA(1-e^{-t/0.04C})\)

Therefore, the charge q(t) on the capacitor is:

\(q(t)= 0.25c(1-e^{-t/0.04s})\)

And the current I(t) in the circuit is:

\(I(t)= 0.25mA(1-e^{-t/0.04s})\)

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Answer:

Explanation:

To solve this problem we need to know the direction in which the ball was moving to start with.

The answer will be different depending n the original angle of the ball's movement.

It might be reasonable to assume that the ball is meant to approach along the x-axis,

but if so, the initial speed of 6.42m/s would be irrelevant to the answer.

So I will solve the problem for the general case of two objects colliding at arbitrary angles, and

tell you how to specialize it for any assumption about the initial conditions.

Let

m1 = 7.5 kg be the mass of the ball,

m2 = 1.6 kg be the mass of the pin,

v1 = 6.42 m/s be the velocity of the ball before the strike,

v2 = 0 m/s be the velocity of the pin before the strike,

α1 be the angle of v1,

α2 be the angle of v2,

w1 be the velocity of the ball after the strike,

w2 = 14.8 m/s be the velocity of the pin after the strike,

β1 be the angle of w1,

β2 = -47° be the angle of w2.

By conservation of momentum:

m1v1 + m2v2 = m1w1 + m2w2

Since the velocities are vectors, the addition is vector addition, and the equality is vector equality.

"Vector equality" means that the x-coordinates are equal and the y-coordinates are equal.

The problem cares only about y-coordinates, specifically the y-coordinate of w1, which is w1sin(β1).

(In general, the y-coordinate of any vector is obtained by multiplying the vector's norm by the sine of its angle.)

Conservation of momentum in the y-coordinate is then

m1v1sin(α1) + m2v2sin(α2) = m1w1sin(β1) + m2w2sin(β2)

Expressing the sought quantity

w1sin(β1) = (m1v1sin(α1) + m2v2sin(α2) - m2w2sin(β2))/m1

Substituting known quantities:

w1sin(β1) = (7.5×6.42×sin(α1) + 1.6×0×sin(α2) - 1.6×14.8×sin(-47°))/7.5

= (48.15×sin(α1) + 17.3)/7.5

In the above expression we do not know α1.

If we assume that the ball is approaching along the x-axis then α1 = 0, and

w1sin(β1) = 17.3/7.5 = 2.3

Under that assumption the y-component of the ball's final velocity is 2.3 m/s;

being positive, it is opposite the direction of the pin.

Answer~

Both life science and physical science helps to understand the natural habitat of the earth. Life science deals with the living thing; thus, it is a study of the organic world. On the other hand, physical science deals with nonliving things; thus, it is a study of the inorganic world.

Explanation:

A car accelerates from rest at 4.5 m/s² for 3 seconds, the car's velocity after 3 seconds is 13. 5 m/s

The distance the car travel in the 3 seconds is 40. 5m/s

The first equation of motion is v = u + a t. It explains the relationship between velocity and time.

V = Final velocity

u = Initial velocity

a = acceleration

t  = time.

How to determine the velocity

When an object is brought to rest or moves from resting, U, initially velocity equals 0.

From the formula,

V = u + at

V = ?

u = 0, because the object was at rest.

a = 4. 5 m/ s²

t = 3 seconds

Velocity, V = 0 + 4. 5 × 3

Velocity , V = 0 + 13. 5

Velocity, V = 13.5 m/s²

How to determine the distance

Velocity = Distance ÷ time

Velocity, V = 13. 5 m/s²

time = 3 seconds

distance = ?

Distance = Velocity × time

Distance = 13. 5 × 3

Distance = 40. 5 m

Therefore, the car's velocity after 3 seconds is 13. 5 m/s² and the distance the car travels after 3 seconds is 40. 5m

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

\(\frac{1}{f} = 2D\)

Near point,

u = 0.6 m

As we know,

⇒ \(\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2\)

By substituting the values, we get

⇒ \(\frac{1}{v} -\frac{1}{0.6} =2\)

            \(\frac{1}{v}=2+\frac{1}{0.6}\)

            \(\frac{1}{v} =\frac{1.2+1}{0.6}\)

            \(\frac{1}{v}=\frac{2.2}{0.6}\)    

By applying cross-multiplication, we get

          \(0.6=2.2 \ v\)

            \(v = \frac{0.6}{2.2}\)

      \(S_{near} = 0.273 \ m\)

The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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Take the moment Allen sees the moose to be the origin.

First, convert his speed to m/s.

90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s

(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of

(25 m/s) • (0.85 s) = 21.25 m

(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of

x = 21.25 m + (25 m/s) t + 1/2 a t²

and has a speed v of

v = 25 m/s + a t

It takes him 3.5 s to come to a full stop. Use this to find the acceleration:

0 = 25 m/s + a (3.5 s)

a = - (25 m/s) / (3.5 s)

a ≈ - 7.1 m/s²

After 3.5 s, he will have traveled a total distance of

x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²

x = 152.5 m ≈ 150 m

(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be

a ≈ - 7.1 m/s²

A force of 1 n accelerates a 1-kg box at the rate of 1 m/s2. the acceleration of a 2-kg box by a net force of 2 n is the same.

In physics, a force is an influence that could alternate the movement of an item. A force can motivate an item with mass to exchange its pace, i.e., to accelerate. pressure also can be described intuitively as a push or a pull. A pressure has each magnitude and route, making it a vector quantity.

it can also purpose a change in the direction, shape, size, and so forth., of the frame. Pushing or pushing a door with pressure is an example. force is a vector amount, which means it has each magnitude and route. Newton's 2d regulation defines force as the "manufactured from a frame's mass and acceleration."A force is not something that an item consists of or 'has in it'. Pressure is exerted on one object with the aid of another. The idea of a force isn't restrained to residing things or non-residing matters.

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Explanation: Momentum: It is defined as the motion of a moving body. Or it is defined as the product of the mass or velocity of an object.

The formula of momentum is: p=mt

where,  

p = momentum =?

m = mass = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, and we get:

p = 1.5 kg * 10 m/s

p = 15kg.m/s

Therefore, the momentum of the ball is 15 kg.m/s

The best estimate for the ratio of Power radiated per unit Area on Earth and Power radiated per unit Area on Venus is 1/16. The correct option is D.

Power radiated is directly proportional to the fourth power of Temperature at the surface and also to the cross-section area.

P = σ A T⁴

For 600 K at Venus, σ A T⁴ = σ A (600)⁴

For 300 K at Earth , σ A T⁴ = σ A (300)⁴

Comparing, we get

[σ A (300)⁴/ σ A (600)⁴ ] = 1/16

Thus, the the best estimate for the ratio is 1/16. The correct option is D.

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Answer:

Dear user,

Answer to your query is provided below

Acceleration is zero because of no change in velocity.

Explanation:

Remember that velocity is a vector quantity and a vector can change in 3 ways

•Magnitude only

•Direction only

•Both magnitude and direction.

Now the magnitude of velocity (speed) can stay constant while the direction is changing. This is the case in circular motion.

In the question above, it is mentioned that the girl is moving along a straight road. Therefore no change in direction of velocity.

The horizontal distance of the ball is 2.41 m.

The time of motion of the ball is calculated by applying the following kinematic equation as shown below.

h = vt + ¹/₂gt²

where;

h = 0  +   ¹/₂gt²

h = ¹/₂gt²

t = √ ( 2h / g )

t = √ ( 2 x 0.9 / 9.8 )

t = 0.43 seconds

The horizontal distance of the ball is calculated as follows;

X = Vt

X = 5.6 m/s  x  0.43 s

X = 2.41 m

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Heya!!

For calculate final velocity, lets applicate formula

                                                \(\boxed{V=V_o+a*t}\)

                                                 Δ   Being   Δ

                                            V = Final Velocity = ?

                                         Vo = Initial velocity = 4,3 m/s

                                          a = Aceleration = 3 m/s²

                                                 t = Time =5 s

⇒ Let's replace according the formula:

\(\boxed{V=4,3\ m/s +3\ m/s* 5\ s}\)

⇒ Resolving

\(\boxed{V=19,3\ m/s}\)

Result:

The velocity after 5 sec is 19,3 meters per second (m/s)

For calculate distance, lets applicate formula

                                              \(\boxed{x=V_o*t+\dfrac{a*t^{2}}{2}}}\)

                                                 Δ   Being   Δ

                                               x = Distance = ?

                                         Vo = Initial velocity = 4,3 m/s

                                          a = Aceleration = 3 m/s²

                                                 t = Time =5 s

⇒ Let's replace according the formula:

\(\boxed{x=4,3 * 5 +\dfrac{3*5^{2}}{2}}}\)

⇒ Resolving

\(\boxed{x = 59 \ m}\)

Result:

The distance is 59 meters.

Good Luck!!

Answer:

f = 4.67 cm

Explanation:

Here, we can use the thin lens formula, as follows:

\(\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}\\\\\)

where,

f = focal length of lens = ?

p = distance of object from lens = 12.4 cm

q = distance of image from lens = 7.5 cm

Therefore,

\(\frac{1}{f} =\frac{1}{12.4\ cm} +\frac{1}{7.5\ cm}\\\\\frac{1}{f} = \frac{1}{4.67\ cm}\)

f = 4.67 cm

A pair of Newton's Third Law effects are represented by gravitational pull of a Earth and Moon. Both the Moon and the Earth are gravitationally pulled in opposite directions by the Earth.

Its surface gravity of a moon is only a sixth that of Earth because an object's mass or size determine its force of gravity now at surface.

The force that now the earth applies to a body is referred to as gravitational force. The downhill and upward movement of water in rivers and streams, as well as the upwards motion of the a ball as it is thrown, are examples of motion brought on by gravitational force.

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Solution:

The block size = 16 words

Tag Block Offset

11 bits 5 bits 4 bits

The memory address 0DB63 will map to the 22nd block and 3 bytes of a block.

cache advantage is, when the cache block is made larger or bigger then, there are fewer misses. The disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful.

After analyzing the question, the solution is:

As given The Cache of 32 blocks and the memory is word addressable:

The main memory size.

The block size.

The total number of blocks = main memory size/ block size:

Says it is 1 M words in size and it requires 20 bits. Now, from which 32 cache block requires 2^20, 20 bits, and block size means offset requires 2^4, 4 bits. Now the sizes of the bag, and, block and word fields are following:

Tag Block Offset

11 bits 5 bits 4 bits

The hexadecimal address value is 0DB63, its binary equivalent is

The tag = ( first 11 bits) = 0000 1101 101

block = ( next 5 bits ) = 10110

offset = ( next 4 bits ) = 0011

Therefore, the memory address 0DB63 will map to the 22nd block and 3 bytes of a block.

The main favored position or circumstance of the cache happens that, when the cache block is created best or considerably therefore, there happen hardly any misses. this takes place when the information in visible form in the block exists secondhand.

One of the loss exists that if the data exist not secondhand before the cache block exist detached from the cache, then it exist not any more valuable. in this place there exist an addition to the best miss punishment.

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Answer:

1, 2, 3, 4 on edge 2020

Explanation:

Answer:

1234

Explanation:

If A 5kg objects is sliding across a floor at 10m/s.then the work done by friction to bring the 5 kg object to a stop is -250 Joules.

To calculate the work done by friction to bring the object to a stop, we need to determine the change in kinetic energy.

Given:

Mass of the object, m = 5 kg

Initial velocity, u = 10 m/s

Final velocity, v = 0 m/s (object comes to a stop)

The work done by friction can be calculated using the equation:

Work = Change in Kinetic Energy

The change in kinetic energy (ΔKE) can be calculated as:

ΔKE = (1/2) * m * (v^2 - u^2)

Plugging in the values:

ΔKE = (1/2) * 5 kg * (0 m/s)^2 - (10 m/s)^2

= (1/2) * 5 kg * (0 - 100 m^2/s^2)

= (1/2) * 5 kg * (-100 m^2/s^2)

= -250 J

The negative sign indicates that the work done by friction is in the opposite direction of the displacement of the object.

Therefore, the work done by friction to bring the 5 kg object to a stop is -250 Joules.

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Answer:

Endurance, or aerobic

Strength, or resistance training

Balance

Flexibility

Explanation:

Answer:

Walking. Walking is simple, yet powerful. It can help you stay trim, improve cholesterol levels, strengthen bones, keep blood pressure in check, lift your mood, and lower your risk for a number of diseases (diabetes and heart disease, for example).

Answer:

the frequency of the wave is 13mm

Explanation:

15mm divided by 3 = 5 x 2= 10+3= 13mm so the answer is 13

Answer:

A) A negative charge of value Q is induced on sphere B

B) there is an attraction between sphere

C) The charge of sphere A is distributed between the two spheres,

Explanation:

This is an electrostatic problem, in general charges of the same sign attract and repel each other.

with this principle let's analyze the different situations

A) The sphere A that is insulating has a charge on its surface and zero charge is its interior

   The conducting sphere B has zero charge, but the sphere A creates an attraction in the electrons, therefore a negative charge of the same value as the charge of the sphere A is induced in the part closest and in the part farther away than one that a positive charge.

A negative charge of value Q is induced on sphere B

B) In this case there is an attraction between sphere A with positive charge and sphere B with negative induced charge

C) When the two spheres come into contact, the charge of sphere A is distributed between the two spheres, therefore each one has a positive charge of value half of the initial charge, as now we have net positive charges in the two spheres charges of the same sign repel each other so the spheres separate

Answer:

Photovoltaic detection is a technique that converts light into electrical energy. It is a process that involves the use of a photovoltaic cell, which is made up of semiconductor materials, to generate an electric current when exposed to light.

The photovoltaic cell absorbs the photons of light, which then knock electrons out of their orbits, creating a flow of electricity. The amount of electricity produced is proportional to the intensity of the light. The photovoltaic cell is commonly used in solar panels to generate electricity from sunlight. The efficiency of the photovoltaic cell is dependent on several factors, including the type of semiconductor material used, the purity of the material, and the thickness of the cell.

The photovoltaic cell has many applications, including in solar power generation, telecommunications, and remote sensing. The technique of photovoltaic detection is an important area of research, as it has the potential to provide a clean and renewable source of energy that can help mitigate climate change.

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The measure of the indicated angle to the nearest degree is Sin 42°.

An angle is formed when two immediate traces or rays meet at a not-unusual endpoint.

The names of primary angles are Acute angle, Obtuse perspective, proper attitude, straight attitude, reflex angle, and full rotation. A perspective is a geometrical form shaped by using joining rays at their endpoints. An angle is normally measured in stages. there are various varieties of angles in geometry.

A right triangle or proper-angled triangle, or greater officially an orthogonal triangle, formerly called a rectangle triangle, is a triangle wherein one angle is a proper angle or facets are perpendicular. The relation among the perimeters and different angles of the proper triangle is the premise for trigonometry.

calculation:-

perpendicular = 43

hypoteneuse = 64.4

Sin θ = perpendicular / hypoteneuse

            = 43 / 64.4

Sin θ   =   0.66770

Sin θ = 41.889

Sin θ = 42° ( nearest degree)

Sin α = 90 - 42

        = 48°

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Answer:

Newton

Explanation:

Newton's are the standard unit of force.

The resultant moment produced by all the weights about point a is 1700 lb * g1A + 195 lb * g2A + 185 lb * g3A.

To determine the resultant moment produced by all the weights about point A, we need to calculate the moments of each weight about the point A and then sum them. The moment of a weight about a point can be calculated using the following equation:

M = W * d

where M is the moment, W is the weight, and d is the perpendicular distance from the weight to the point in question.

Let's first find the moments of the boom, cage, and man about point A:

M1 = 1700 lb * g1A

M2 = 195 lb * g2A

M3 = 185 lb * g3A

where g1A, g2A, and g3A are the distances between points g1, g2, and g3, and point A.

Now, the resultant moment can be calculated by summing all the moments:

M_resultant = M1 + M2 + M3

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Explanation:

To determine the force F of interaction between the half ring and the point charge, we can use the principle of superposition, which states that the total force on a point charge due to a collection of other charges is the vector sum of the individual forces that each of those charges would exert on the point charge if it were the only charge present.

First, we need to find the electric field at the center of curvature due to the charged half ring. The electric field at a point on the axis of a uniformly charged ring is given by:

E = kqz / (z^2 + R^2)^(3/2)

where k is Coulomb's constant, q is the linear charge density, z is the distance from the center of the ring to the point on the axis, and R is the radius of the ring.

At the center of curvature of the half ring, z = R, so the electric field is:

E = kq / (2R)

Next, we can use the electric field to find the force on the point charge q:

F = qE

Substituting the given values, we get:

F = (20 x 10^-9 C) x (9 x 10^9 N·m^2/C^2) x (1/20 cm)

F = 9 x 10^-3 N

Therefore, the force of interaction between the half ring and the point charge is 9 x 10^-3 N.

This force can also be interpreted as the force required to hold the point charge at the center of curvature against the electric field due to the charged half ring. It is an attractive force because the point charge is opposite in sign to the charged half ring.

Answer:

5.65N.

Explanation:

Solution Given:

radius of R = 10 cm=10/100=0.1 m

linear density λ = 1 Mikrokulon/m= 10^-6 Coulomb/m

force F=?

q1 = 20nC=20*10^-9 C

we have

Coulomb's law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this can be expressed as:

F = k*(q1*q2)/r^2

since q1 is located at the center of curvature of the half ring , so the half ring is uniformly charged with a linear density of λ= 1 μC/m.

again

equation becomes.

F=k*(q1*λL)/r^2

Since the half ring is a semicircle,

we have L=πr

F=k*(q1*λ*πr)/r^2

substituting value

F=9 x 10^9 N*m^2/C^2 *(20*10^-9 C* 10^-6 m^3*π*0.1 m)/(0.1m^2)

F=5.65 N

Therefore, the force of interaction between the half ring and the point charge is 5.65N.

The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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The value of acceleration is 3 m/s2.

Our body makes every effort to open up our airways. When sneezing, the diaphragm, abdomen, vocal cord, and chest muscles all work together. As a result, the air leaving our lungs accelerates rapidly.

Change in velocity, Δv

Change in time, Δt

Note that

1 km = 1000 m

1 h = 3600 s

Therefore

Δv = (125,000 m/h)*(1/3600 h/s) = 3 m/s2.

Therefore, The value of acceleration is 3 m/s2.

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Answer:

29.4 kW

Explanation:

time t = 10 s, height h = 30 m, mass m = 1,000 kg

power output = work done/time

work done = increase in potential energy = mgh = 1,000 * 9.8 * 30 = 294,000 J = 294 kJ

power output = 294/10 = 29.4 kW