The amount of deformation a material experiences due to an applied force is called

Answer:

Explanation:

Given parameters

initial velocity u = 4.9m/s

height of drop H = 100m

acceleration due to gravity g = -9.8m/s²(since the body is ascending i.e moving against gravity)

Required

time taken by the packet to reach the ground 't'

Using the equation of motion

S = ut+1/2gt²

0-100 = 4.9t+1/2(-9.8)t²

-100 = 4.9t-1/2(9.8)t²

-100 = 4.9t-4.9t²

4.9t-4.9t² +100 = 0

4.9t²-4.9t² -100 = 0

Multiplying through by 10

49t²-49t-1000 = 0

Using the general equation t = (-b±√b²-4ac)/2a

t = -(-49)±√(-49)²-4(49)(-1000)/2(49)

t = 49±√(2401+196000)/2(49)

t = 49±√(198401)/98

t = 49±√(198401)/98

t = 49±449.42/98

t = (49+449.42)/98

t = 494.42/98

t = 5.04secs

Hence, it took the packet 5.04secs to reach the ground

The final velocity is calculated using the formula v = u + at

v = u-gt

v = 4.9-(9.8)(5.04)

v = 4.9-49.392

v = -44.492m/s

This shows that the final velocity is 44.49m/s(moving downwards)

It will take 3.07 meter long to reach at its maximum value.

The maximum static friction force that can be applied to the string before it slips is the product of the normal force and the coefficient of static friction as

Fmax = μs × N

The volleyball will reach at the maximum height at 0.673 s after it is served.

Calculating for the height the volleyball  will reach before going to  downwards.

7.2 m/s / 9.81

= 0.77945 s

The distance traveled is given by the formula as follows;

d = 1/2 AT^2,

Thus, substitute the known value for A and T,

d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2

d = 4.905 m/s^2 0.607542 s^2

d = 2.979995

Thus, the volleyball will travel 2.979 meters straight up from the point upon which it was launched.

Thus we need to add the 0.80 meters initial height.

d = 2.979995 m + 0.8 m

d = 3.779995

The answer is 3.78 m.

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In order for a single magnifying glass to magnify an object, the lens must have a convex shape.

A convex lens curves outward and is thicker at the center than at the edges. When light passes through a convex lens, it refracts or bends inward, converging at a focal point. This allows the lens to create a magnified image of the object being viewed.

The distance between the object and the lens, as well as the distance between the lens and the viewer's eye, will also affect the magnification. Additionally, the lens must be positioned at the correct distance from the object to produce a clear, focused image.

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The CRO (Cathode Ray Oscilloscope) will display a triangular wave that is vertically stretched and horizontally compressed.

The vertical deflection plates will cause the triangular wave to be displayed with a peak-to-peak amplitude of\(100 cm (50 V * 0.1 cm/V)\), while the horizontal deflection plates will cause  sawtooth wave to be displayed with a peak-to-peak amplitude of \(5000 cm (100 V * 0.02 cm/V).\) The synchronization of the two inputs will ensure that the triangular wave and the sawtooth wave are displayed in a coordinated manner, with each cycle of the sawtooth wave corresponding to five cycles of the triangular wave. The resulting display will show a pattern of diagonal lines that gradually rise and then quickly drop back to the starting position, with each line representing a cycle of the sawtooth wave.

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If we push to wheelbarrow at exactly the same speed then the heavier wheelbarrow will require more force to push at that constant speed.

According to galileo's Law of Inertia and the first law of motion proposed by Newton,

If a body is in the state of rest or in the state of motion it will continue to do so, until and unless an external force is applied on the body to change it state of motion or state of rest.

According to Galileo, everybody possess some inertia.

Inertia can be defined as the property of the body that allows the body to regain its current state of rest or motion.

Inertia of a body is directly proportional to the mass of the body.

Hence we can conclude here, that the heavier wheelbarrow will require more force to push at a constant speed because we are trying to change the state of the wheelbarrow.

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The airplane's airspeed is 120.0 km/h, with a heading of 30.0° east of north. We can break this down into northward and eastward components. The northward component is 120.0 km/h * cos(30.0°) ≈ 103.9 km/h, and the eastward component is 120.0 km/h * sin(30.0°) ≈ 60.0 km/h.

The wind is blowing towards the west at 30.0 km/h, so it has a westward component of 30.0 km/h.

To find the airplane's ground speed, we need to combine these components. The northward component remains unchanged at 103.9 km/h, while the eastward component gets reduced due to the westward wind, giving us 60.0 km/h - 30.0 km/h = 30.0 km/h.

Now, we can find the resultant ground speed using the Pythagorean theorem: √((103.9 km/h)^2 + (30.0 km/h)^2) ≈ 107.9 km/h.

So, the approximate speed of the plane relative to the ground is 107.9 km/h.

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Answer:

simltinoues divide two eqn El÷E2 Kq\ r^2\kq \2r^2 then you get it 7.5 N\c the answer is A

a) The cart's linear velocity at the bottom of the hill is approximately 4.95 m/s.

b) The cart's final linear kinetic energy is approximately 61.31 J.

c) The cart's linear momentum at the bottom of the hill is approximately 24.75 kg·m/s.

d) The angular velocity of a wheel at the bottom of the hill is 49.5 rad/s.

e) The cart's gravitational potential energy when it is halfway down the hill is approximately 30.625 J.

We can use principles of conservation of energy and motion.

a) The potential energy the cart possesses at the top of the hill is converted into kinetic energy at the bottom, neglecting friction and rolling inertia. We can use the conservation of energy to find the linear velocity.

The potential energy at the top of the hill (PE) is given by:

PE = m * g * h

where

m = mass of the cart = 5.00 kg

g = acceleration due to gravity = 9.8 m/s²

h = height of the hill = 1.25 m

PE = 5.00 kg * 9.8 m/s² * 1.25 m

= 61.25 J

The kinetic energy at the bottom of the hill (KE) is given by:

KE = 0.5 * m * v²

where v is the linear velocity of the cart.

Since the potential energy is converted into kinetic energy,

PE = KE

61.25 J = 0.5 * 5.00 kg * v²

v² = (61.25 J) / (0.5 * 5.00 kg)

v² = 24.5 m²/s²

v = √(24.5 m²/s²)

v ≈ 4.95 m/s

Therefore, the cart's linear velocity at the bottom of the hill is approximately 4.95 m/s.

b) To find the final linear kinetic energy (KE), we can substitute the linear velocity (v) into the kinetic energy formula:

KE = 0.5 * m * v²

KE = 0.5 * 5.00 kg * (4.95 m/s)²

KE ≈ 61.31 J

Therefore, the cart's final linear kinetic energy is approximately 61.31 J.

c) The linear momentum (p) of an object is given by the product of its mass (m) and linear velocity (v):

p = m * v

p = 5.00 kg * 4.95 m/s

p ≈ 24.75 kg·m/s

Therefore, the cart's linear momentum at the bottom of the hill is approximately 24.75 kg·m/s.

d) Since we neglected rolling inertia, the angular velocity of a wheel at the bottom of the hill can be found by dividing the linear velocity (v) by the radius (r) of the wheel:

ω = v / r

ω = 4.95 m/s / 0.10 m

ω = 49.5 rad/s

Therefore, the angular velocity of a wheel at the bottom of the hill is 49.5 rad/s.

e) To find the gravitational potential energy (PE) when the cart is halfway down the hill, we need to determine the height at that point.

Given that the hill's height is 1.25 m, halfway down the hill would be at a height of 1.25 m / 2 = 0.625 m.

PE = m * g * h

PE = 5.00 kg * 9.8 m/s² * 0.625 m

PE ≈ 30.625 J

Therefore, the cart's gravitational potential energy when it is halfway down the hill is approximately 30.625 J.

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The water budget equation for the Creek drainage basin is: Precipitation = Stream discharge + Evapotranspiration ± Change in storage.

To develop the water budget equation for this problem, we need to account for the inputs and outputs of water in the Creek drainage basin. The water budget equation can be expressed as,

P = Q + ET ± ΔS

P = Precipitation input (mm)

Q = Stream discharge (m³/s)

ET = Evapotranspiration (mm)

ΔS = Change in storage (mm)

First, let's convert the units of the given values,

Precipitation input (P) = 550 mm

Stream discharge (Q) = 1.8 m³/s

Now, let's determine the evapotranspiration (ET) and change in storage (ΔS) terms. However, the problem doesn't provide information about ET, so we cannot calculate it accurately. The problem doesn't provide information about ΔS, so we cannot calculate it accurately.

Given these limitations, we can write the simplified water budget equation for this problem,

550 mm = 1.8 m³/s + ET ± ΔS

Please note that without information about evapotranspiration and changes in storage, we cannot fully determine the water budget for the Creek drainage basin.

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Answer:

by using formula,

wavelength= velocity/frequency

= 120/6

= 20 meter

Ans: 20 meter

Answer:

wavelength= velocity

= 120/6= 20 meter

Explanation:

The approximate internal rate of return (IRR) for installing the device is estimated to be above 20%. Indicating that the actual IRR must be higher than 20% to move the NPV closer to zero.

To calculate the NPV for installing the device, we need to consider the cash flows and discount rate. The initial cost of the device is £500,000, followed by net revenues of £100,000 after one year and £500,000 after two years. The nuclear company's high discount rate of 20% reflects its limited borrowing capacity and alternative productive uses for its cash.

We can start by assuming an IRR higher than 20% and calculate the corresponding NPV. At a 20% discount rate, the NPV is already -£70,000. If the IRR were lower than 20%, the NPV would be even more negative, moving further away from zero.

Therefore, to move the NPV closer to zero, the IRR must be higher than 20%. By using trial and error to adjust the discount rate, we can find an approximate IRR that results in an NPV closer to zero. However, without knowing the exact timing and magnitude of the cash flows, it is challenging to provide an exact IRR estimate.

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Answer:

moon lo

Explanation:

It is the most volcanically active world in the solar system.

RS is a composition of rotation and reflection in the Euclidean plane E². The precise description of RS depends on the specific properties of the rotation R and reflection S.

In general, if R and S have the same axis or line of symmetry, the composition RS results in a translation. If R and S have intersecting lines of symmetry, RS yields a glide reflection. If R and S have perpendicular lines of symmetry, RS produces a rotation.

It is important to consider special cases, such as parallel lines of symmetry, coinciding axes, or perpendicular lines of reflection, as they may lead to different outcomes. The classification of isometries in E² involves understanding how rotations and reflections combine to create different transformations in the plane.

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a) Highest to lowest mechanical advantage:

Machine C

Machine A

Machine B

b) Highest to lowest efficiency:

Machine A

Machine C

Machine B

Mechanical advantage is described as a measure of the force amplification achieved by using a tool, mechanical device or machine system.

Mechanical advantage = Output force / Input force = 50 N / 6 N = 8.33 is machine C

Mechanical advantage = Output force / Input force = 20 N / 5 N = 4 is machine A

Mechanical advantage = Output force / Input force = 25 N / 10 N = 2.5 is machine B

b) Highest to lowest efficiency:

Efficiency = Output work / Input work = (20 N)(2 m) / (5 N)(4 m) = 0.8 or 80% is machine A

Efficiency = Output work / Input work = (50 N)(27 m) / (6 N)(20 m) = 0.5625 or 56.25% is machine C

Efficiency = Output work / Input work = (25 N)(200 m) / (10 N)(3.5 m) = 0.35714 or 35.71% is machine B.

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When the crest of one wave passes through the trough of another wave, destructive interference takes place.

The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.

When the crest of one wave passes through the trough of another wave, interference takes place.

This interference can be constructive, where the amplitudes of the waves add up, resulting in a larger wave, or destructive, where the amplitudes of the waves cancel out, resulting in a smaller wave or no wave at all.

The type of interference that occurs depends on the relative amplitudes, wavelengths, and phases of the two waves.

Therefore, destructive interference takes place when the crest of one wave passes through the trough of another wave.

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Answer:he is correct

Explanation: when Donald runs all his weight is moving at the same speed so when all that weight hits a stopped object the weight will transfer its force into that object Kinda like how a truck hits a car

If the report stated that the Starship Enterprise had just returned from a 5-year voyage while traveling at 0.84c, and it meant five Earth years, we can calculate the elapsed time on the ship using time dilation.

Time dilation occurs when an object is moving at high speeds relative to another object, causing time to pass more slowly for the moving object.To calculate the elapsed time on the ship, we can use the formula for time dilation:t_ship = t_earth / sqrt(1 - (v^2/c^2))where t_ship is the elapsed time on the ship, t_earth is the time measured on Earth, v is the velocity of the ship (0.84c in this case), and c is the speed of light.Plugging in the values, we get:t_ship = 5 / sqrt(1 - (0.84^2))Evaluating the equation gives us:t_ship ≈ 5.93 Earth yearsTherefore, approximately 5.93 Earth years would have elapsed on the ship during the reported 5-year voyage.b. If the report meant five years of ship time, we can calculate the corresponding time passed on Earth. Since time dilation occurs at high speeds, the time experienced by a moving object is dilated or stretched compared to the time experienced by a stationary observer.To calculate the time passed on Earth, we can rearrange the time dilation formula:t_earth = t_ship * sqrt(1 - (v^2/c^2))Using the given values:t_earth = 5 * sqrt(1 - (0.84^2))Calculating the equation gives us:t_earth ≈ 2.72 Earth yearsTherefore, approximately 2.72 Earth years would have passed during the reported five years of ship time.

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When air at 1000 mb is lifted to 700 mb, it will be colder than the surrounding air.

This represents an unstable condition, as the air is not able to stay in its current state and is forced to move up and away from its original position. This creates a situation where the colder air is more dense and is more likely to sink back down towards its original position.

In an unstable condition, the air is constantly moving and shifting, making it difficult to predict what the weather conditions will be.

This condition is considered to be unstable because the air has become less dense than the air around it. As the air rises and cools, the air molecules condense and form clouds which can lead to the development of storms.

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Answer:

7.34 meters

Explanation:

If the entire Kinetic energy is converted to Potential energy

1/2 mv^2 = mgh          didvide both sides by ' m '

1/2 v^2 = gh                 solve for 'h'

1/2 v^2 / g = h              plug in the given values  ,   g = 9.81 m/s^2

1/2 (12)^2 / 9.81 = h = 7.34 m

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

The energy of the photon of blue light is  4.55 × 10^-19 J. Blue light is in the visible spectrum.

Using the formula;

E = hc/λ

E = energy of the photon = ?

h = Plank's constant = 6.6 × 10^-34 Js

c = Speed of light = 3 × 10^8 m/s

λ = wavelength of light = 435nm or 435  × 10^-9 m

Substituting values;

E =  6.6 × 10^-34 Js ×  3 × 10^8 m/s/435  × 10^-9 m

E = 4.55 × 10^-19 J

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The magnitude of the work done by the car's brakes to slow the car down is 235,537.5 Joules.

The work done:

W = ΔKE

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

ΔKE = (1/2) × m × (vf² - vi²)

Where m = mass of the car,

vf = final velocity, and

vi = initial velocity.

Let's replace the given values into the equation:

m = 1,500 kg

vf = 28.6 m/s

vi = 45.9 m/s

ΔKE = (1/2) × 1,500 kg  × ((28.6 m/s)² -  (45.9 m/s)²)

Now, determine the magnitude of the work done:

W = ΔKE

ΔKE = (1/2) ×  1,500 kg ×  ((28.6 m/s)² -  (45.9 m/s)²)

= (1/2) ×  1,500 kg ×  (-314.05 m²/s²)

= -235,537.5 J

The amount of work done by the car's brakes to slow it down is 235,537.5 Joules since the change in kinetic energy is negative (indicates a drop in kinetic energy).

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Answer:

A) Vo = 137.34 [m/s]

B) Y = 961.38 [m]

C) y = 367.8[m] for t = 3 [s] and y = 470.88[m] for t = 4[s]

Explanation:

To solve this problem we must use the following equation of kinematics. We must keep in mind that the negative sign of the equation means that the acceleration of gravity acts in the opposite direction to the movement of the balloon.

A)

\(v_{f} =v_{o} -g*t\)

where:

Vf = final velocity = 0 (the maximum hang time, maximum elevation)

Vo = initial velocity [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time = 14 [s]

0 = Vo - (9,81*14)

Vo = 137.34 [m/s]

B)

Now we use the following equation.

\(y=y_{o}+v_{o} *t -(\frac{1}{2} )*g*t^{2}\)

where:

Yo = initial position = 0

Y = final position [m]

Vo = initial velocity = 137.34 [m/s]

Now replacing

Y = (137.34*14) - (0.5*9.81*14²)

Y = 961.38 [m]

C)

With the above equation, we can calculate the distances between t = 3 and t = 4 [s]

\(y=y_{o} +v_{o} *t - (1/2)*g*t^{2}\)

y = (137.34*3) - (0.5*9.81*3²)

y = 367.8[m] for t = 3 [s]

for t = 4[s]

y =  (137.34*4) - (0.5*9.81*4²)

y = 470.88[m]

Answer:

yes you can find a motor in a circuit

Answer:

neutral

Explanation:

Answer:

The time required is 0.45 s.

Explanation:

Height, h = 1 m

initial velocity, u = m/s

Let the time is t.

Use second equation of motion

\(h = u t + 0.5 at^2\\\\1 = 0 +0.5 \times 9.8 \times t^2\\\\t = 0.45 s\)

Cricket bat uses one common law of physics

When a ball hits the bat with some velocity ,the cricket bat should hit with some more velocity to get greater impact velocity to move the ball somewhere .

Answer:

The mass of the apple is 0.49kg

Explanation:

Potential energy=mgh

P=mgh

6=m×1.22×10

6=12.2m

divide both sides by 12.2

m=6/12.2

m=0.49kg