the toddler now bounces high enough to lose contact with the mattress once each cycle, like a trampoline. what is the minimum amplitude of oscillation (in cm) required for this to occur?

The angular velocity of link BD is 2.89 rad/s and the angular acceleration is 0 rad/s².

The expression for angular acceleration can be calculated by applying the vector method.

Here, the vector method of angular velocity will be applied to get the expression for angular velocity ω2 and angular acceleration α2 as follows:

Step 1:Velocity of the collar relative to B and velocity of point A of the link AB and collar can be written as VB = VA + AB X ω1

Here, AB is perpendicular to the plane of rotation, and the velocity of A relative to the collar is ω1 * AB.

Therefore, AB X ω1 = 0 as AB is parallel to AB. Thus, VB = VA.

Then, the velocity of point B can be given as:

VB = BD X ω2Hence, BD X ω2 = VA, we get: ω2 = VA / BD

Step 2: Acceleration of point A relative to collar: AB X α1 = 0 (AB parallel to the plane of rotation)

AA = AB X ω1 X ω1=0

Acceleration of point A relative to C: AC X α1 = AC X 10α2=0

Since, AC=AB=1m

α2=0

Acceleration of collar C relative to point B: BD X α2=VA

Here, α2 = 0, and VB = VA = 5m/s.

Hence:BD * 0 = 5m/sω2 = VA / BD= 5 / (1.732 x 1)=2.89 rad/s

Thus, the angular velocity of link BD is 2.89 rad/s and the angular acceleration is 0 rad/s².

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120m per minute

If you take 960 and divide it by 8 you get 120 then you have the answer 120m per minute

Answer:

The charge on each end of the wire produces what we call an electric field. The electric field makes negative electrons want to travel towards the positive end of the wire (the end with not enough electrons).

the answer is d because it's going to positive.

is am alright? let me know.

Note that the age of material on a tectonic plate depends in part on what type of crust it is composed of. Oceanic crusts tends to be younger because it is constantly recycled at the Mid-oceanic ridges. On the other hand, the Continental crust tends to be older because it can often remain intact for longer periods.

Tectonic plates are massive chunks of the Earth's crust and upper mantle. They are composed of oceanic and continental crust. Earthquakes occur around mid-ocean ridges and major faults that indicate plate boundaries.

It is to be noted that Tectonic plates are important because they form the Earth's crust and determine the location and type of earthquakes, volcanoes, and mountain ranges.

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Full Question:

The age of material on a tectonic plate depends in part on what type of crust it is composed of. ________________tends to be younger because it is constantly recycled at the __________________ . On the other hand, _______________tends to be older because it can often remain intact for longer periods.

1. (continental crust/ ocean floor/ oceanic crust)

2. (mid-oceanic ridges/ hot spots/deep sea vents)

3. (continental crust/ ocean floor/ oceanic crust)

The distance from the axis of the force applied is 2.05 m.

The distance from the axis of the force applied is calculated as follows;

The formula for torque;

τ = Fr

where;

Another formula for torque is given as;

τ = Iα

where;

τ = (mr²)α

τ = (25 kg x (0.925 m)²) x (1.84 rad/s²)

τ = 39.36 Nm

The distance is calculated as;

r = τ/F

r = ( 39.36 Nm ) / (19.2 N)

r = 2.05 m

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Answer:

What is 37 °C in °F?

98.6

According to calculations, 37 °C is calculated to be equal to 98.6 °F. The formula below can be used to convert temperatures from Celsius (°C) to Fahrenheit (°F): °F = (°C * 9/5) + 32.

We may convert  temperature measured in degrees Celsius to degrees Fahrenheit by multiplying it by 1.8 and then by 32 since one degree Celsius is equal to 33.8 degrees Fahrenheit.

Use the following formula to convert temperature from Celsius (°C) to Fahrenheit (°F): °F = (°C * 9/5) + 32

The given temperature is 37 °C.

°F = (37 * 9/5) + 32

So, °F = 98.6 °F

Thus, 37 °C is equivalent to 98.6 °F.

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The statement which is false is as follows:

Thus, the correct option for this question is D.

Friction may be defined as a type of force that resists the sliding or rolling of one solid object over another through the surface that acts on it.

Friction is a force that takes place as the resistance of motion when one object rubs over another. Friction force always opposes the motion along with air resistance.

Friction can also be somehow reduced with the impact of lubrication over the rubbing surface. It is true that air resistance is greater the faster the object is traveling.

Air resistance does not depend on the blowing of the wind. It does not mean that if there is no wind blowing, the air resistance is negligible.

Therefore, the correct option for this question is D.

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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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The average speed of a particle in the position X=3t-3t² is 18m/s

At the position X = 3t - 3t²

t₁ = 1s = X = 3(1) - 3(1)²

X = (3 x 1) - (3 x 1²)

X = 3 - 3 = 0m

At t₂  = 3s = X = 3(3) - 3(3)²

X = (3 x 3) - (3 x 3²)

X = 9 - 27 = -18m

Displacement, Δx = X₂ - X₁

= -18 - 0 = -18m

The change in time would then be

Δt = t₂ - t₁

= 2 - 1 = 1

The Average velocity can be calculated by dividing the change in displacement by the change in time, Vₐ = Δx/Δt

Vₐ = -18m/1s = -18m/s

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Answer:

water seeping through a pot of soil

Explanation:

All the others permanently change the chemical structure of the element so they are chemical reactions

Answer: 20.29m/s

Explanation:

v^2 = v^2o + 2a(x-xo)

v^2 = velocity

v^2o = initial velocity

a = acceleration

x = final position/distance

xo = initial position/distance

In this case, the initial velocity is 0 since the ball wasn't moving before it was dropped. The final position is 21 as the motion ended after the ball traveled 21m. The initial position is 0. The acceleration is 9.8m/s (free fall). Plug these numbers into the formula:

v^2 = 0 + 2(9.8)(21)

v^2 = 20.28792744

Round to get 20.29m/s

The correct answer is 29.41 * 10⁻⁴ m/s

What is the work function?

From the photoelectric theory of Einstein, photons are emitted when light of appropriate wavelength is incident on a clean metal surface. The minimum energy that the incident light must possess is called the  work function of the metal.

We know that;

Kinetic energy = Energy of incident photon - Work function

Energy of incident photon = hf = 6.6 * 10⁻³⁴ Js * 1.51×10¹⁵ s−1

= 9.9 * 10⁻¹⁹ J

Work function = Energy of incident photon - Kinetic energy

8.17 * 10⁻¹⁹ = 9.9 * 10⁻¹⁹ - KE

KE = 8.17 * 10⁻¹⁹ - 9.9 * 10⁻¹⁹ = -1.73 *10⁻¹⁹ J

Kinetic energy = 0.5mv²

But since photons are massless so Kinetic energy will be equal to 0.5v².

KE = 0.5v²

-1.73*10⁻¹⁹ = 0.5v²

v² = -865*10⁻¹⁶

v = 29.41 * 10⁻⁴ m/s

Therefore, the velocity of ejected photons from a gold surface having work function 8.17x10⁻¹⁹ and frequency 1.51 x 10¹⁵  is 29.41 * 10⁻⁴ m/s.

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The magnitude of the magnetic field inside the solenoid is approximately 0.0754 Tesla.

The magnitude B of the magnetic field inside the solenoid can be calculated using the formula B = μ₀ * n * I / l, where μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A), n is the number of turns, I is the current, and l is the length of the solenoid.

B = (4π x 10⁻⁷ Tm/A) * (900 turns) * (200 x 10⁻³ A) / (30 cm)
Convert the length to meters:
l = 30 cm * (1 m / 100 cm) = 0.3 m
Now, calculate B:
B = (4π x 10⁻⁷ Tm/A) * (900 turns) * (200 x 10⁻³ A) / (0.3 m)
B ≈ 0.0754 T

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Answer:

The value of tension in the two ropes are 354.2 N and 553.8 N

Explanation:

Find the free body diagram attached below.

For the system to be in mechanical equilibrium, the following conditions have to be satisfied.

1. Upward forces must be equals to the downward forces,

T1 + T2 =210N + 698N

\(T1 + T2 =908 ------------------ equation 1\)

2. Clock-wise turning moments = anti-clockwise turning moments

\(T1 \times 2.8 = 210N \times 1.4m + 698N\times (2.8 - 1.8)\\2.8T1 =294Nm + 698Nm\\2.8T1 =992Nm\\T1 = \frac{992}{2.8}=354.2 N\)

From this, we were able to evaluate the value of T1 by taking our moments about T2 as the turning point

We can input the value of T1 into equation 1 to give us the value of T2 as

T2 = 908 - T1

T2 = 908 - 354.2N= 553.8 N

There fore, the value of tension in the two ropes are 354.2 N and 553.8 N

Basically, light waves hit different parts of the spoon at different angles, so they all bend a little differently. By the time they come back to you, they're all crooked, so you're looking upside down. It has a concave side.

A glossy spoon is like a mirror, only curved. Since the surface is perfectly smooth, it creates a mirror image, or specular reflection. However, since the spoon is curved, the direction of the reflected light depends on where it hits. This outwardly curved surface is called a convex surface. Basically, light waves hit different parts of the spoon at different angles, so they all bend a little differently. By the time they come back to you, they're all crooked, so you'll look upside down. It has a concave side.

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The force experienced by a cosmic ray electron that moves perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.0 x 10⁻⁵ T is   -1.0025 x 10^-11 N.

The problem asks us to find the force experienced by a cosmic ray electron that moves perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.0 x 10⁻⁵ T. We can use the formula F = qvB, where F is the force experienced by the electron, q is its charge, v is its velocity, and B is the magnetic field strength.

In this case, we know the velocity of the electron is 6.25 x 10⁷ m/s and the magnetic field strength is 1.0 x 10⁻⁵ T. However, we don't know the charge of the electron. We can assume it is the same as the charge of an electron, which is -1.602 x 10⁻¹⁹ C.

So, plugging in the values into the formula, we get:

F = (-1.602 x 10⁻¹⁹ C) x (6.25 x 10⁷ m/s) x (1.0 x 10^-5 T)

F = -1.0025 x 10⁻¹¹ N

Therefore, the force experienced by the cosmic ray electron is -1.0025 x 10⁻¹¹ N. Note that the negative sign indicates that the force is in the opposite direction to the velocity of the electron, which is consistent with the fact that the electron is moving perpendicular to the magnetic field.

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Solution :

A vector is defined as an element that has magnitude of some measure and direction.

It is given there is vector 'd' which has magnitude 2.6 m and its direction is towards north.

a). -d

    The magnitude of the vector '-d' is 2.6 m and its direction is reversed, i.e. its direction is towards south.

b). d/2.0

  The magnitude of the vector 'd/2.0' is 1.3 m and its direction is towards north.

c). - 2.5d

  The magnitude of the vector increases by 2.5 times i.e. 2.5 x 2.6 = 6.5 m and the direction is towards south.

d). 5.0d

    The magnitude of the vector increases by 5 times i.e. 5 x 2.6 = 13 m and the direction is towards north.

To find the magnitude of the drag force of the wind on the canopy, we can use the drag force formula:

Drag Force (Fd) = 0.5 × Drag Coefficient (Cd) × Air Density (ρ) × Velocity^2 (v^2) × Area (A)

We are given:
- Velocity (v) = 6.5 m/s
- Drag Coefficient (Cd) = 0.50
- Air Density (ρ) = 1.2 kg/m³
- Area (A) is not provided in the question.

Since we don't have the canopy's area, we cannot find the exact magnitude of the drag force. However, we can provide a general equation:

Fd = 0.5 × 0.50 × 1.2 kg/m³ × (6.5 m/s)² × A

Fd = 0.3 × 42.25 × A

Fd = 12.675 × A

The magnitude of the drag force of the wind on the canopy is 12.675 times the canopy's area (A). To find the exact value, you'll need to know the canopy's area in square meters.

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Since acceleration is negative, the skater will travel a positive distance before coming to a stop. Therefore, the skater will travel a distance of 16 / acceleration + 8 / acceleration.

The distance the skater will travel before coming to a stop can be determined using the equation of motion: distance = initial velocity × time + (1/2) × acceleration × time². In this case, the skater is coasting to a stop, so their acceleration is negative, opposing their initial velocity.

Given that the skater is moving at 4 m/s initially and neglecting air resistance, we know the initial velocity is 4 m/s. Since the skater is coasting to a stop, their final velocity will be 0 m/s.

To find the distance traveled, we need to determine the time it takes for the skater to stop. Since the acceleration is constant, we can use the equation final velocity = initial velocity + (acceleration × time). Solving for time, we get time = (final velocity - initial velocity) / acceleration.

Substituting the given values, we have time = (0 - 4) / acceleration. Since the skater is stopping, the acceleration is negative, so we have time = -4 / acceleration.

Now, substituting this value of time into the equation of motion, we have distance = 4 m/s × (-4 / acceleration) + (1/2) × acceleration × (-4 / acceleration)².

Simplifying the equation, we get distance = -16 / acceleration - 8 / acceleration.

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The resultant of the forces is 29 N. Option D

The resultant force is the force that acts in a given direction. Now we have two forces as enumerated in the question.

Thus;

Resultant = √(10)^2 + (20)^2 - [2 * 10 * 20 * cos (60 - 30))

Resultant = 29 N

Thus, the resultant of the forces is 29 N.

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Missing parts;

Two forces of magnitude 10N and 20N act on a body in directions making angles 30° and 60° respectively with x-axis what is the resultant force

A.17N

B. 19N

C. 23N

D. 29N

E. 37N​

Answer:

• Go Native. Use native plants in your landscape. ...

• Reduce Chemical Use. Use fewer chemicals around your home and yard, and make sure to dispose of them properly - don't dump them on the ground!

• Manage Waste. ...

Explanation:

There are some easy ways to conserve water to help protect our aquifers, watersheds, and ground water. First, we can all conserve water, by not running the water while we brush our teeth and making sure we fix all leaking or dripping faucets. Another was to conserve water is to collect rain water to use to water your plants. Speaking of plants, you can also plant more zeroscape type landscapes that require less water. We can also update our shower heads and toilets to be low-flow and more efficient. Can't afford a new toilet? Simply add a brick to your tank and this will cut down on the water used for every flush.

The option that would NOT suggest an interaction effect is the "Two parallel lines." interaction effect. The correct answer is option(a).

When one independent variable's effect on the dependent variable varies according to the value of another independent variable, this is known as the interaction effect. In other words, the level of the other independent variable determines the impact of one independent variable on the dependent variable. For example, in a study on the effect of a new medication on blood pressure, the interaction effect would occur if the impact of the medication varies depending on the age of the patients.

Age would be the moderating variable in this example. According to the given options, two parallel lines would represent that the two independent variables being analyzed have no effect on the dependent variable, meaning that there is no interaction effect present. Therefore, option A would NOT suggest an interaction effect. The remaining options suggest an interaction effect as they indicate that there is an impact on the dependent variable based on the level of the independent variables.

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The wavelength of the emitted photon is approximately 2.17 um.

The energy of the emitted photon can be calculated using the formula:

Given that the energy of the emitted photon is 0.57 eV, we can rearrange the formula to solve for the wavelength:

Substituting the given energy value, we get:

Therefore, the wavelength of the emitted photon is approximately 2.175 um.

The frequency of the emitted photon can be calculated using the equation:

Given the wavelength of 2.175 um, we need to convert it to meters by multiplying by 10⁻⁶:

Now we can calculate the frequency:

Therefore, the frequency of the emitted photon is approximately 1.38 × 10^14 Hz.

(c) Wavelength and frequency are related by the speed of light equation:

Since the speed of light is a constant, we can rearrange the equation to solve for wavelength:

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The X₅ impulsive flare has a total X-ray flux that is four times greater than the X1 long-duration flare.

The total X-ray flux for the X5 impulsive flare is:

5x10-4e⁻² W/m² * 3600 s/hour * 1 hour

= 5.4 J/m²

The total X-ray flux for the X1 long-duration flare is:

1x10⁻⁴ e-t/3 W/m² * 3600 s/hour * 1 hour

= 1.2 J/m²

As you can see, the X₅ impulsive flare has a total X-ray flux that is four times greater than the X1 long-duration flare. Therefore, the X impulsive flare is more likely to signal the beginning of a significant space weather event.

The reason for this is that the X₅ impulsive flare is a much more powerful event. It releases a much larger amount of energy in a much shorter period of time.

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Answer:

The first law is that if no force acts on an object, its motion will not change. In the second law, the force acting on an object is equal to its mass and acceleration. The third law is that when two objects interact, they act on each other with equal magnitude and opposite forces.

Answer:

a). Average speed = 26 km per hour

b). Average velocity = 10 km per hour

Explanation:

a). Distance between A to B = 5 km

    Distance between B to C = 4 km

    Total distance traveled between A to B to C and then back to B

    = 5 + 4 + 4

    = 13 km

Time taken to cover this distance of 13 km = \(\frac{1}{2}\) an hour

Therefore, average speed during this trip = \(\frac{13}{\frac{1}{2}}\)

                                                                      = 26 km per hour

b). Average velocity = \(\frac{\text{Displacement}}{\text{Duration}}\)

   Displacement = (Distance between A to B) + (Distance traveled between B to C) - (Distance traveled from C to B)

                           = 5 + 4 - 4

                           = 5 km towards A

Average velocity = \(\frac{5}{\frac{1}{2}}\)

                             = 10 km per hour

mark as brain list answer

Answer:

Approximately \(14\; {\rm m\cdot s^{-1}}\).

Explanation:

The gravitational potential energy \(\text{GPE}\) of this roller coaster is proportional to the height \(h\) of the roller coaster.

The kinetic energy \(\text{KE}\) of this roller coaster is proportional to the square of speed \(v\).

The question states that the track is frictionless. Thus, during the descent, the \(\text{GPE}\) of this roller coaster is turned into \(\text{KE}\) without any energy loss.

When the roller coaster was at \((1/2)\) of the the initial height, only\(1 - (1/2) = (1/2)\) of the original \(\text{GPE}\) was turned into \(\text{KE}\). The \(\text{KE}\!\) of this roller coaster at that height would be \(1 - (1/2) = (1/2)\!\) of the \(\text{KE}\!\!\) when the roller coaster is at the ground level.

The \(\text{KE}\) of the roller coaster is proportional to \(v^{2}\) (the square of speed \(v\).) Thus, since the \(\text{KE}\!\) at \((1/2)\) the initial height is \(1 - (1/2) = (1/2)\!\) the \(\text{KE}\!\!\) at the ground level, the \(v^{2}\) at \((1/2)\!\) the initial height would also be \(1 - (1/2) = (1/2)\!\) the \(v^{2}\!\) at the ground level.

Since \(v = 20\; {\rm m\cdot s^{-1}}\) at the ground level, \(v^{2} = (20\; {\rm m\cdot s^{-1}})^{2}\) at the ground level. The \(v^{2}\) at \((1/2)\) the initial height would then be:

\((1 - (1/2))\times (20\; {\rm m\cdot s^{-1}})^{2}\).

Thus, the speed \(v\) at \((1/2)\) the initial height would be:

\(\begin{aligned}& \sqrt{(1 - (1/2))\times (20\; {\rm m\cdot s^{-1}})^{2}} \\ =\; & \sqrt{\frac{1}{2} \times (20\; {\rm m\cdot s^{-1}})^{2}} \\ =\; & \sqrt{200}\; {\rm m\cdot s^{-1}} \\ \approx\; & 14\; {\rm m\cdot s^{-1}}\end{aligned}\).