If you're under stress what is the best way to recharge and calm the brain

Answers

Answer 1

Answer:

Take slow, deep breaths. Or try other breathing exercises for relaxation. ...

Soak in a warm bath.

Listen to soothing music.

Practice mindful meditation. The goal of mindful meditation is to focus your attention on things that are happening right now in the present moment. ...

Write. ...

Use guided imagery.

Answer 2
Take deep breaths and close your eyes

Related Questions

9. Two spheres, one of mass 4.0 x 107 kg and the other of mass 8.0 x 107 kg are separated by a
distance of 400.0 m. The force of gravity between them is
a) 1.3 N b) 530 N c) 1.8 x 104 N
d) 5.1 x 1021 N
e) 5.3 x 1022 N

Answers

Hi there!

We can use the gravitational force equation:

[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]

Where:

G = Gravitational Constant

m₁ = mass of one sphere (kg)

m₂ = mass of the other sphere (kg)

r = distance between the spheres (m)

Plug in the given values into the equation:

[tex]F_g = \frac{(6.67*10^{-11})(4.0*10^7)(8.0*10^7)}{400^2} = \boxed{1.334 N}[/tex]

Thus, the correct answer is A.

What are last two commandments? Please write them down.

Answers

[tex]⇒[/tex]

The Second Commandment refers to: The second of the Ten Commandments, which is: "Thou shalt have no other gods before me. Thou shalt not make unto thee any graven image" under the Talmudic division of the third-century Jewish Talmud.

Answer:

Thou shalt love the Lord thy God with all thy heart, and with all thy soul, and with all thy mind. This is the first and great commandment. And the second is like unto it, Thou shalt love thy neighbor as myself.

Explanation:

Why do heavier objects roll down slopes faster than lighter objects? (Assuming they are the same shape)

Answers

Answer:

because of the friction against the object

Explanation:

sjkdvsndjkvnisdvnuisbvibsdv.

Answers

Answer:

hello!

Explanation:

what is the magnitude of its velocity just before it strikes the ground, if the bottle was thrown straight out horizontally from the tower with a speed of 11.8 m/s?

Answers

There's not enough given information to find the answer.

The horizontal velocity has no effect on anything that happens vertically. To find the object's vertical speed when it hits the ground, we need to know what height it was dropped from, and that's not given in the question.

Since the bottle is falling vertically, the vertical displacement s is the height of the tower from which it was thrown.

If a bottle is thrown horizontally from a tower, its initial vertical velocity is 0 m/s since there is no vertical component to its initial velocity. The only force acting on the bottle in the vertical direction is gravity, which causes it to accelerate downward at a rate of approximately 9.81 m/s² (assuming Earth's gravity).

Since the initial vertical velocity is 0 m/s and the acceleration is 9.81 m/s², you can use the following kinematic equation to find the magnitude of its velocity just before it strikes the ground:

v2=u2+2as

Where:

v is the final vertical velocity (which we're trying to find)

u is the initial vertical velocity (0 m/s)

a is the acceleration due to gravity (-9.81 m/s², negative because it's acting downward)

s is the vertical displacement (which is the height of the tower)

Since the bottle is falling vertically, the vertical displacement

s is the height of the tower from which it was thrown.

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Select all of the following that are involved in translation:

amino acids
tRNA
mRNA
ribonucleotides
deoxyribonucleotides
RNA Polymerase

Answers

Answer:

Amino Acids

tRNA

mRNA

ribonucleotides

Explanation:

Amino Acids

- During translation, an mRNA sequence is read using the genetic code, which is a set of rules that defines how an mRNA sequence is to be translated into the 20-letter code of amino acids, which are the building blocks of proteins.

tRNA

- At the beginning of translation, the ribosome and a tRNA attach to the mRNA. The tRNA is located in the ribosome's first docking site. This tRNA's anticodon is complementary to the mRNA's initiation codon, where translation starts. The tRNA carries the amino acid that corresponds to that codon. ransfer ribonucleic acid (tRNA) is a type of RNA molecule that helps decode a messenger RNA (mRNA) sequence into a protein. tRNAs function at specific sites in the ribosome during translation, which is a process that synthesizes a protein from an mRNA molecule.

mRNA

- Translation is the process by which a protein is synthesized from the information contained in a molecule of messenger RNA (mRNA). ... Then a transfer RNA (tRNA) molecule carrying the amino acid methionine binds to what is called the start codon of the mRNA sequence.

ribonucleotides

- During transcription, a ribonucleotide complementary to the DNA template strand is added to the growing RNA strand and a covalent phosphodiester bond is formed by dehydration synthesis between the new nucleotide and the last one added.

The weight of the atmosphere above 1 m- of
Earth's surface is about 100.000 Ņ. Density, of
course, becomes less with altitudd- But suppose
the density of air were a constant 1.2 kg/m?

Answers

Answer:

1.09 kg.m

Explanation:

no need

A 25kg cannon ball is shot out of a 1600 kg cannon that is initially at rest . When fired the velocity of the cannon ball is 200m/s. What is the recoil velocity of the cannon?

Answers

The correct answer is the recoil velocity of the cannon is equal to 90m/s.

The recoil velocity of the cannon is equal to 3.125m/s.

What is law of conservation of linear momentum?

According to this law, the sum of the momentum before and after the collision must be equal.

m₁ u₁ + m₂u₂ = m₁v₁ + m₂ v₂

where m₁ and m₂ is the mass of the collided bodies, u₁ and u₂ are initial speed while v₁ & v₂ is final speed.

Conservation of momentum is a characteristic held by an object where the total amount of momentum never changes.

Given the mass of the cannonball, m₁ = 25 Kg

The velocity of the cannonball, v₁ = 200 m/s

The mass of the cannon, m₂ = 1600 Kg

Assume that v₂ is the recoil velocity of the cannon.

From the law of conservation of momentum: P(initial) = P (final)

m₁ v₁ = m₂ v₂

25 × 200 = 1600 × v₂

v₂ = 3.125 m/s

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What is the gravitational potential energy of a circus performer with a mass of 50 kg walking on a tightrope 10 m above the ground

Answers

Answer:

5000 J

Explanation:

GPE = mgh

50*10*10

= 5000 J

The gravitational potential energy of the circus performer walking on the tightrope above the ground is 4900J.

Given the data in the question;

Mass of circus performer; [tex]m = 50kg[/tex]Distance from ground level; [tex]h = 10m[/tex]Gravitational potential energy; [tex]U = ?[/tex]

Gravitational Potential Energy

Gravitational potential energy is simply the potential energy an object possesses relative to another object due to gravity.

It is expressed as;

[tex]U = mgh[/tex]

Where m is mass, h is height or distance from the other massive object and g is the gravitational field ( [tex]g = 9.8m/s^2[/tex] )

To determine the gravitational potential energy of  circus performer, we substitute our values into the expression above.

[tex]U = mgh\\\\U = 50kg\ *\ 9.8m/s^2\ *\ 10m\\\\U = 4900kgm^2/s^2\\\\U = 4900J[/tex]

Therefore, the gravitational potential energy of the circus performer walking on the tightrope above the ground is 4900J.

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9) How many valence electrons does this atom have?
A.2
B. 3
C.8
D. 13

Answers

Answer:

I think it's 2

Explanation:

Within each column, or group, of the table, all the elements have the same number of valence electrons. This explains why all the elements in the same group have very similar chemical properties. For elements in groups 1–2 and 13–18, the number of valence electrons is easy to tell directly from the periodic table.

The first bag you load into your vehicle from the pick-stage cart is the first bag you need to open on your route.

Answers

The given statement is false because the first bag you load into the vehicle will be at the bottom during the trip.

During a pick-stage cart, the first bad you load into the vehicle is always placed on the bottom, while subsequent bags are arranged on top of the first bag, all inside the vehicle.

At any point on your route, any bad you tend to open becomes the last bag you load into the vehicle because the first bag is far below the bottom.

Thus, we can conclude that the given statement is false because the first bag you load into the vehicle will be at the bottom during the trip.

"Your question is not complete, it seems to be missing the following information";

The first bag you load into your vehicle from the pick-stage cart is the first bag you need to open on your route. False/True

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A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 when the ball is at its lowest point (θ = 0).
a) What is the gravitational potential energy, in joules, of the ball before it is released?
b) What will be the speed of the ball, in meters per second, when it reaches the bottom?

Answers

Explanation:

a) The height of the ball h with respect to the reference line is

[tex]h = L - L\cos{31°} = L(1 - \cos{31°})[/tex]

so its initial gravitational potential energy [tex]U_0[/tex] is

[tex]U = mgh = mgL(1 - \cos{31°})[/tex]

[tex]\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})[/tex]

[tex]\:\:\:\:\:=0.23\:\text{J}[/tex]

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

[tex]\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U[/tex]

We know that the initial kinetic energy [tex]K_0,[/tex] as well as its final gravitational potential energy [tex]U[/tex] are zero so we can write the conservation law as

[tex]mgL(1 - \cos{31°}) = \frac{1}{2}mv^2[/tex]

Note that the mass gets cancelled out and then we solve for the velocity v as

[tex]v = \sqrt{2gL(1 - \cos{31°})}[/tex]

[tex]\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}[/tex]

[tex]\:\:\:\:\:= 1.3\:\text{m/s}[/tex]

Answer:

a. 0.23J

b. 1.35 m/s

Explanation:

a. U = mgh where where m = mass of the object, g = acceleration due to gravity, and h = height

h = L - Lcos(θ) where L = length of the rope, and θ = angle with respect to vertical.

Therefore, U = mg(L - Lcos(θ))

U = 0.25 * 9.8 (0.65m - 0.65cos(31° ))

U = 0.2275 ≈ 0.23J

The gravitational potential energy of the ball before it is released = 0.23J

b. To determine the velocity of the object at the bottom of its motion, all of the energy has gone from gravitational potential into kinetic since at the bottom, the problem says that U = 0. The kinetic energy of an object is given by the following equation:  

[tex]K.E=\frac{I}{2}*mv^{2}[/tex]

where m = mass of the object and v = velocity of the object. Since we know that all of the energy was transferred into kinetic energy at the bottom, we can conclude that:

[tex]0.2275=\frac{1}{2} *0.25*v^{2}[/tex]

[tex]v^{2}=\frac{2*0.2275}{0.25}[/tex]

[tex]v^{2}=1.82[/tex]

[tex]v=\sqrt{1.82}=1.3491\\[/tex] ≈ [tex]1.35m/s[/tex]

Therefore, the speed of the ball when it reaches the bottom = 1.35m/s

 

A 100-g block slides back and forth on a frictionless surface between two springs, as shown in Fig. 7.18. The left-hand spring has k = 110 N/m and its maximum compression is 21 cm.The right-hand spring has
k = 240 N/m. Find (a) the maximum compression of the right-hand spring and (b) the speed of the block as it moves between the springs.

Answers

(a) Energy is conserved at every point in the block's motion, so the potential energy P stored in the first spring at its maximum compression is the same as is stored in the second spring.

The total work performed on the block by the first spring is

W = -1/2 (110 N/m) (0.21 m²) = -2.4255 J

The work performed by the second spring is the same, so

W = -1/2 (240 N/m) x²

Solve for x :

x² = -2W/(240 N/m) = 0.0202125 m²

x ≈ 0.14 m = 14 cm

(b) By the work-energy theorem, the total work performed by either spring on the block as the spring is compressed is equal to the change in the block's kinetic energy. The restoring force of the spring is the only force involved. At maximum compression, the block has zero velocity, while its kinetic energy and hence speed is maximum just as it comes into contact with either spring.

W = 0 - K

W = -1/2 (0.10 kg) v²

v² = -2W/(0.10 kg) = 48.51 m²/s²

v ≈ 7.0 m/s

The Israelites grumbled 11 times against God. True or False?

Answers

Answer:

This is not a question about physics.

Anyway. Eleven times seems underrated.

Answer:

True

Explanation:

Because they didn't always appreciate the work of God

If a block is sitting in a shelf 5 foot high and the same size block is sitting on the shelf above it at 6 feet high. Which has more potential energy? Explain your reasoning.

Answers

Answer:

Explanation:

Assuming that not only is the size of the two blocks the same but the MASS is also the same, then the higher block will have more potential energy.

Potential energy is equal to the amount of work required against gravity to position a mass

Work is a force times a distance. PE = W = Fd

As each block has the same mass, they each create the same force which would be their weight mg

PE = W = Fd = mgd

so potential energy is directly proportional to the height of each block

the lower block has potential energy PE = Fd = mg(5)

while the upper block has potential energy PE = mg(6)

above a reference frame origin, possibly the floor.

It is possible that the higher block has less potential energy then the lower block if only the size is considered. If the higher block is a one foot cube of ice at 57.2 lbs and the lower block is a one foot cube of copper at 559 lbs then the lower mass would have much more potential energy because of its much greater mass.

559(5) = 2795 ft•lb of potential energy

57.2(6) = 343.2 ft•lb of potential energy

How much force is required to swing a 3 kg mass in a circle with a radius of 1.5 m at a speed of 5 m/s?

Answers

The force that is required to swing the mass is 50 Newton.

Given the following data:

Mass = 3 kgRadius = 1.5 metersSpeed = 5 m/s

To find the amount of force that is required to swing the mass, we would use centripetal force formula:

Mathematically, centripetal force is given by the formula:

[tex]F_c = \frac{mv^2}{r}[/tex]

Where:

m is the mass.v is the speed.r is the radius.

Substituting the parameters into the formula, we have;

[tex]F_c = \frac{3 \times 5^2}{1.5}\\\\F_c = \frac{3 \times 25}{1.5}\\\\F_c = \frac{75}{1.5}\\\\F_c = 50[/tex]

Centripetal force = 50 Newton

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how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure change

Answers

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

[tex]\rho = \frac{PM}{RT}[/tex]

[tex]\frac{dP}{dz} = -\rho g\\\\[/tex]

where;

ρ is densityT is temperaturedP is rate of change of vertical  pressure

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

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In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H .

Answers

We have that for the Question "In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H ." it can be said that the time required is

[tex]T=0.745s[/tex]

From the question we are told

In the circuit, the capacitor is fully charged when switch is closed. Calculate the time needed for the potential energy stored by the circuit to be equally distributed between the capacitor and inductor. The capacitance is =20.0 mF and inductance is =45.0 H .

Generally the equation for the Time   is mathematically given as

[tex]T=\sqrt{LC}*\frac{\pi}{4}[/tex]

[tex]T=\sqrt{20*10^{-3}*45}*\frac{\pi}{4}[/tex]

[tex]T=0.745s[/tex]

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The LANDSAT C Earth resources satellite has a nearly circular orbit with an eccentricity of 0.00132. At perigee the satellite is at an altitude (measured from the earth's surface) of 417km. a) Calculate its altitude at apogee. b) Calculate its period. c) Calculate the velocity at perigee.

Answers

We have that for the Question it can be said that

altitude at apogee = [tex]434.96km[/tex]period = [tex]5585.41s[/tex]velocity at perigee = [tex]7.67km/s[/tex]

From the question we are told

The LANDSAT C Earth resources satellite has a nearly circular orbit with an eccentricity of 0.00132. At perigee the satellite is at an altitude (measured from the earth's surface) of 417km.

Generally the equation for perigee radius is mathematically given as

[tex]r_p = R_E+Z_P\\\\=6378+417\\\\=6795km[/tex]

the equation for orbit is

[tex]r_p = \frac{h^2}{u}*\frac{1}{1+e}\\\\6795 = \frac{h^2}{398600}*\frac{1}{1+0.00132}\\\\h = 52077.5km^2/s[/tex]

the equation for velocity at perigee is mathematically given as

[tex]V_p = \frac{h}{r_p}\\\\= \frac{52077.5}{67.95}\\\\=7.67km/s[/tex]

the equation for Apagee radius is mathematically given as

[tex]r_a = \frac{h^2}{u} * \frac{1}{1-e}\\\\= \frac{52077.5^2}{398600} * \frac{1}{1-0.00132}\\\\= 6812.97km[/tex]

the equation for altitude at Apagee is mathematically given as

[tex]Z_a = r_a - R_E\\\\=6812.97-6378\\\\=434.96km[/tex]

the equation for period

[tex]Period T = \frac{2\pi}{u^2} * (\frac{h}{\sqrt1-e^2})^3\\\\= \frac{2\pi}{398600^2} * (\frac{52077.5}{\sqrt1-0.00132^2})^3\\\\= 5585.41s[/tex]

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An object of mass is used to provide tension in a 6.0 ‑m‑long string that has a mass of 0.252 kg, as shown. A standing wave that has a wavelength equal to 1.5 m is produced by a source that vibrates at 24 Hz.

Answers

The tension on the string at the given length and mass is 54.43 N.

The given parameters:

length of the string, L = 6.0 mmass of the string, m = 0.252 kgwavelength, λ = 1.5 mfrequency of the source, F = 24 Hz

The speed of the wave is calculated as follows;

[tex]v= f\lambda\\\\v = 24 \times 1.5\\\\v = 36 \ m/s[/tex]

The tension on the string is calculated as follows;

[tex]v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{T}{M/l} } \\\\v = \sqrt{\frac{Tl}{M} } \\\\v^2 = \frac{Tl}{M} \\\\T = \frac{v^2 M}{l} \\\\T = \frac{(36)^2 \times 0.252}{6} \\\\T = 54.43 \ N[/tex]

Thus, the tension on the string at the given length and mass is 54.43 N.

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An object of mass is used to provide tension in a 6.0 ‑m‑long string that has a mass of 0.252 kg, as shown. A standing wave that has a wavelength equal to 1.5 m is produced by a source that vibrates at 24 Hz. Determine the tension on the string.

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demonstrate how souni wave works​

Answers

Answer:

Explanation:

Sound waves exist as variations of pressure in a medium such as air. They are created by the vibration of an object, which causes the air surrounding it to vibrate. The vibrating air then causes the human eardrum to vibrate, which the brain interprets as sound.

during the spin cycle of a washing machine the cloth stick to the outer wall of the barrel as it spins at a speed of 65 m per second. the radius of the barrel is 0.35 m. what is the magnitude and direction of centripetal acceleration of the clothes which are located on the wall of the barrel​

Answers

Hi there!

Centripetal acceleration can be written as:

[tex]a_c = \frac{v^2}{r}[/tex]

v = tangential velocity (m/s)

r = radius (m)

Plug in the given values:

[tex]a_c = \frac{65^2}{0.35} = \boxed{12071.43 \text{ m}/s^2}}}[/tex]

The centripetal acceleration vector ALWAYS points towards the CENTER POINT of the barrel.

A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant

Answers

516.154 megawatts of heat are exhausted to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ([tex]Q_{out}[/tex]), in megawatts:

[tex]Q_{out} = Q_{in} - W[/tex]

[tex]Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](1)

Where:

[tex]\eta[/tex] - Efficiency.[tex]W[/tex] - Electric power, in megawatts.

If we know that [tex]\eta = 0.39[/tex] and [tex]W = 330\,MW[/tex], then the energy rate exhausted to the river is:

[tex]Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)[/tex]

[tex]Q_{out} = 516.154\,MW[/tex]

516.154 megawatts of heat are exhausted to the river that cools the plant.

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3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *

Answer is 1.1 x 10^4 J

Answers

Answer:

WD = 11328 J → 1.1 × 10⁴ (to 2 sig figs)

Explanation:

WD = work done (J)

F = force (N), i.e. the normal force on the person

W = weight (N)

s = displacement (m)

m = mass (kg) = 60

a = acceleration (m/s²) = 2

g = gravity (m/s²) = 9.8

t = time (s) = 4

u = initial velocity (m/s) = 0

The formulas or equations that are relevant ate:

WD = F × s

F = m × a

W = m × g → note: this is simply a variation of the formula F = ma

s = ut + ¹/₂.at²

We want to find WD, so we need to know the force and the displacement (or distance);

Firstly, we want to find the normal force on the person in the elevator;

The downwards force the person produces on the elevator is the weight of the person, which is:

W = 60 × g

W = 60 × 9.8

W = 588

We are told the elevator is accelerating upward, so we can construct an equation to find F with the formula F = ma:

F - W = 60 × 2

F - 588 = 120

F = 708

Note: illustrations may be helpful to understand this, as has been shown in the picture

We also need displacement, which we use the formula s = ut + ¹/₂at²:

s = 0(4) + ¹/₂(2)(4)²

s = 16 m

Now we have F and s, we can calculate WD:

WD = 708 × 16

WD = 11328 J → This can be rounded to 1.1 × 10⁴ to 2 significant figures

answers please
5 pointss​

Answers

Answer:

1.a shadow is caused by an object blocking light so that it does not reach a surface. The area in shadow appears black because there is no light falling on it----it appears dark. In fact most shadows are not totally black because light usually bounces around the obstruction off other objects.

2.Shadow gives us the shape about the object and we can know the nature of that image by it's shadow. ... As shadow is complete with the all features of objects like it's shape, size and everything; This helps to know whether the shape of the object. Hence, Shadow tells us about the shape of the object.

3.The bird or an airplane flying high in the sky do not cast their shadow on the ground because they are at very high altitude due to which the ground which acts as screen for the shadow is far away and in absence of the screen the shadow cannot be formed.

Explanation:

BY Roses_are_Rosie

A trapeze artist performs an aerial maneuver. While in a tucked position, as shown in figure A, she rotates about her center of mass at a rate of 6.27 rad/s. Her moment of inertia about this axis is 17.9 kg⋅m2. A short time later, the aerialist is in the straight position, as shown in figure B. If the moment of inertia about her center of mass in this position is now 33.1 kg⋅m2, what is her rotational speed in revolutions per minute (rpm)?

Answers

The rotational speed of the trapeze artist is 32.372 revolutions per second.

In this question we should apply the principle of angular momentum conservation to determine the final angular speed of the trapeze artist, since there are no external forces acting on the trapeze artist:

[tex]I_{A}\cdot \omega_{A} = I_{B}\cdot \omega_{B}[/tex] (1)

Where:

[tex]I_{A}[/tex], [tex]I_{B}[/tex] - Initial and final moments of the trapeze artist, in kilograms-square meter.

[tex]\omega_{A}[/tex], [tex]\omega_{B}[/tex] - Initial and final angular speeds, in radians per second.

If we know that [tex]I_{A} = 17.9\,kg\cdot m^{2}[/tex], [tex]I_{B} = 33.1\,kg\cdot m^{2}[/tex] and [tex]\omega_{A} = 6.27\,\frac{rad}{s}[/tex], then the final angular speed is:

[tex]\omega_{B} = \frac{I_{A}}{I_{B}}\times \omega_{A}[/tex]

[tex]\omega_{B} = \frac{17.9\,kg\cdot m^{3}}{33.1\,kg\cdot m^{2}}\times 6.27\,\frac{rad}{s}[/tex]

[tex]\omega_{B} = 3.390\,\frac{rad}{s}[/tex] ([tex]32.372\,rpm[/tex])

The rotational speed of the trapeze artist is 32.372 revolutions per second.

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When converted to a household measurement, 9 kilograms is approximately equal to a) 9000 grams. b) 9000 ounces. 19.8 ounces. d) 19.8 pounds.

Answers

Answer:

9000 grams which is also very close to 19.8 pounds

so answers a) and d) are correct

Explanation:

9 kg(1000 gm/kg) = 9000 gms

9 kg(2.205 lb/kg) = 19.845 lb

I guess it rather depends on whose household you are in. One in the US would probably use 19.8 lb while one in Canada might use 9000 gm.


Sin is the ____________________________ of the law. 1 John 3:4.

Answers

Answer:

transgression

Explanation:

Một xe có khối lượng 20000kg chuyển động dần đều dưới tác dụng của lực 600N,vận tốc ban đầu của xe bằng 15m/s.hỏi :
a,gia tốc của xe
b,sau bao lâu xe dừng lại
c,đoạn đường xe đã chạy được kể từ lúc hàm cho đến khi dừng hẳn.

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when 15 newtons of force is applied to the 0.50 kg book, the friction keeps the book from sliding down the wall

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The minimum force to keep the book from sliding is 15 N.

The minimum coefficient of friction to keep the book from sliding is 0.33.

The given parameters;

applied force, F = 15 Nmass of the book, m = 0.5 kg

The normal force on the book is calculated as;

[tex]F_n = mg\\\\F_n = 0.5 \times 9.8\\\\F_n = 4.9 \ N[/tex]

The minimum force to keep the book from sliding is calculated as;

[tex]F - F_s = 0\\\\F_s = F\\\\F_s = 15 \ N[/tex]

The minimum coefficient of friction to keep the book from sliding is calculated as;

[tex]\mu_s F = F_n\\\\\mu_s = \frac{F_n}{F} \\\\\mu_s = \frac{4.9}{15} \\\\\mu_s = 0.33[/tex]

"Your question is not complete, it seems to be missing the following information";

When a force of 15 newtons is applied to the 0.50 kg book, the friction keeps the book from sliding down the wall. What is the minimum force and minimum coefficient of friction to keep the book from sliding?

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