The surface of the exit road is horizontal, not banked. (See figure.) If the static
friction between the tires and the surface of the road is us = 0.688 and the
maximum speed with which the car can exit the highway safely without sliding is
25.2 m/s, what is the radius of curvature of a highway exit, r?

Answer:

a) A Helium nucleus with two protons and one neutron

b) 0.324 of the original amount

Explanation:

a) When Tritium decays, the nucleus emits an electron and an antineutrino, which then changes it from a Triton with one proton and two neutrons, to a Helium nucleus with two protons and one neutron.

The decay constant for the decay of Tritium is gotten from

\(t_{1/2}\) = 0.693/k

where

\(t_{1/2}\) is the half life = 12.3 yrs

k is the decay constant = ?

substituting, we have

12.3 = 0.693/k

k = 0.693/12.3 = 0.0563

The fraction that will remain after 20 years can be calculated from

\(N\) = \(N_{0} e^{-kt}\)

where

\(N\) is the final remaining amount of Tritium after 20 years

\(N_{0}\) is the original amount of Tritium

k is the decay constant = 0.0563

t is the time = 20 years

Equation can be re-written as

\(N/N_{0}\) = \(e^{-kt}\)

where

\(N/N_{0}\) is the fraction of the tritium that will be remaining after 20 years.

substituting values, we have

\(N/N_{0}\) = \(e^{-0.0563 * 20}\)

\(N/N_{0}\)  =  \(e^{-1.126}\)

\(N/N_{0}\) = 0.324 of the original amount

Answer: 112 meters

Explanation:

Use the equation: d = vi+vf/2 x t which turned into: 28/2 times 8 = 112 meters

Answer:

a. one line down one line to the right one live to the northwest from the object

b. t1=190 t2=310

Explanation:

The ways to solve this problem is through knowledge dissemination, citizens who were outraged by the article about the negative statements and malicious stories about Filipinos, could submit positive articles about the culture of the Filipino people.

Another way could be through open dialogue, with Filipino internet users, publicly apologizing seeking to become more involved with the Filipino community.

Therefore, the best way to solve the problem is through ethically-based freedom of expression, avoiding stereotypes of a culture you don't belong to, and looking for a way to disseminate respectful and empathetic information about a people.

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Answer:

Explanation:

We shall find the final velocity of  aircraft with respect to aircraft carrier using the following relation.

v² = u² + 2as

v² = 0² + 2 x 59 x 97

v² = 11446

v = 107 m /s

velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h

= 48.152 x 1000 / (60 x 60) m/s

= 13.37 m /s

This velocity of aircraft carrier will be added to the velocity of aircraft .

So absolute velocity of aircraft = 107 m /s + 13.37 m/s

= 120.37 m/s

Answer:

48 meters north

Explanation:

If you measure the distance between the point Marus started and the point where they stopped after traveling 1,250 meters back south, you get 48 meters. To make this a displacement quantity, you add 'north' to the end and get 48 meters north, considering this was the initial direction of the motorcycle.

Answer:

4161 J/kg·°C

Explanation:

We can use the principle of conservation of energy to solve this problem, which states that the total heat energy in a closed system is constant. The heat lost by the tea is equal to the heat gained by the milk.

Let's first calculate the heat lost by the tea:

Q(tea) = mcΔT

Q(tea) = (0.27 kg)(3910 J/kg·°C)(90.0°C - 85.0°C)

Q(tea) = 6555 J

where m is the mass of tea, c is the specific heat of tea, and ΔT is the change in temperature.

Next, let's calculate the heat gained by the milk:

Q(milk) = mcΔT

Q(milk) = (0.02 kg)(c)(85.0°C - 6.0°C)

Now we can equate the two expressions:

Q(tea) = Q(milk)

6555 J = (0.02 kg)(c)(79.0°C)

Solving for c, we get:

c = 4161 J/kg·°C

Therefore, the specific heat of milk is approximately 4161 J/kg·°C.

Answer:

Any situation that has a path that stops at the same position that it started from has a displacement of zero

Explanation:

hope it helps

The gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

The gravitational force between two objects depends on their masses and the distance between them. This force is given by the formula:

F = (G × m₁ × m₂) / r² where F is the gravitational force, m₁ and m₂ are the masses of the two objects, r is the distance between them, and G is the gravitational constant, which has a value of 6.7 × 10⁻¹¹ N m²/kg².

Using this formula, we can find the gravitational force that the moon has on a person with a mass of 50 kg.

The mass of the moon is 7 × 10²² kg, and the distance between the moon and the person is 380,000,000 m.

Therefore, we have:

m₁ = 50 kg

m₂ = 7 × 10²² kg

r = 380,000,000 m

G = 6.7 × 10⁻¹¹ N m²/kg²

Substituting these values into the formula, we get:

F = (G × m₁ × m₂) / r²

F = (6.7 × 10⁻¹¹ × 50 kg × 7 × 10²² kg) / (380,000,000 m)²

F = 1.15 N

Therefore, the gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

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The error in her thinking is that oxygen has has six electrons and a negative charge is acquired by nitrogen when it gains two electrons.

Oxygen is a member of group 16. The elements in group 16 has six valence electrons. This means that they need an extra two electrons to complete their octet.

If an atom gains two electrons, it will have a charge of -2 and not +2, a positive charge means that the atom lost electrons. Nonmetals like oxygen do not loose electrons rather they gain electrons.

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If the coefficient of kinetic friction between Josh's sled and the snow is 0.08, he slides 6.97 meter from the base of the hill.

To find how far from the base of the hill Josh's sled ends up, we need to first find the speed of the sled at the bottom of the hill using the conservation of energy principle,

mgh = (1/2)mv², plugging in the values given in the problem, we get,

m(9.81 m/s²)(3.5 m) = (1/2)mv²

Simplifying and solving for v, we get,

v = √(2gh)

v = √(2(9.81 m/s²)(3.5 m))

v = 8.29 m/s

Now we can use the kinematic equation,

d = vt - (1/2)at, to find how far the sled slides on the horizontal patch of snow before coming to a stop, where d is the distance traveled, v is the initial velocity (8.29 m/s), a is the acceleration due to friction (-μg), and t is the time it takes to come to a stop (which we can find by setting v = 0 and solving for t),

0 = 8.29 m/s - μg*t

t = 8.29 m/s / μg

Substituting this value of t back into the kinematic equation, we get,

d = (8.29)(8.29/μg) - (1/2)μg(8.29/μg)²

d = 6.97 m

Therefore, Josh's sled ends up 6.97 meters from the base of the hill.

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Explanation:

Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.

So, we have:

P1/T1=P2/T2

Solving for P2, we have:

P2=(P1*T2)/T1

NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!

Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:

P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]

P2=228930.7156 N/m^2

Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2

Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!

Hope this helps!

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Answer:

Distance = 40 meters.

Explanation:

Given the following data;

Force = 25N

Time = 10 seconds

Power = 100 Watt

To find the distance;

First of all, we would determine the work done;

Workdone = power * time

Workdone = 100 * 10

Workdone = 1000 Nm

Next, we use the following formula to find the distance;

Workdone = force * distance

1000 = 25 * distance

Distance = 1000/25

Distance = 40 meters.

Answer:

11948.64 Newtons

Explanation:

Answer:

Differences

microscopic refers to substances visible to the naked eye

macroscopic are substances invisible to naked eye

Similarities

both refer to different scales that are useful to determining the size to different compounds.

Explanation:

Ima find more

The magnitude of the force on the 89 kg person who is decelerating at 30 m / s² is 272.44 N

According to Newton's second law of motion,

F = m a

m = W / g

F = Force

m = Mass

a = Acceleration

W = Weight

g = acceleration due to gravity

a = 30 m / s²

W = 89 kg  

F = 89 * 30 / 9.8

F = 272.44 N

According to Newton's second law of motion, Acceleration of an object is dependent upon the mass of the object and the net force acting on it.

Therefore, the magnitude of the force on the 89 kg person who is decelerating at 30 m / s² is 272.44 N

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Answer:

Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

Explanation:

Answer:

B The car has come to a stop and has zero velocity

Explanation:

I just did the Question

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

Answer:

trueeeeeeee..........mmmm...........

Answer:

Vy = 4.43 m/s

Explanation:

given data

velocity = 9 m/sec

angle horizontal = 29.5 degrees

solution

we get here ball initial vertical velocity will be here

Vy = v sinθ       ...........................1

put here value and we get

Vy = 9 × sin(29.5 )  

Vy = 4.43 m/s

Answer:

The strength of the magnetic field created by current is:

B=μ₀I / 2 π R

where I is the current

R is the shortest distance to the wire

The constant μ₀ is 4π * 10^-7 T * m / s

Explanation:

Explanation:

Potential energy is a scalar quantity because it has only magnitude and no direction. Its ability to be negative does not affect its scalar nature.

Please mark brainliest

Answer: Insulator

Explanation:

An insulator is a material that restricts the flow of electricity or thermal energy.

Answer:

plastic or wood

Explanation:

Answer:

Introduction:

In this report, we will examine whether the 100 square meter parachutes with a drag coefficient of 1.5 are suitable for dropping 700-kilogram parcels from a plane at a refugee camp. The main concern is that the parcels cannot hit the ground at a velocity of more than 5 meters per second without damaging the contents.

Calculation:

To determine whether the 100 square meter parachutes with a drag coefficient of 1.5 are suitable, we need to calculate the terminal velocity of the parcels. The terminal velocity is the maximum velocity that the parcels can reach when they are falling through the air. We can calculate the terminal velocity using the following equation:

Vt = sqrt((2mg)/(ρACd))

Where:

Vt is the terminal velocity

m is the mass of the parcel (700 kg)

g is the acceleration due to gravity (9.8 m/s^2)

ρ is the air density (1.22 kg/m^3)

A is the area of the parachute (100 m^2)

Cd is the drag coefficient (1.5)

Substituting these values into the equation, we get:

Vt = sqrt((2 x 700 x 9.8)/(1.22 x 100 x 1.5)) = 52.0 m/s

This means that without any parachute, the parcel would hit the ground with a velocity of 52 m/s. However, the parachutes are designed to provide air resistance, which will slow down the parcels.

To determine whether the parachutes are suitable, we need to calculate the velocity at which the parcels will hit the ground when they are attached to the parachutes. We can use the following equation to calculate the force of air resistance:

F = (1/2)ρAv^2Cd

Where:

F is the force of air resistance

ρ is the air density (1.22 kg/m^3)

A is the area of the parachute (100 m^2)

v is the velocity of the parcel

Cd is the drag coefficient (1.5)

When the force of air resistance is equal to the weight of the parcel, the parcel will stop accelerating and will reach its terminal velocity. Therefore, we can set the force of air resistance equal to the weight of the parcel:

F = mg

Substituting the values into the equation, we get:

(1/2)ρAv^2Cd = mg

Solving for v, we get:

v = sqrt((2mg)/(ρACd))

Substituting the values into the equation, we get:

v = sqrt((2 x 700 x 9.8)/(1.22 x 100 x 1.5)) = 25.9 m/s

This means that the velocity at which the parcels will hit the ground when they are attached to the parachutes is 25.9 m/s.

Conclusion:

Based on our calculation, the 100 square meter parachutes with a drag coefficient of 1.5 are suitable for dropping 700-kilogram parcels from a plane at a refugee camp. The velocity at which the parcels will hit the ground when they are attached to the parachutes is 25.9 m/s, which is below the maximum velocity of 5 m/s specified by the U.S. Army. However, if the mass of the parcels or the area of the parachutes changes, the velocity at which the parcels hit the ground will also change. Therefore, it is important to recalculate the velocity for different scenarios to ensure the safety of the parcels.

Explanation:

Answer:

5 cm - 53 degrees

Explanation:

\( \sqrt{ {4}^{2} + {3}^{2} } = 5cm\)

\( \tan ^{ - 1} (4 \div 3) \) = 53°

Answer: The answer is B; Electrons move from areas of high to low electric potential

Explanation:

I had this question on a quiz and got it correct.

The required Electrons move from areas of high to low electric potential best describes Jodi’s error. Option B is correct.

Electric current is the flow of electric charge through a material. It is typically expressed in units of amperes (A), which represent the rate at which charge flows through a conductor.

Here,
Jodi's error is in statement 2. Electrons move from areas of high electric potential to areas of low electric potential, not from low to high electric potential. This is because electric potential is a measure of the energy per unit charge, so electrons will naturally flow from areas of higher potential to areas of lower potential to balance out the potential difference.

Thus, the required, Electrons move from areas of high to low electric potential best describes Jodi’s error. Option B is correct.

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Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

The substance that would be most likely to provoke a response similar to the barium dance is c. lead.

Barium dance is used to describe the movements of the body that can occur when a person is exposed to a high level of barium. Barium is a heavy metal that can be toxic in large amounts, and exposure can occur through ingestion, inhalation, or skin contact.

Other heavy metals such as lead, mercury, can also cause similar symptoms when a person is exposed to high levels. Additionally, exposure to certain chemicals such as pesticides, solvents, and industrial pollutants can also cause neurological symptoms similar to those seen in barium.

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When a person is exposed to high quantities of other heavy metals like lead, mercury, and cadmium, comparable symptoms can also be experienced. Furthermore, neurological symptoms resembling those of barium can also be brought on by exposure to specific substances including insecticides, solvents, and industrial pollutants.

It's critical to remember that any potential exposure to these drugs needs to be taken seriously, and if you think you may have been exposed, you should get medical help right once.

based on the information in the passage, which of these substances would be most likely to provoke a response similar to the barium dance

a. oxygen

b. nitrogen

c. lead

d. sulphur

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s