Which statement best describes the air in Earth's atmosphere?
- a pure substance made of one kind of gas
- a solution of gases that are physically mixed
- a compound made of two or more kinds of gases
- a mixture of two gases that are chemically combined

To determine the minimum uncertainty in the position of a 30.1 g ball moving at 27.2 m/s with a speed accuracy of 0.15%, follow these steps:

1. Convert the mass of the ball from grams to kilograms: 30.1 g = 0.0301 kg.

2. Calculate the uncertainty in the ball's speed: 0.15% of 27.2 m/s = 0.0015 × 27.2 m/s ≈ 0.0408 m/s.

3. Apply the Heisenberg uncertainty principle: Δx * Δp ≥ ħ/2, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and ħ is the reduced Planck constant (approximately 1.055 × 10^-34 Js).

4. Calculate the uncertainty in momentum: Δp = m * Δv = 0.0301 kg * 0.0408 m/s ≈ 0.00123 kg m/s.

5. Solve for the minimum uncertainty in position: Δx ≥ ħ/(2 * Δp) ≈ (1.055 × 10^-34 Js) / (2 * 0.00123 kg m/s) ≈ 4.28 × 10^-32 m.

The minimum uncertainty in the position of the 30.1 g ball moving at 27.2 m/s with a speed accuracy of 0.15% is approximately 4.28 × 10^-32 meters.

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Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

Assuming electrical energy costs 0.080 dollars per KW.h, calculate the cost of running each of the following appliances for 24 h if 115 V is supplied to each then A 75 W of power stereo will cost and an electric oven that draws 20 A of current will cost . A television with a resistance of 60 ohms will cost

For a

Power P = Work/time

putting values,

75 W= Work/24h

Work = 1.800 kW-hr.

Cost = 0.080×1.800 W-hr.

Cost = 0.114 Doller.

b)

P = VI = 115*20 = 2300 W = 2.3 kW

Work = 2.3*24 = 55.2 kW-hr

cost = 0.08*55.2 = 4.4 Doller

c )

P = V²/R

P = 115²/60

P = 220.6 W = 0.22 kW

Work = 0.22kW * 24 = 5.28 Doller

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If ball B rolls off the shelf with the same initial speed as ball A and both experience the same gravitational acceleration, then ball B will also land at the same distance away from the shelf.

Ball B rolls off a shelf with the same initial speed as ball A, but the distance it travels after landing varies on a number of variables, including the angle of projection and the presence of any air resistance. Let's assume the following circumstances in order to give a generic response:

1. The height above the ground of the two shelves is the same.

2. The initial speed at which the two balls are released is the same.

3. No air resistance exists.

Ball B will land at the same distance from ball A in these circumstances. This is because the initial velocities of both balls have the same horizontal component, leading to equal horizontal displacements. The path and landing distance without air resistance under the identical initial conditions.

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(a) The maximum frequency of revolution is 4.0 rad/s.

b (i) The tension in the string is maximum when the object is at the top of its path and minimum when it is at the bottom of its path.

b (ii) The speed of the mass can be found using the minimum tension in the string is 2 m/s.

The maximum frequency of revolution is given by the maximum tension in the string, which is 60 N.

T = mv² / r

where;

The linear speed is given by:

v = 2πrf

where;

f = √T / (mr²))

f = √(60 / (0.60 x 1.20²))

f = 4.0 rad/s

(i) The tension in the string is maximum when the object is at the top of its path and minimum when it is at the bottom of its path. This is because the tension provides the centripetal force that keeps the object moving in a circular path.

(ii) The speed of the mass can be found using the minimum tension in the string, which is 2.1 N:

T = mv² / r = 2.1 N

Rearranging and solving for v, we get:

v = √(Tr / m)

v = √(2.1 x 1.20 / 0.60)

v = 2.0 m/s

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The value that represents the mass of block C that would keep crate D in equilibrium is 196.2 kg (Option D).

To determine of value that represents the mass of block C that would keep crate D in equilibrium, ee can begin by determining the tension in the ropes. Since the crate is held in a certain location by two blocks, the tension in the ropes should be equal. We can use this fact to calculate the tension in the ropes by assuming that they have the same magnitude.

In this case, let T be the tension in each rope segment. The forces acting on the system can be represented in the following free-body diagram:

For the x-direction: T sin 30° = T sin 60° + µg = 0

where µ is the coefficient of friction and g is the acceleration due to gravity. Since the pulleys are frictionless, µ = 0. Therefore, we can simplify the equation as follows:

T sin 30° = T sin 60°

Dividing both sides by T:

sin 30° = sin 60°/2

This gives us: T = (2/√3) mg where m is the total mass of the system and g is the acceleration due to gravity.

Substituting the given values, we obtain:

T = (2/√3) × 150 × 9.81 ≈ 1962 N

Next, we can use the law of the lever to calculate the mass of block C required to keep the crate in equilibrium. According to this law, the sum of the moments acting on the system must be zero. We can choose the point where the ring is attached to the rope as the pivot. This gives us the following equation:

(135 kg)(9.81 m/s²)(3 m) + (mC)(9.81 m/s²)(6 m) - (150 kg)(9.81 m/s²)(4.5 m) - (1962 N)(3 m) = 0

Simplifying: (135)(3) + (6mC) - (150)(4.5) - (1962)(3) = 0

Rearranging: 6mC = 150(4.5) + 1962 - 135(3)

6mC = 2127

mC ≈ 196.2 kg

Therefore, the value that represents the mass of block C required to keep crate D in equilibrium is approximately 196.2 kg. Thus, the correct option is D.

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Answer:

Your answers are Continental crust and Mountain range.

Explanation:

Continental crust is the first blank, and Mountain range the second.

I hope I helped, and I always appreciate Brainliest!!! :)

Answer:

3.33 A

Explanation:

Equalent Resistance=30 ohms

I = V / R

I = 100 / 30

I = 3.33 A

The current expected to measure when the heat strips are energized is 14.1A.

Current can be defined as the rate of flow of electric charge round a circuit.

To calculate the current that will be measured when the heat strips are energized, we use the formula below.

Formula:

Where:

Make I the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, the current expected to measure when the heat strips are energized is 14.1A.

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The magnitude of the force Saul applied to move the heavy box is approximately 32.56 N. We can use the work-energy principle.

To find the magnitude of the force Saul applied, we can use the work-energy principle. The work done on an object is equal to the force applied multiplied by the displacement of the object in the direction of the force. Mathematically, it can be expressed as:

Work = Force x Displacement x cos(theta),

where theta is the angle between the force and the displacement vectors.

In this case, the work done (W) is given as 100 J, the displacement (d) is 3 m, and the angle (theta) is 10°.

100 J = Force x 3 m x cos(10°).

To solve for the force (F), we rearrange the equation:

Force = 100 J / (3 m x cos(10°)).

Using a calculator, we can find the magnitude of the force Saul applied by substituting the values:

Force ≈ 32.56 N.

Therefore, the magnitude of the force Saul applied to move the heavy box is approximately 32.56 N.

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Answer:

The basic construction, unchanged since Torricelli's experiment in 1643, is a glass tube about three feet long, closed at one end, filled with mercury, and inverted with the open end immersed in a cistern of mercury.

Answer:

18.9 seconds

Explanation:

See the attached worksheet.  Since the desired information is outside the range of the graph, develop the equation of the line that is drawn.  From that equation [y = (9/10)x+3] the time required to travel 20 meters is foind to be 18.9 seconds.

By using the concept of work and energy, the below are the answers

1. 100J

2. 100J

3. 0

4. Total energy = K.E + P.E

5. 1.02 kg

6. 14 m/s

Given that 100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.

1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?

Work is the product of force and the distance.

The potential energy gained by the rock will be equal to work done in lifting  up the rock which is 100 J

2. At point C, the rock's potential energy will be equal to the work done.

That is, at point C, P.E = 100J

3. The rock's kinetic energy at point A is Zero.

That is, at point A, K.E = 0

4. At point B, some of the rock's potential energy will be changed to Kinetic energy. Yes. Because the Total energy = K.E + P.E

5. What is the mass of the rock?

Weight W = mg

where g = 9.8m/s^2

W = 10N

Substitute both into the formula

10 = 9.8m

make m the subject of the formula

m = 10/9.8

m = 1.02 kg

6. What is the rock's velocity just before it hits the ground

Work done = Energy

That is, Work done = maximum P.E = maximum K.E

100 = 1/2m\(v^{2}\)

Substitute mass into the formula

200 = 1.02\(V^{2}\)

V = \(\sqrt{196}\)

V = 14 m/s

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The definition of moment of inertia allows to find the results for the with respect to each axis are:

1-Iₓ = m r²

2-Iy=9m r²

3-Iz=10m r².

What is moment of inertia ?

The moment of inertia is a scalar quantity is obtained by the expression.

Sol-

I =∫ r² dm

Where I is the moment of inertia , m the mass of the body and r the distance from the axis of rotation.

In the case of point particles, the expression reduces to

    I = m r²

In this case we calculate with respect to each coordinate axis.

a) With respect to the x axis the distance is x = r, the moment of inertia is

      Iₓ = m r²

with respect to the y-axis, indicate that the distance is y = 3 r.

We calculate

        I_y = m (3r) ²

        I_y = 9 m r²

c) Regarding the z-axis.

We look for the distance.

         z =√x^2+y^2

         z = √ 1^2+3^2

We calculate the moment of interia  Iz = m (10r²)

        Iz = 10 m r²

In conclusion using the definition of moment of inertia we can look for the results of the moment of inertia for each axis are:

  a) Iₓ = m r²

  b) I_y = 9 m r²

  c) I_z = 10 m r²

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The intensity of the cell phone radiation is 1.8 times greater than the maximum allowable microwave oven leakage. The radiation intensity of a cell phone can be calculated using the formula:

I = P/4πr²

where I is the radiation intensity, P is the power output, and r is the distance from the source. Substituting the given values, we get:

I = (2.0 W)/(4π(2.00 in/39.37 in/m)²) = 9.0 mW/cm²

(b) Comparing this intensity with the maximum allowable microwave oven leakage of 5.0 mW/cm² at a distance of 2.00 in, we get:

9.0 mW/cm² / 5.0 mW/cm² = 1.8

The intensity of the cell phone radiation is 1.8 times greater than the maximum allowable microwave oven leakage.

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The magnitude of the external magnetic field is approximately 0.194 Tesla.

To solve this problem, we can use the equation for the magnetic force on a current-carrying wire:

F = BILsinθ

where F is the magnetic force, B is the magnitude of the magnetic field, I is the current, L is the length of the wire, and θ is the angle between the wire and the magnetic field.

Given:

Length of the wire, L = 0.19 m

Current, I = 7.6 A

Angle, θ = 17°

Force, F = 6.2 x \(10^{-3}\)N

Substituting the given values into the equation, we have:

6.2 x \(10^{-3}\) N = B * 0.19 m * 7.6 A * sin(17°)

Simplifying the equation, we can solve for B:

B = (6.2 x \(10^{-3}\) N) / (0.19 m * 7.6 A * sin(17°))

Calculating this expression, we find:

B ≈ 0.194 T

Therefore, the magnitude of the external magnetic field is approximately 0.194 Tesla.

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Answer:

Δt=48 s

Explanation:

v: Barry's speed

v.: speed of the walkway

Δx=30 m

Δt1=30 s ,  Δt2=120 s

|Δx1|=|Δx2|

Δx=v*Δt

=> (v+v.)*30=(v-v.)*120

v=v.*5/3

30=(v+v.)*30

=> 30=(5v./3 +v.)*30

v.=3/8 m/s

v=5v./3 , v.=3/8

=> v=5/8 m/s

Δx=v*Δt

30=5/8 *Δt

Δt=48 s

Answer:

v = λf = (4 m)(3 Hz) = 12 m/s

Explanation:

Answer

Clinometers measure both inclines and declines using three different units of measure: degrees, percentage points, and topo . Astrolabe is example of an inclinometer that was used for celestial navigation and location of astronomical objects from ancient times to the Renaissance.

Answer:

A: It decreased trade because soldiers no longer protected the oases.

I took it on E2020. I hope I've helped. If you need further explanation, please comment below.  

Answer:

Yes it is A

It decreased trade because soldiers no longer protected the oases.

Explanation:

When trying to establish economical hauling, hauling distance is usually the most important factor to consider. Hauling distance refers to the  point of origin and the destination of the goods being transported.

The farther the distance, the higher the cost of transportation, as it requires more time, fuel, and resources to haul the goods. While other factors such as hauling time, means of hauling, and what is being hauled can also impact the cost and efficiency of transportation, the distance typically has the most significant impact on the overall cost. Therefore, it is crucial to consider the hauling distance when trying to establish an economical hauling plan. To minimize the cost of transportation, various strategies can be employed, such as optimizing the routing to reduce the distance traveled, utilizing more efficient means of transportation, such as rail or water transport, and using technology to improve logistics and reduce waste. Ultimately, the goal of establishing an economical hauling plan is to minimize transportation costs while ensuring that goods are delivered safely, reliably, and on time. By prioritizing hauling distance and utilizing efficient transportation strategies, businesses can achieve a more cost-effective and sustainable transportation plan.

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To determine the magnification of the object's image, we can use the formula: magnification = image height / object height

First, we need to find the image height. We can use the thin lens equation:

1/f = 1/d_o + 1/d_i

where f is the focal length, d_o is the object distance, and d_i is the image distance.

We know that f = 23.9 cm and d_o = -1.6 cm (since the object is in front of the lens and we take the distance as negative). We can solve for d_i:

1/23.9 = 1/-1.6 + 1/d_i
d_i = 2.2 cm

Now we can find the image height using similar triangles:

image height / object height = d_i / d_o
image height / 1 = 2.2 / -1.6
image height = -1.375 cm

(Note that the negative sign indicates that the image is inverted.)

Finally, we can calculate the magnification:

magnification = image height / object height = -1.375 / 1 = -1.375

So the magnification of the object's image is -1.375.

To determine the magnification of the object's image when a small statuette sits at a distance of 1.6 cm from a converging lens with a focal length of 23.9 cm, you can follow these steps:

Step 1: Use the lens formula:
1/f = 1/do + 1/di
where f is the focal length (23.9 cm), do is the object distance (1.6 cm), and di is the image distance.

Step 2: Solve for the image distance (di):
1/23.9 = 1/1.6 + 1/di

Step 3: Rearrange the equation and solve for di:
1/di = 1/23.9 - 1/1.6
di = 1 / (1/23.9 - 1/1.6) ≈ 1.64 cm

Step 4: Calculate the magnification using the formula:
M = -di/do
M = -1.64/1.6 ≈ -1.025

The magnification of the object's image is approximately -1.025, which means the image is slightly larger than the object and inverted.

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Answer:

Periods indicate that a new period at a new principal energy level which is filled with electrons. Groups indicate that the elements share similar properties because their shells are filled with the same number of valence electons.

Explanation:

e see that starting at period 2 Li has a  +1 charge , He has +2 charge, B has +3 charge, C has the potential to have a +4 charge or -4 charge (because it can gain or loose electrons) and N has a -3 charge, etc. Groups however, have similar properties and every element in a 1A - 8A has the same number of valence electrons. All except the noble gases in which He only needs 2 electrons to be in a perfect state. This is its octet as it would be a completely different element if it had more than 2 electrons in its shell. An example of the groups sharing similar properties is if we were to look at group 1A. In group 1A we find H, Na, K, etc. These are all highly unstable elements when alone (not in a compound) and they are very reactive. If you were to put Na for instance in water, it would light up and you'd see a flame sitting on top of the water. A scientist could do that with the rest of the elements in the group and they would see this increasing in reactivity as you go down the periods. Francium isn't visible long enough to be able to see this happen though. It's more of that kind of element that scientists prove that it exists, but not a very common element to get to learn much about.

The velocity of the rollercoaster going down the hill is equal to 5.270 m/s.

Given the following data:

To calculate the velocity of the rollercoaster going down the hill:

Mathematically, kinetic energy is calculated by using the formula;

\(K.E = \frac{1}{2} MV^2\)

Where:

Making V the subject of formula, we have:

\(V =\sqrt{\frac{2K.E}{M} }\)

Substituting the given parameters into the formula, we have;

\(V =\sqrt{\frac{2\times 25000}{1800} }\\\\V=\sqrt{27.7778}\)

Velocity, V = 5.270 m/s

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Answer:

   I = 2 A

Explanation:

Electric current is defined as the amount of face per unit of time

         I = Q / t

where Q is the charge and t is the time

let's reduce the time to the SI system

         t = 1 min (60s / 1 min) = 60 s

let's calculate

         I = 120/60

         I = 2 A

A hammock on earth is tied with ropes between two trees. a person lies in it. If ropes are stretched tightly rope likely to break.

Tension:

Tension is described as the pulling force it get transmitted axially by the means of a string, a rope, chain, or similar object, or by each end of a rod, truss member, or similar three-dimensional object.

The formula to find the tension of a string:

T = ma + mg

T = tension

m= mass

a = acceleration

g = acceleration due to gravity

As it is given that the hammock is tied with ropes between two trees. So when the ropes it stretched tightly they rope more likely to break.

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In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].