A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?

Answer:

op

Explanation:

At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away

Answer:

(i) 0.6s (ii) 8.42m

Explanation:

U² = 6.02² + 7²

U = 9.23

angle of projection

tanø = 6.02/7

ø = 40.7

Time of fligt

t = Usinø/g

t = 9.23 sin 40.7/10

t = 0.6

H range = U²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

The time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.

Velocity ​​is defined as the directional motion of an object which is indicated by the rate of change of position as observed from a particular frame of reference. It is measured by a particular standard of time. It is a vector quantity as it has both magnitude and direction.

It can be expressed as:

v= d/t

Where. v is the velocity in m/s

d is the displacement measured in meter 'm'

t is the time measured in seconds 's'

For above given information,

v = 6.02 +7.0j, so the initial velocity will be u

u² = 6.02² + 7²

u = 9.23m/s

Angle of projection, tanø = 6.02/7

ø = 40.7

Time taken=  t = u sinø/g

t = 9.23 sin 40.7/10

t = 0.6s

Horizontal range  = u²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

Thus, the time taken reach the maximum height is 0.6s and the horizontal range is 8.42m.

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Answer:

C. crests

don't trust me though

Answer:

The less dense areas created as a sound wave propagates are called rarefactions.

Explanation:

Hoped this helped.

Answer:

No - It is connected in parallel instead of series

Explanation:

The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.

An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.

The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).

The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.

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The equilibrium conditions allow to find the results for the balance forces are:

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0[tex]\frac{W_1 \ 1 + W_2 \ 1.5}{2}[/tex]

Let's calculate F₂

         F₂ = [tex]\frac{W_1 \ 1 + W_2 \ 1.5 }{2}[/tex]  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = [tex]\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}[/tex]  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

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Answer:

Explanation:

What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.

horizontal velocity component

vx = 55.0/2.50 = 22 m/s

vertical initial velocity component

(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²

vy₀ = 13.0625 m/s

θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°

The velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.

The given parameters;

The resultant vertical velocity of the boat is calculated as;

[tex]v_y = 8 \ - \ 5 sin(30)\\\\v_y = 8 - 2.5\\\\v_y = 5.5 \ km/h[/tex]

The resultant horizontal velocity of the boat is calculated as;

[tex]v_x = -2 \ + 5cos(30)\\\\v_x = 2.33 \ km/h[/tex]

The resultant boat velocity is calculated as follows;

[tex]v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.33^2 + 5.5^2} \\\\v = 5.97 \ km/h[/tex]

The direction of the velocity;

[tex]\theta = tan^{-1} (\frac{v_y}{v_x} )\\\\\theta = tan^{-1}(\frac{5.5}{2.33} )\\\\\theta = 67 \ ^0[/tex]

Thus, the velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.

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No,the poping of microvave is not same there time is different

Answer:

F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]

Explanation:

In pic

_________________

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Answer:

0.25

Explanation:

That is the right answer.

The fraction of the initial kinetic energy is lost in the collision is 35.3%.

The given parameters:

The initial kinetic energy of the system collision is calculated as follows;

[tex]K.E_i = \frac{1}{2} mv_1_i^2 + \frac{1}{2} mv_2_i^2\\\\K.E_i = \frac{1}{2} (m)(8.3)^2 + \frac{1}{2} (m) (0)^2\\\\K.E_i = 34.445 m[/tex]

The final velocity of ball A after collision is calculated as follows;

[tex]u_1 + v_1 = u_2 + v_2\\\\8.3 + v_1 = 0 + 6.4\\\\v_1 = 6.4 - 8.3\\\\v_1 = -1.9 \ m/s[/tex]

The final kinetic energy of the system after collision is calculated as follows;

[tex]K.E_f = \frac{1}{2} m(-1.9)^2 + \frac{1}{2} m(6.4)^2\\\\K.E_f = 22.285 \ m \[/tex]

The fraction of the initial kinetic energy is lost in the collision is calculated as follows;

[tex]= \frac{K_i - K_f}{K_i} \\\\= \frac{34.445 - 22.285}{34.445} \\\\= 0.353\\\\= 35.3\%[/tex]

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Answer:

D. Scientist back up their statements with facts.

Explanation:

The majority of the laws, theorems and thesis are base on with experiments. The follow different strategies for confirm the hypothesis with facts, one of the most famous models is the scientific method. That is a way of proof something in a trustworthy.

In the image is the resume about this method (graphical organizer).

Answer:

Groundwater Returning to earths Surface come Up Though A drought

Explanation:

THIS DOES NOT NEED TO BE EXPLAINED IT'S EZZZ

The correct statement is, the system performed 800 J of work on its surroundings.

The given parameters:

Apply first law of thermodynamics;

ΔU = Q + W

where;

In adiabatic process no heat is gained or lost by the system.

Q = 0

ΔU = W  (work is positive when it is done on the system by its surrounding)

If work is done by the system to the surrounding, the new equation becomes;

ΔU = - W

W = - ΔU

W  = -800 J

This implies that the system performed 800 J of work on its surroundings.

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

In a thermodynamic process, the internal energy of a system in a container with adiabatic walls decreases by 800 J. Which statement is correct?

a. The system lost 800 J by heat transfer to its surroundings.

b. The system gained 800 J by heat transfer from its surroundings.

c. The system performed 800 J of work on its surroundings.

d. The surroundings performed 800 J of work on the system.

e. The 800 J of work done by the system was equal to the 800 J of heat transferred to the system from its surroundings.

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Answer:

Our atmosphere has five different layers. They are:

1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.

2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.

3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.

4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.

5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.

Answer:

Attitude, motivation and determination

Explanation:

Let me know if I'm right.

The factors leading to the success through the challenging physical activities are Attitude, motivation and determination.

Although the information regarding the live session is not mentioned, but as a general point of view, the major factors that need to considered while undergoing through the challenging physical activities are as follows:

Thus, we can conclude that the factors leading to the success through the challenging physical activities are Attitude, motivation and determination.

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Answer:

Explanation:

Consult the attached free body diagram.

If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces

• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0

• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0

since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).

We only really need the first equation. Simplifying it, we get

F [archer] - T cos(θ) - T cos(θ) = 0

F [archer] - 2T cos(θ) = 0

F [archer] = 2T cos(θ)

cos(θ) = F [archer] / (2T)

We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say

T = 0.842 F [archer]

and from this we have

cos(θ) = F [archer] / (2 • 0.842 F [archer])

cos(θ) = 1/1.684

cos(θ) ≈ 0.593

Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.

The heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].

The given parameters:

The heat released by the aluminum during the process is calculated as follows;

[tex]Q = mc \Delta t \\\\Q = 1.5\times 900 \times (2700 - 23)\\\\Q = 3.6 \times 10^6 \ J[/tex]

Thus, the heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].

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Answer:

Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.

Answer:

Capacitor is a device which used to store energy .

Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.

Explanation:

Answer:

c

Explanation:

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

Answer:

option B is the correct answer

Explanation:

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An example of a point source water pollutant is a sewer pipe draining directly into a river.

According to the United States Environmental Protection Agency, a  point source pollution is defined as;  “any single identifiable source of pollution from which pollutants are discharged, such as a pipe, ditch, ship or factory smokestack.”

Hence, an example of a point source water pollutant is a sewer pipe draining directly into a river. The pollution is coming from an easily identifiable source which is a sewer pipe connected to a river directly.

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