A container with 3 kg of pure water at a temperature of 12 degrees Celsius is placed in a refrigerator where the air temperature is kept at 4 degrees Celsius. how much heat (in joules) is gained by the air when the water cools to 4 degrees Celsius?

The value of parameter D is -3.

Angular frequency is  the angular displacement of any element of the wave per unit time.

f(x) = Acos(x) + D

where;

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

From the blue colored graph, y = -2 at x = 0

-2 = cos(0) + D

-2 = 1  +  D

-2 - 1 = D

-3 = D

Thus, the value of parameter D is -3.

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Explanation:

getting me up to date with recent mission development

The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

From the graph, the amplitude of the red colored wave is 1 unit.

From the graph, the amplitude of the red colored wave is 2.1 unit.

Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.

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Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the

Explanation:

a) The original momentum of the fullback is 331.2 kgm/s.

b) The impulse exerted on the fullback is - 331.2 kgm/s

c) The impulse exerted on the tackler is 331.2 kgm/s

d) The average force exerted on the tackler is 389.64 N

Mass of the fullback, m = 92 kg

Initial velocity of the fullback , u = 3.6 m/s

Time of the motion of the fullback , t = 0.85s

The original momentum of the fullback ;

[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]

The impulse exerted on the full-back;

J = ΔP = [tex]m v_f - m v_i[/tex]

J = [tex]m ( v_f - v_i)[/tex]

J = 92 ( 0- 3.6)

J = - 331.2 kgm/s

The impulse exerted on the tackler;

[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]

The average force exerted on the tackler;

F= Δmv/t

[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]

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The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N

Sphere 1 losses 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C

Sphere 2 gains 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C

Using the Coulomb's law equation, the force can be obtained as illustrated below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²

F = 4.59×10⁻⁴ N

Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N

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The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

The independent variable which is can be changed by increasing the rate of salt added to the equation.

The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.

Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.

Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.

If  the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.

Therefore, The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

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Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   = 1/9  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

It will take you 0.23 seconds to hear your echo

Echo can be simply defined as the reflection of sound wave.

Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.

Speed V is the distance per time

V = D / T

Substitute all the parameters into the formula

340 = 77.2 / T

Make T the subject of formula

T = 77.2 / 340

T = 0.2270

T = 0.23 s to two significant digits

Therefore, it will take you 0.23 seconds to hear your echo.

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Answer:

lowering the ambient temperature causes chemical reaction to proceed more slowly,so a battery used in a low temperature produces less current than that at high temperature . As cold batteries run down they quickly reach the point where they cannot deliver enough current to keep up the demand.. hence discharge more quickly in cold weather..

(1) If the ship moves upward with a constant velocity, the period increases, false.

(2) If the length of the pendulum is doubled, the new period will be: square root of 2 times T₀. true.

(3) If the ship accelerates upward, the period increases, true.

(4) If the ship accelerates downward at 9.81 m/s² , the pendulum will no longer oscillate, false.

(5) If the mass of the pendulum is halved, the period decreases, false.

The period of a simple pendulum is given as;

T = 2π√(L/g)    ------- (1)

where;

When the ship moves upward with a constant velocity, the period will not change.

T ∝√L

when, L = 2L₀

T = T₀√2

Thus, if the length of the pendulum is doubled, the new period will be: square root of 2 times T₀.

t = √2h/g

T = √I/mgd

where;

when the mass is halved, the period increases.

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Answer:

Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].

Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.

The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].

Apply the Pythagorean Theorem to find the initial speed of this projectile:

[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].

Answer:

44.7 seconds

Explanation:

D = 500 m

   500 = .25  t^2

    500/.25 = t^2

          2000 = t^2

                t = 44.7 seconds

Answer:

Vo (vertical) = Vo sin 30 = 30 m/s / 2 = 15 m/s

Vo (horizontal) = Vo cos 30 = .866 30 m/s = 26 m/s

From the calculation, the force constant is 192 N. Also, friction would decrease the extension.

We know that the force constant can be obtained by the use of the relation;

F = Ke

F = applied force

K = force constant

e = extension

We know from Hooks law that the force applied is directly proportional to the extension.

We can write;

F = mgcosθ

F = 43 Kg * 9.8 m/s^2 * sin31°

F = 217 N

K = 217 N/1.13 m

K = 192 N/m

If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.

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B. As the resistance increases, the slope on the current vs voltage graph gets flatter.

Resistance is the opposition to the follow of current in electrical material.

The mathematical relationship between resistance, current and voltage is given by Ohm's law.

R = V/I

where;

From the equation above, increase in resistance means an increase in voltage or decrease in current.

In Current vs voltage graph, the following will occur;

Thus, as the resistance increases, the slope on the current vs voltage graph gets flatter.

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The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

                        [tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex]    We substitute d for to represent density.

                         [tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

                      [tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]    

                       [tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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Answer:

See below

Explanation:

29 mile/hr * 5280 f/mile / 3600  s/hr = 42.53  ft/ sec

   42.53 ft / sec    /   12 feet =  3.54  cycles / sec = 3.54 Hz

      for frequency I suppose....where is the referred to figure?

The distance of separation between the lenses is determined as 12 cm.

The distance of the image formed by the first lens is calculated as follows;

1/d₁ = 1/f - 1/di₀

1/di₁ = 1/4 - 1/12

1/di₁ = 2/12

1/di₁ = 1/6

di₁ = 6 cm

Use magnification formula as shown below;

M = (di₁di₂)/(d₀₁d₀₂)

1 = (di₁di₂)/(d₀₁d₀₂)

di₂ = (d₀₁d₀₂)/(di₁)

di₂ = (12 x d₀₂)/(6)

di₂ = 2d₀₂

1/f = 1/ d₀₂ + 1/di₂

1/f =  1/ d₀₂ + 1/2d₀₂

1/f = 1.5/d₀₂

d₀₂ = 1.5f

d₀₂ = 1.5(4 cm) = 6 cm

D = di₁ + d₀₂

D = 6 cm + 6 cm

D = 12 cm

Thus, the distance of separation between the lenses is determined as 12 cm.

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The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

The resistance of a wire is the opposition to the flow of current. It depends on the following;

Given that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A

The given parameters are;

The formula to use to get R will be

R = ρL / A

Substitute all the necessary parameters into the formula

R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]

R = 8.6 x [tex]10^{-3}[/tex]  Ω

From Ohm's law, V = IR

Substitute all the necessary parameters into the formula

V = 0.2 x 8.6 × [tex]10^{-3}[/tex]

V = 1.72 x [tex]10^{-3}[/tex] V

Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

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Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no  external torque acting on the object.

Angular momentum of a system is conserved as long as there is no net external torque acting on the system.

Thus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

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The maximum rate at which energy can be added to the circuit element mathematically given as

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

Generally, the equation for P is  mathematically given as

[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]

Therefore

[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]

[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]

[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]

Max temp Change

[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]

[tex]\text { time }=3 \times 60[/tex]

t=180s

In conclusion, Max Energy Rate

[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

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Answer: The answer is A.

Explanation:

Just because u have gay parents doesnt mean theyre bad lol XD

Answer:

M V R = constant      angular momentum is constant because  no forces act in the direction of V

Since M (mass) = constant

V R = constant

The force is directed along the gravitational force vector (towards the center of rotation)

Answer:

No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be

The maximum force constant of the spring Kmax is 2337.9 N/m.

The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.

Force constant is denoted by K and its unit is N/m.

Where;

K = spring constant

x = displacement

The work done by the spring is given below as follows:

Work done = Fx/2

Kinetic Energy = mv²/2

Force on an inclined plane = mgsinθ

Total force, F = mgsinθ + frictional force

F = 1390 * sin 22° + 515

F = 1035.7 N

Work done = change in KE

Fx/2 = mv²/2

Fx = mv²

m  = 1390/9.81 = 141.692

Solving for x;

x = mv²/F

x = 141.692 * 1.8²/1035.7

x = 0.443 m

The maximum force constant of the spring Kmax = 1035.7/0.443

Kmax = 2337.9 N/m

In conclusion, the maximum force constant of the spring  is the ratio of the total force and displacement.

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Note that the complete question is given below:

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria

a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N

b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N

c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N

Tension force refers to a pulling force that is exerted by a string or cable about an axis.

a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:

Taking moment about the vertical axis

T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°

Solving for T₁ by taking the left end as the pivot;

T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0

T₁ * 1.285 = 681.5563

T₁ = 530.4 N

Therefore;

T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°

T₂ = 655.62 N

b. From calculation, the magnitude of the tension in the rope in the right end is T₁.

T₁ = 530.4 N

c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:

Taking moments about the left end in the horizontal direction;

T₃ = T₁ * cos 40°

T₃ =  530.4 N * cos 40°

T₃ = 406.3 N

In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.

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The correct answer is 5828.675 J.

Given combined mass 4kg and mass of bullet 150gm=0.150kg.

Total mass= 4+0.150=4.150kg

Velocity=53 m/s

Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to  [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.

a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision ​

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(a) The velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.

(b) The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.

Apply the principle of conservation of linear momentum to determine the velocity of the balls;

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

0.2(150) + (0.3)(-0.4) = 0.2v₁ + 0.3v₂

29.88 = 0.2v₁ + 0.3v₂

Apply one directional linear velocity

u₁ + v₁ = u₂ + v₂

v₁ = u₂ + v₂ - u₁

v₁ = -0.4 + v₂ - 150

v₁ = v₂ - 150.4

Substitute the value of v₁ into the first equation;

29.88 = 0.2(v₂ - 150.4) + 0.3v₂

29.88 = 0.2v₂ - 30.08 + 0.3v₂

59.96 = 0.5v₂

v₂ = 59.96/0.5

v₂ = 119.92 m/s

v₁ = 119.92 - 150.4

v₁ = -30.48 m/s

V(cm) = (0.2 x 150  +  0.3 x 0.4) / (0.2 + 0.3)

V(cm) = 60.24 m/s

V(cm) = (0.2 x 30.48 +  0.3 x 119.92) / (0.2 + 0.3)

V(cm) = 84.14 m/s

Thus, the velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.

The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.

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