Complete the missing information from the circuit

We push a car to the distance of 33.939m if we do 28000J work , exerting a force of 825N.

F = applied force

S = distance moved

=28000J/825N

=33.939meter

Thus, we can conclude that the distance moved by the car is 33.939m.

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⇒ Acids taste sour, react with metals, react with carbonates, and turn blue litmus paper red. Bases taste bitter, feel slippery, do not react with carbonates and turn red litmus paper blue.

The study of acids and bases is crucial to chemistry. The Lewis acid/base motif, which broadens the concept of an acid and base beyond H+ and OH- ions, is one of the most relevant theories.

Acids are ionic compounds, which means they have a positive or negative charge. In water, these ionic compounds separate to create hydrogen ions, or H+.

The quantity of H+ ions in the solution determines how strong an acid is. Acid is stronger the more H+ there is. Bases are ionic substances that separate in water to produce the negatively charged hydroxide ion (OH-). The quantity of Hydroxide ions in a base determines its strength (OH-). The strength of the base increases with OH- ion concentration.

Hope this helps,

- Eddie.

The information is missing here but data that may support the existence of life in an exoplanet may include the discovery of atmospheric biological gases.

An exoplanet is any planet that surrounds another star beyond the sun, and therefore they are discovered by analyzing other galaxies in the Universe.

The existence of life in these exoplanets can be indirectly supported by the presence of gases that are only produced by organisms, which are detected by suitable telescopes that analyze emission spectra.

In conclusion, the information is missing here but data that may support the existence of life in an exoplanet include biological gases.

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Answer: Downstroke: When the handle of a water pump is lifted up, the piston is pushed down. It is downstroke of the piston. In this condition, the volume of part A decreases and water pressure increases in this part. The increased pressure shuts the valve ‘A’ and opens the valve ‘B’ It makes the water of part A into pass part B of the cylinder.

Upstroke: When the handle is pushed down, it lifts the piston up. It is upstroke. During upstroke, the volume of part A increases and water Downstroke Upstroke pressure at that part decreases. It makes valve A opened and valve B closed. Thus, the water from the source enters into part A of the cylinder. The water of part B is lifted up and it comes out from the spout.

Explanation:

Answer:

Approximately [tex]28^{\circ}[/tex].

Explanation:

The refractive index of the air [tex]n_{\text{air}}[/tex] is approximately [tex]1.00[/tex].

Let [tex]n_\text{glass}[/tex] denote the refractive index of the glass block, and let [tex]\theta _{\text{glass}}[/tex] denote the angle of refraction in the glass. Let [tex]\theta_\text{air}[/tex] denote the angle at which the light enters the glass block from the air.

By Snell's Law:

[tex]n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})[/tex].

Rearrange the Snell's Law equation to obtain:

[tex]\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}[/tex].

Hence:

[tex]\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}[/tex].

In other words, the angle of refraction in the glass would be approximately [tex]28^{\circ}[/tex].

Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]

Solve for V of the given expression

Firstly, recall the velocity-time equation

[tex]v = u + at[/tex]

plug in known values to the equation

[tex]v = (4) + (0.21)(3.8)[/tex]

solve for final velocity

[tex]v = 4.792ms[/tex]

Hope it helps!

Assuming point K is held in equilibrium, by Newton's second law we have

• net horizontal force

[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]

• net vertical force

[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]

where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies

[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]

Solve the first equation for [tex]F_C[/tex].

[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)

Substitute this into the second equation and solve for [tex]F_B[/tex].

[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)

[tex]F_B = \dfrac57 F_A[/tex]

[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]

A list of planet-star radius ratios is an example of evidence. I agree with this statement.

I agree with A list of planet-star radius ratios is an example of evidence because the Sample records that are taken for planet-star radius ratio are known to often gives a kind of an improved approximate of the stellar characteristics in regards to the smallest stars in the Kepler target list.

The planetary radius is known to be the distance that exist between a what we see as planet's center and its surface.

Therefore, one can say that planetary radius is known to be a measure of a given planet's size.

So, A list of planet-star radius ratios is a sample evidence and I do affirm the above  statement.

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The required orbital speed of the ice cube is 355,358m/s

The force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. This can be expressed mathematically as;

Fr = GMm/r²

The distance is calculated as;

s = Gm/r²

Solving both equation, we will have:

v²/r = Gm/r²

v² = Gm/r

Take the square root of both sides

v = √Gm/r

Solve the required orbital speed

V =  √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s

Hence the required orbital speed of the ice cube is 355,358m/s

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The tension on box A is 117.6 N

Equilibrium is a state of an object with respect to a given observable quantity during the time for which there is no change in that quantity.

A body is said to be in equilibrium when:

Given that box A has a mass of 20.0kg and Box B has a mass of 30.0kg. Box A is placed on a horizontal surface that is frictionless and Box B is hanging from a pulley by a rope connected to Box A.

The tension in the rope will be the same.

Given that Box A is placed on a horizontal surface that is frictionless, that is,

T = ma .... (1)

Also,  Box B is hanging from a pulley by a rope connected to Box A. That is,

W - T = ma ...... (2)

Where W = mg

Since the acceleration of the system is 5.88m/s² which will be the same for the two equations, then the Tension T on Box A can be calculated by substituting acceleration and mass into equation (1)

T = 20 x 5.88

T = 117.6 N

Therefore, the tension on box A is 117.6 N

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Answer:

Electrical force

Explanation:

The electrons aren't actually touching each other. Normal is something on something pushed by gravity by they are touching, frictional is rubbing, applied is pushing

wind (b), also some others are:

- steam

- gas

- liquid fules

Based on the given data about, volume, mass, and density of the spheres;

The density, mass, volume, and shape of an object determines whether it will float or sink in water.

For an object to float, it will displace its own weight of the fluid in which it floats. This is according to Archimedes' principle which states that the upward force acting on a body immersed fully or partially in a fluid, is equal to the weight of the fluid displaced.

Considering the given scenarios:

1. The new sphere has a density of ρ > ρ0 and a mass of m = m0.

This means that the volume of new sphere is smaller than the volume of the original sphere, therefore, the water level will fall,

2. The new sphere has a mass of m > m0 and a radius of r = r0.

This means that the new sphere will have a greater volume and mass. However, the density may be greater or less.. Therefore, the water level may rise or fall.

3. The new sphere has a radius of r = r0 and a density of ρ > ρ0.

The new sphere has equal volume, greater density and mass than the original sphere. Therefore, the water level will rise.

In conclusion, the level of the water in the container depends on the volume, mass and density of the object placed in it.

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Answer:

Density of a substance is the ratio of mass of the substance to its volume.

Answer:

Density of a substance is the ratio of mass of the substance to its volume.

Answer:

When an object is immersed in water. it is pulled downwards due to gravitational pull of earth. Water exerts upward force on the object. This makes object rise up. This upward force is called buoyancy or upthrust.

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

a)There is this impulse-momentum change equation.

[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

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Solids have very small intermolecular space and when heated, do not expand as per the container volume. Basically, their expansion is not good.

Liquids, on the other hand, have a definite volume and their expansion is ideal. Also, they fill the container only to the desired level, unlike gases.

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:

Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:

i₃ = i₁ + i₂    

- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0      

ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0      

- i₁ - i₂ + i₃ = 0                                                      (1)

(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂                (2)

(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄        (3)

If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ =  0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:

- i₁ - i₂ + i₃ = 0

25.1 · i₁ + 40.5 · i₂ = 71

25.1 · i₂ + 78.25 · i₃ = 26.5

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

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a 10 kg block reaches a point with a velocity of 15 m per second and slides down a rough track my the coefficient of the kinetic energy between the two surface ab and the block iis0.52

[tex]\huge\underline{\underline{\boxed{{ \mathbb{SOLUTION:}}}}}[/tex]

[tex]\leadsto[/tex] Total internal reflection occurs when a ray of light is traveling through a medium and, when that medium changes, it does not refract into the second medium but instead it reflects back into the first medium.

By Snell's law, we know that

[tex]\longrightarrow \sf{n_1 sin \emptyset_1 = n_2 sin \emptyset_2}[/tex]

If this happens, then [tex]\sf{n_2 \angle n_1}[/tex], and the ray does not lose intensity due to refraction.

[tex]\huge\underline{\underline{\boxed{\mathbb{ANSWER:}}}}[/tex]

[tex]\large \bm{ A.}[/tex] A ray of light has the same intensity both entering and exiting a fiber optic cable.

The rate at which the person is now losing thermal energy is 99 J/s.

The thickness of the person's skin is calculated as follows;

Q = k(ΔT)h

where;

h = Q/k(ΔT)

h = (120) / (0.22 x 303)

h = 1.8 m

Q = k(ΔT)h

where;

Q = (0.18)(306)(1.8)

Q = 99 J/s

Thus, the rate at which the person is now losing thermal energy is 99 J/s.

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Answer:

No

Explanation:

When a substance changes its state , there will not be any change in the mass of substance . For example , if we change 100 grams of water (liquid) to ice(solid) , the mass of ice will be same i.e., 100 grams . This shows there is no change in the mass even if a substance changes its state.

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The correct answer is Self-critical.

Why is self-improvement?

Self-critical:

Self-critical causes them to always strive for self-improvement and never to see accomplishing a goal as good enough.

The characteristic of a turbulent person causes them to always strive for self-improvement and to never see accomplishing a goal as good enough is Self-critical.

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The work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.

In Physics, we define the work done as the product of the of force and distance. Hence, we generally define the work done as that which occurs when the force applied moves a distance in the direction of the force. This implies that the work done is not a vector but a  scalar quantity.

In relation to the man and the bag, the work done is the product of the force that the man applies and the displacement of the bag. As such, the reason for the tiredness of the man is that the internal energy that he possesses is transferred to moving the bag.

Thus, the work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.

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The force F, necessary to support an object with a mass of 991 kg placed on piston 2 is 379.5 N.

Apply Pascal principle to calculated the force needed to support the mass as shown below;

F₁/A₁ = F₂/A₂

F₁/d₁² = F₂/d₂²

F₁ = F₂d₁²/d₂²

where;

Substitute the given parameters and solve for the  force needed to support the mass given.

F₁ = (991 x 9.8 x 0.0187²)/(0.0946)²

F₁ = 379.5 N

Thus, the force F, necessary to support an object with a mass of 991 kg placed on piston 2 is 379.5 N.

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Answer:

Andromeda galaxy is the correct answer.

Answer:

24.2 m/s

Explanation:

Mass is irrelevant in this situation....

Displacement:  ( to find time)

 x = xo + vo t - 1/2 at^2

 30= 0   + 0  - 1/2 (9.8)t^2

              t = 2.47 seconds

Velocity:

vf = a t   = 9.8 (2.473)  = 24.2 m/s

The height of the light house is 17.86m.

To find the answer, we have to know about the simple pendulum.

                       [tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]

where, g is acceleration due to gravity, and l is the length of the pendulum.

              [tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]

Thus, we can conclude that, the height of the light house is 17.86m.

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(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

Note: At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

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