An 84 N force has been applied to a block to the right but the block is moving vertically! How weird. How
much work has been done to the block? Explain your reasoning.

a. Contact forces, friction, air resistance

The question is incomplete as it does not have the options which are:

The waves have a trough.

Particles move only small distances.

particles travel the length of the wave, moving in an up-and-down motion.

The waves have a crest.

The energy is transferred perpendicular to the direction of the wave motion.

Answer:

Explanation:

Transverse waves are the waves which can be characterised easily by their oscillation in a perpendicular direction to the path of the wave propagation.

The particle follows the wave nature and moves only to a small distance like ripples in the water and these waves possess the trough, the lower part and the crest the upper part.

These waves can be observed in ripples of the water, in a string of the guitar while playing and the electromagnetic waves.

Thus, the selected options are correct.

The resulting displacement is approximately 9.22 blocks at an angle of approximately 26.57 degrees north of west.

To find the displacement and total distance covered by the newspaper delivery person, we can use the Pythagorean Theorem and trigonometric functions.

We can see that the delivery person moves 3 blocks west and then 6 blocks east. These two movements cancel each other out, so the net displacement in the east-west direction is 0.

Next, we can see that the delivery person moves 4 blocks north. Using the distance formula, the distance covered in the north-south direction is:

\($$\sqrt{(0-0)^2+(4-0)^2}=\sqrt{16}=4$$\)

Therefore, the total distance covered is 3 + 4 + 6 = 13 blocks.

To find the resulting displacement in magnitude and angle form, we can use trigonometric functions. The resulting displacement is the vector sum of the individual movements.

To start, let's use the inverse tangent function to find the angle between the resulting displacement and the positive x-axis:

\($$\tan^{-1}\left(\frac{4}{3+6}\right)\approx 26.57^\circ$$\)

This angle corresponds to the angle that the vector makes with the positive x-axis. To find the magnitude of the vector, we can use the Pythagorean Theorem:

\($$\sqrt{(3+6)^2+4^2}\approx 9.22$$\)

Therefore, the resulting displacement is approximately 9.22 blocks at an angle of approximately 26.57 degrees north of west.

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If the reaction has a negative ΔG (Gibbs free energy), it indicates that the reaction is spontaneous and thermodynamically favorable. The correct statement is "Keq > 1" when ΔG < 0.

In this case, the following statement must be true:

Keq > 1.

Keq represents the equilibrium constant of the reaction, which is a ratio of the concentrations (or pressures) of the products to the concentrations (or pressures) of the reactants, each raised to the power of their stoichiometric coefficients. When Keq is greater than 1, it implies that the concentration of products is higher than the concentration of reactants at equilibrium, indicating that the reaction favors the formation of products.

The terms "exothermic" and "endothermic" refer to the heat transfer of a reaction, not the Gibbs free energy change. The sign of ΔG does not provide direct information about whether the reaction is exothermic or endothermic. The exothermic or endothermic nature of a reaction is determined by the overall energy change (enthalpy change, ΔH) of the reaction.

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Answer:

hippocrates

I guess..........

At 4:00 p.m., the speed difference between the ships was 12.87 km/h.

Navigation is indeed a technical skill required for ship management. B: The process of selecting the route to take to arrive at a specific location. 2a: sailing as a sport, including handling or sailing. B: a ship leaving a port. Examples of Verbs and Phrases with Sail More information on sailing

If you read how-to books & boating periodicals, you could believe that sailing is difficult, but this isn't the truth. An experienced instructor can educate you the fundamentals of sailing in just one afternoon. After just just few days of lessons, the majority of new students strike off on their own.

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When  the core of a star reaches a temperature of about 100 million degrees (k), the new event named  triple-alpha process happens where the helium will begin to fuse into carbon.

The triple-alpha mechanism, which turns three helium nuclei into one carbon atom, will start to fuse helium into carbon when the temperature in the core reaches roughly 100 million degrees. There is a lot of heat produced by this.

The triple-alpha process is unusually sensitive to temperature; when the reaction's temperature is doubled, it proceeds nearly a trillion times faster! Therefore, while the fusing helium heats the core, which cannot expand to cool down, the increasing temperature prompts the helium fusion to abruptly proceed millions of times faster, which very quickly heats the core even more, which in turn prompts the helium to fuse way, way faster.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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Answer:

Alkali metals

Explanation:

Elements in this group are highly reactive, soft, lustrous and highly conductive.

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

Examples of alkali electrons include:

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

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Answer: Thunderstorm.

Explanation: When a cold, dry air mass collides with a warm, humid air mass over the land, that's a tornado, but on top of that the air rises and it cools and condenses that's a thunderstorm

When a cold, dry air mass collides with a warm, humid air mass over the land, that's a tornado, but on top of that the air rises and it cools and condenses that's a thunderstorm.

A thunderstorm, sometimes referred to as an electrical storm or a lightning storm, is a type of storm that is distinguished by the presence of lightning and the thunder that results from the lightning's acoustic effect on the Earth's atmosphere.  

Thundershowers are the name for relatively weak thunderstorms. A cumulonimbus is a form of cloud where thunderstorms develop.  They frequently produce heavy rain.  and occasionally snow, sleet, or hail and they are typically accompanied by strong wind.

However, some thunderstorms result in little or no precipitation. A squall line, often known as a succession of thunderstorms, can form or become a rainband.

Therefore, When a cold, dry air mass collides with a warm, humid air mass over the land, that's a tornado, but on top of that the air rises and it cools and condenses that's a thunderstorm.

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The energy per photon associated with a frequency of 1260 kHz is 2.10×10^-25 J.

To calculate the energy per photon, we can use the equation: E = hf, where E represents the energy, h is the Planck's constant (6.62607015 × 10^-34 J·s), and f is the frequency of the photon. Given that the frequency is 1260 kHz, we need to convert it to hertz (Hz) by multiplying it by 10^3:

Frequency = 1260 kHz × 10^3 = 1.26 × 10^6 Hz

Now, we can substitute the values into the equation:

E = (6.62607015 × 10^-34 J·s) × (1.26 × 10^6 Hz)

E = 8.33929859 × 10^-28 J

The answer is given in scientific notation as 8.34 × 10^-28 J. However, the question specifically asks for the answer in the format of 0.00×10^0. To achieve this, we can multiply the result by 10^3 and adjust the exponent accordingly:

E = (8.33929859 × 10^-28 J) × (10^3)

E = 8.33929859 × 10^-25 J

Thus, the energy per photon associated with a frequency of 1260 kHz is 2.10×10^-25 J.

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A): a) The velocity of the stone is -7.96 m/s b) The velocity of the stone after rising 15 m is 8.92 m/s. B) : a) The marginal cost function is C'(x) = 3 + 0.06x + 0.0009x².

A) a) To find the velocity of the stone after 6 seconds, we take the derivative of the given height equation h(t) with respect to time t to obtain the velocity equation v(t): v(t) = dh/dt = 13 - 1.66t

Substituting t = 6 seconds into the equation, we get:

v(6) = 13 - 1.66(6) = -7.96 m/s

Therefore, the velocity of the stone after 6 seconds is -7.96 m/s, which means that the stone is moving towards the surface of the moon.

b) To find the velocity of the stone after it has risen 15 m, we need to solve the height equation h(t) for the time t at which h(t) = 15 m. This gives us a quadratic equation: 0.83t² - 13t + 15 = 0

Solving for t using the quadratic formula, we get:

t = 7.85 s (approximately) or t = 2.68 s (approximately)

Since the stone is rising, we choose the positive value of t, which means that it takes 7.85 seconds for the stone to reach a height of 15 m.

To find the velocity at this point, we use the velocity equation v(t) that we obtained earlier: v(t) = 13 - 1.66t

Substituting t = 7.85 seconds, we get:

v(7.85) = 13 - 1.66(7.85) = 8.92 m/s

Therefore, the velocity of the stone after it has risen 15 m is 8.92 m/s, which means that it is still moving upwards.

B) (a) The marginal cost is the derivative of the cost function with respect to the number of units produced, i.e., C'(x) = dC/dx.

Taking the derivative of the cost function C(x) with respect to x, we get:

C'(x) = 3 + 0.06x + 0.0009x²

Therefore, the marginal cost function is C'(x) = 3 + 0.06x + 0.0009x².

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The direction cosines of a→+b→, are

Given data:

a = 4i^ + 7j^ - 5k^

b= 3i^ + 4j^ + k^

a→+b→ = 7i + 11j - 4k

( a→+b→ ) / | a→+b→ | = ( 7i + 11j - 4k ) / \(\sqrt{7^2 + 11^2 + 4^2 }\)

                                = ( 7i + 11j - 4k ) / 13.64

                                = 0.5232 i + 0.8065 j + 0.2933 k

Therefore the direction cosines are

cos ∝ = 0.5232

cos β = 0.8065

cos γ = 0.2933

Hence we can conclude that the direction cosines of a→+b→ are as listed above.

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Hypotheses

Null Hypothesis (H₀): The current ozone level is not significantly different from the normal level of 5.8 ppm.

Alternative Hypothesis (H₁): The current ozone level is significantly lower than the normal level of 5.8 ppm.

Z-test statistic

To calculate the z-test statistic, we use the formula:

z = (sample mean - population mean) / (standard deviation / √sample size)

Given:

Sample mean (x) = 5.7 ppm

Population mean (μ) = 5.8 ppm

Standard deviation (σ) = 1.1

Sample size (n) = 990

Plugging in the values:

z = (5.7 - 5.8) / (1.1 / √990)

z ≈ -2.28

Type of test

This is a one-tailed test. We are interested in whether the ozone level is significantly lower, not higher, than the normal level.

Decision rule

Since the significance level is 0.02, we reject the null hypothesis if the z-test statistic is less than the critical value corresponding to a 0.02 (2%) one-tailed significance level.

Conclusion

The z-test statistic value of -2.28 falls in the critical region and is less than the critical value for a 0.02 (2%) one-tailed test. Therefore, we reject the null hypothesis. The data supports the claim that the current ozone level is significantly lower than the normal level at the 0.02 level.

In summary, the data provides evidence to support the researcher's claim that the current ozone level is insufficient at a significance level of 0.02. The calculated z-test statistic of -2.28 falls in the critical region, leading to the rejection of the null hypothesis.

This suggests that there is a significant difference between the observed sample mean of 5.7 ppm and the expected population mean of 5.8 ppm. The ozone level appears to be lower than normal, warranting further attention and investigation.

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If the Sun takes 233 million years to orbit once around the Milky Way, how many orbits had the Sun made when it was 1.1 billion years old

To determine how many orbits the Sun had made when it was 1.1 billion years old,

1. Convert 1.1 billion years to million years: 1.1 billion years = 1100 million years.
2. Divide the age of the Sun by the time it takes to complete one orbit around the Milky Way: 1100 million years / 233 million years = 4.72 orbits.

Since the Sun cannot complete a partial orbit, it had made 4 orbits around the Milky Way when it was 1.1 billion years old.

Regarding how many times the Sun has orbited around the Milky Way since it first formed, we need to know the current age of the Sun. The Sun is approximately 4.6 billion years old.

Following the same steps as above:
1. Convert 4.6 billion years to million years: 4.6 billion years = 4600 million years.
2. Divide the age of the Sun by the time it takes to complete one orbit around the Milky Way: 4600 million years / 233 million years = 19.74 orbits.

Similar to the previous case, the Sun cannot complete a partial orbit, so it has made 19 orbits around the Milky Way since it first formed.

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Answer:C car

Explanation:

A toy car is being pushed to the right 6.63 N force and moving at a constant speed  the force of friction acting on the car less than 6.63 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question car is moving with a constant velocity so force of friction is less than the applied force 6.63 N. Frictional force is opposite force to the motion.

A toy car is being pushed to the right 6.63 N force and moving at a constant speed  the force of friction acting on the car less than 6.63 N.

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Answer:

once a week

Explanation:

once a week so answer B

Answer:

B-Once a week

Explanation:

You answer is B-Once a week

The induced emf in the loop is  -1500 μ V or  - 0.0015 V  .

According to the question

A conducting loop in the form of a circle is placed perpendicular to a magnetic field of 0. 50 t.  

i.e

Magnetic field (B) = 0. 50 T

Area of circle or loop = \(\pi r^{2}\)  

Now,

The area of the loop decreases at a rate of 3. 0 × 10⁻³ m/s

i.e

dA = 3. 0 × 10⁻³ meter²

dt = 1 sec  

As per the formula of Induced e.m.f in the loop

emf is dependent on number of turns of coil, shape of the coil, strength of magnet and speed with which magnet is moved. Emf is independent of resistivity of wire of the coil.

\(e=-B*\frac{dA}{dt}\)

where A is the area of the loop.

Now ,

Substituting the values in the formula

\(e=-B*\frac{dA}{dt}\)

\(e= - 0.50 *\frac{ 3 * 10^{-3}}{1}\)    

e = - 0.0015 V

OR

e = -1500 * 10⁻⁶ V

e =  -1500 μ V

Negative just signifies emf will such be induced that current induced will oppose change in magnetic field though it

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Answer:

Wind Turbines are used in a variety of applications from harnessing offshore wind resources to generating electricity for a single home.

Explanation:

Answer:

Suppose any object with mass in our planet.

The object will be affected by the gravitational force, that pulls the object down. Now, a table in your dining room is also affected by this force, but the table is in a surface (the ground) and it does not move.

We know that if an object does not move, then the net force acting on it is equal to zero, this means that there is a force equal and opposite ot the gravitational force.

This force is the normal force, that the ground applies on the table. This force comes as a "response" to the table pushing the ground (By 3rd Newton's law).

So two forces that are always equal and opposite are the forces caused by this law.

The distance covered by the object is given as,

Plug in the known values,

Thus, the distance covered by the object is 81 m.

The given function of the particle's position as a function of time is x(t) = (0.02625 m)sin(6πt). Given that a particle moving in simple harmonic motion passes through the equilibrium point (x=0)3 times per second, the time period T of the particle is given as; T = 1/f = 1/(3 s^-1) = 0.333 s The amplitude A of the particle is given as; A = 0.02625 mGiven that the particle's velocity at x = 0.02625 ma is negative at time t = 0.

To find out the velocity of the particle we differentiate the given function of position with respect to time. The velocity v(t) of the particle is given by;v(t) = dx/dtv(t) = (0.02625 m)cos(6πt) × 6πThe velocity of the particle at time t = 0 when x = 0.02625 ma is negative is given as;v(0) = (0.02625 m)cos(0) × 6πv(0) = 0Negative velocity means that the velocity is directed in the negative direction.

This implies that the phase angle θ is π radians or 180°.Hence, the function describing the particle's motion is;x(t) = A sin (ωt + θ)where;ω = 2πf = 6πrad/sθ = 180°x(t) = A sin (ωt + θ)x(t) = 0.02625 sin (6πt + 180°)

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I don't think you finished writing down the question, I think there needs to be more information.

The mass of the block on the frictionless horizontal surface is 5 kg.

The mass of the block is determined by applying Newton's second law of motion:

F = ma

The given parameters are

the magnitude of the first force, F₁ = 12 N

the magnitude of the second force, F₂ = 2 N

acceleration of the block, a = 2 m/s²

So, determining the mass of the block is

F = ma

∑F = ma

F₁ - F₂ = ma

12 - 2 = 2m

10 = 2m

m = 5 kg

Thus, the mass of the block on the frictionless horizontal surface is 5 kg.

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Explanation:

look up the coordinates or if they provide you with a map use that

Answer:

find each point in and draw a straight line until they touch and whatever is there is the country

Answer:

C. Cables transmit infrared waves over longer distances.

Explanation:

A.pex

Answer:

The answer is C.Cables transmit infrared waves over longer distances.

Explanation:

I took AP EX quiz.

If the percent difference is larger than 5%, it indicates that there may be some experimental error or that the theoretical model is not accurate enough to predict the behavior of two lenses in contact.

To compare the agreement between experimental and theoretical values of fab, we can calculate the percent difference between the two values. If the percent difference is small, it suggests that equation (4) is a valid model for the equivalent focal length of two lenses in contact.

First, we need to calculate the theoretical value of the fab using equation (4). Then, we can measure the focal lengths of lenses a and b experimentally and combine them to get the experimental value of fab. We can then calculate the percent difference between the two values using the formula:

% difference = |(theoretical - experimental) / theoretical| x 100%

If the percent difference is less than 5%, it suggests that equation (4) is a valid model for the equivalent focal length of two lenses in contact.

Overall, comparing the agreement between experimental and theoretical values of the fab is important in determining the validity of equation (4) as a model for the equivalent focal length of two lenses in contact.

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Answer:

sorry, I do not know :(

Explanation:

Do not be angry!!!