Which one of the following expressions may be used to correctly find the angle ? in the drawing? 2 km ?DT - 2. 5 km ?=cos-112 ?=tan-1 sin- A) B)/ ?=tan"(5) D) 9-sin-i-

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

The weight of the box is 1471.5 N.

In science and engineering, an object's weight is defined as the force of gravity acting on the thing. Weight is sometimes described as a vector quantity in common textbooks, denoting the gravitational force acting on the item.

Given parameters:

Mass of the box, m = 150 kg.

Acceleration due to gravity, g = 9.8 m/s².

Hence, weight of the box is, W = mg = 150 kg × 9.81 m/s² =1471.5 N.

SO, weight of the  box with a mass of 150 kg on Earth is 1471.5 N.

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The correct answer is 4/19 value will be obtained for r2.

An effect size is an estimate of that number based on a sample that measures the magnitude of the association between two variables in a population. It can refer to the value of a parameter for a fictitious population, the value of a statistic generated from a sample of data, or the equation that operationalizes how statistics or parameters result in the impact size value. The correlation between two variables, the regression coefficient in a regression, the mean difference, or the likelihood that a specific event (like a heart attack) will occur are examples of effect sizes. Effect sizes are an important component of power analyses, sample size planning, and meta-analyses. They are a complement to statistical hypothesis testing.

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Answer: 660 N.

Explanation: Force on a free falling body is F=mg.

Therefore, Force =66×10 =660

(g is gravitational acceleration, taking it as 10)

Answer: The minimum reaction time required to stop the car before hitting the object is approximately 6.94 seconds.

Explanation: To determine the minimum reaction time to stop the car before hitting the object, we can use the kinematic equations of motion. Let's denote the initial velocity of the car as v₀ = 90 km/h = 25 m/s, the distance to the object as x = 60 m, and the maximum deceleration as a = -7.2 m/s².

We can use the following kinematic equation:

x = v₀t + (1/2)at²

Since the car needs to come to a stop, the final velocity will be zero (v = 0 m/s). Plugging in the values, we have:

0 = v₀t + (1/2)at²

Rearranging the equation, we get:

(1/2)at² = -v₀t

Simplifying further, we have:

(1/2)(-7.2)t² = -25t

Dividing both sides of the equation by t, we get:

(1/2)(-7.2)t = -25

Simplifying again, we have:

-3.6t = -25

Dividing both sides by -3.6, we find:

t = 25 / 3.6 ≈ 6.94 seconds

Therefore, the minimum reaction time required to stop the car before hitting the object is approximately 6.94 seconds.

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Answer:

The kinetic energy is 16 joules.                        

Explanation:

Kinetic energy (E) is given by:

\( E = \frac{1}{2}mv^{2} \)

Where:

m: is the mass

v: is the speed

When the kinetic energy is 4 joules, the speed of the object is:

\( v_{1}^{2} = \frac{2E_{1}}{m} = \frac{8}{m} \)                         

Now, if the speed is increased twice then we have:

\( E_{2} = \frac{1}{2}mv_{2}^{2} = \frac{1}{2}m(2v_{1})^{2} = \frac{1}{2}m[4(\frac{8}{m})] = 16 J \)

Therefore, the kinetic energy is 4 times the initial value.  

I hope it helps you!                                

The sun is Yellow, red, orange, and blue. Therefore option B is correct.

The star is moving toward an observer. Therefore option B is correct.

39. The type of radioactive decay represented by the equation provided Gamma. Therefore option A is correct.

40. The statement is completed correctly by the information which states that the coal in Alberta has a lower carbon content than coal in other parts of Canada, resulting in reduced greenhouse gas emissions.

29. graph B depicts the relationship between gravitational field strength and the mass of the International Space Station orbiting Earth.

30. The correct sequence of colors representing the changing temperature from the surface to the atmosphere of the sun is Yellow, red, orange, and blue. Therefore option B is correct.

1. When a star undergoes a blue shift, it indicates that the star is moving toward an observer. Therefore option B is correct.

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The concentration of PO43- in a 4.71 M solution of phosphoric acid (H3PO4) at equilibrium cannot be determined without additional information about the acid dissociation constants. Since the solution is 4.71 M, the concentration of H3PO4 at equilibrium is also 4.71 M.

The concentration of PO43- in a 4.71 M solution of phosphoric acid (H3PO4) at equilibrium can be determined by considering the dissociation of phosphoric acid in water. Phosphoric acid, H3PO4, is a weak acid that partially dissociates in water.

The balanced equation for the dissociation of H3PO4 is as follows:

H3PO4 ⇌ H+ + H2PO4-

H2PO4- ⇌ H+ + HPO42-

HPO42- ⇌ H+ + PO43-

At equilibrium, a certain amount of H3PO4 will dissociate into H+, H2PO4-, HPO42-, and PO43-. Since we are interested in the concentration of PO43-, we need to determine the concentration of H3PO4 at equilibrium.

Since the solution is 4.71 M, the concentration of H3PO4 at equilibrium is also 4.71 M.

The extent of dissociation depends on the acid dissociation constant, Ka, for each step of the dissociation. Without knowing the values of Ka, we cannot determine the exact concentration of PO43-. We would need more information to calculate the concentration of PO43- accurately.

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Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

Stem-and-leaf plots are an excellent way to display data. The outliers in a stem-and-leaf plot are the observations that are located far away from the other observations.

Stem-and-leaf plots are excellent for displaying numerical data in an orderly manner. They assist in determining the minimum, maximum, and range of data and can assist in determining patterns and variations. They also provide an excellent visual representation of the distribution of data and are a fantastic tool for spotting outliers. Outliers are observations in a dataset that are considerably different from the other observations. They can be identified in stem-and-leaf plots as values that are located far away from the rest of the data.

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You can improve how you handle stress and difficulties (adversity) in your life by using a variety of stress management techniques. You may lead a more well-balanced existence by controlling your stress.

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Answer:

Gravity

Explanation:

That's easy because gravity is the only thing that can pull us down that hard at that fast without anything helping it.

The force constant of the spring is 200.696 N/m & The height the object achieves when the spring releases its energy is 2.5087 m

The spring constant is the force needed to stretch or compress a spring, divided by the compressive or expansive distance. It's used to determine stability or instability in the spring, and therefore the system it's intended for. we know,

F = kx

Therefore,

k = F/x

We also know that the force being exerted on the spring is equal to the mass of the object. Hence, F = mg = 5.13 * 9.8 N = 50.174 N and we know compression due to the mass is 0.25m. Therefore,

K = 50.174/0.25 N/m

K = 200.696 N/m

Therefore, The Spring Constant is 200.696 N/m

On release, the spring potential energy gets converted to kinetic energy. Hence, on release, the height attained by the object is given by:

h = \(1/2 kx^{2}\)

We know that k=200.696 N/m and x=0.25 m. Therefore the height is:

h = \(1/2 (200.696 N/m)(0.25 m)^{2}\)

h = 2.5087 m

Therefore, the force constant of the spring is 200.696 N/m & The height the object achieves when the spring releases its energy is 2.5087 m

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The two equal-mass objects will stick together and move as one object after the completely inelastic collision.

In an inelastic collision, the two objects involved stick together and move as a single object after the collision. Since the objects have equal masses and are moving towards each other at equal speeds, their momenta will be equal and opposite before the collision.

During the collision, the objects come into contact and exert forces on each other. These internal forces cause the objects to deform and redistribute their velocities. However, due to the conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision.

Since the initial momenta of the objects are equal and opposite, the final momentum of the combined object will also be zero. This means the objects will stick together and move as one object after the collision, with a common velocity. This scenario is known as a completely inelastic collision.

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The pressure just inside the natural draft hyperbolic cooling tower must be determined based on an air flow rate of 150,000 mgfs and a flow area of 940 m2.

To calculate the pressure just inside the cooling tower, we can use the equation for mass flow rate:

Mass flow rate = density × velocity × area

Given that the mass flow rate (m) is 150,000 mgfs (mgfs stands for "metric gallons per minute"), and the flow area (A) is 940 m2, we need to convert mgfs to a standard unit of flow rate such as kg/s.

1 mgfs = 0.06308 kg/s

Therefore, the mass flow rate in kg/s is:

m = 150,000 mgfs × 0.06308 kg/s = 9,462 kg/s

Next, we need to consider the density of the air (ρ). At standard conditions (typically 20°C and 1 atm), the density of air is approximately 1.2 kg/m3.

Using the equation for pressure (P) and rearranging the mass flow rate equation, we have:

P = m / (ρ × A)

Substituting the known values:

P = 9,462 kg/s / (1.2 kg/m3 × 940 m2)

Simplifying:

P = 7.917 atm

Therefore, the pressure just inside the cooling tower must be approximately 7.917 atm

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Find the greatest electric field strength in the sun at the earth's position. The highest electric field strength in the sun at the earth's location is 1.056 kV/m.

It is a frequent fallacy that “Terra” is the worldwide genus of the planet, although in reality Earth doesn't even have an important diplomatic name. The conventional English name of a planet, including in academia, is “Earth”

Our homeworld Earth is a stony, terrestrial world. It features a substantial and lively surface with mountains, rivers, gorges, plains and much more. Being an ocean planet, Earth is unique. Water covers 70percent of Earth's crust.

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Answer:

The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.

Explanation:

hope this helps tho i don't quite know what you mean

Answer:

25000 J or 25 kJ

Explanation:

K = 1/2 mv^2

where mass is in kilograms kg

and velocity is in metres per seconds m/s

Here, m = 2000, v = 5

v^2 = 5*5 = 25

1/2 mv^2 = 1/2 * 2000 * 25

= 50000/2

= 25000

Hope it helps!

Answer:

S=D/T

S=2340/3.5

S=668 km/hrs

Explanation:

Answer:

avoid verbal argument and walk away

Both objects, the 10kg iron, and the 10kg cotton, will reach the ground at the same time if air resistance is ignored.

This result is because, in the absence of air resistance, the acceleration of a falling object is independent of its mass. All objects, regardless of their mass, will experience the same acceleration due to gravity. This acceleration is approximately 9.8 m/s^2 near the surface of the Earth. Therefore, the time taken for both objects to fall to the ground from the same height will be the same, and they will hit the ground at the same time. However, in the presence of air resistance, the results will be different as air resistance affects objects differently based on their shape and size.

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Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. ... In reaction, a thrusting force is produced in the opposite direction.

Explanation:

Answer: Often lightning occurs between clouds or inside a cloud. But the lightning we usually care about most is the lightning that goes from clouds to ground.

Explanation:  As the storm moves over the ground, the strong negative charge in the cloud attracts positive charges in the ground.

When both the force and time of contact are doubled, the impulse on an object increases by a factor of four.

When we talk about impulse, we are referring to the change in momentum of an object over a period of time. Impulse is the product of force and time, as given by the formula:
Impulse = Force x Time

the impulse on the ball would be:
Impulse = Force x Time
Impulse = 10 N x 2 s
Impulse = 20 Ns


This would cause the ball to have a momentum of:
Momentum = Mass x Velocity
Momentum = 0.5 kg x (20 N / 0.5 kg) x 4 s
Momentum = 80 Ns
The impulse on the ball increased by a factor of four, and so did the momentum. This shows that the greater the impulse on an object, the greater the change in momentum it will experience.

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Answer:

v_f = -7.5 m / s

Explanation:

Let's analyze this exercise a little, two forces that act on a body for different time intervals are indicated, each force creates an impulse and since this is a vector quantity we must add in the form of vectors. The net momentum is

we assume that the direction to the right is positive

             I = I₁ + I₂

             I = F₁ Δt₁ - F₂ Δt₂

             I = 3  1.5 - 4  3

             I = -7.5   N s

now let's use the relationship between momentum and momentum, suppose the object starts from rest (vo = 0)

             I = Δp

             I = m (v_f - v₀)

             v_f = I / m

              v_f = -7.5 / m

to finish the calculation we must assume a mass m = 1 kg

              v_f = -7.5 m / s

the negative sign in the body is moving to the left

The student should apply a 17V potential differential across the 0.7 mm piece.

The potential difference between places A and B is defined as the change in potential energy (PE) of a charge (q) transported from A to B, divided by the charge (V = VB - VA).

The joules per coulomb, sometimes known as volts (V) in honor of Alessandro Volta, are the units of potential difference.

The change in potential energy (PE) of a charge (q) transported from point A to point B, divided by the charge, is thus used to determine the potential difference between points A and B as V = VB - VA.

Volt is the SI unit for electrical potential. V stands for it.

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The Earth's temperature without an atmosphere, determined by substituting 342 W/m² for outgoing longwave radiation and rearranging the Stefan-Boltzmann equation, is approximately 255 K.

By substituting the given values into the Stefan-Boltzmann equation, we can solve for the Earth's temperature without an atmosphere. Assuming an emissivity of 1, the equation becomes 342 = (5.67 × 10^-8) × T^4. Solving for T yields a temperature of approximately 255 K.

Converting this temperature to degrees Celsius, we subtract 273.15 to obtain approximately -18.15 °C. Similarly, converting to degrees Fahrenheit using the conversion formula, we find approximately -0.67 °F.

This temperature of -18.15 °C (or -0.67 °F) represents the hypothetical temperature of the Earth without an atmosphere. Comparing it to the actual average temperature of the Earth, around 15 °C (or 59 °F), we can see that it is significantly colder. The presence of the atmosphere is crucial for trapping heat through various greenhouse gases, such as carbon dioxide and water vapor, which maintain a habitable temperature range on Earth. Without the atmosphere's greenhouse effect, the Earth's temperature would be much colder, emphasizing the vital role played by our atmosphere in sustaining life on the planet.

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A, C, D, E, and F are all potential results of an overweight aircraft.

A. Reduced thrust: An overweight aircraft will require more thrust to maintain a given airspeed and altitude, which can lead to reduced thrust available for other purposes, such as takeoff or climb performance.

C. Decreased rate of climbing: An overweight aircraft will have a reduced rate of climb due to the increased weight, which can be particularly problematic in high-altitude or hot and humid conditions.

D. Failure to complete the flight: An aircraft that is too heavy may be unable to take off or complete the flight due to safety concerns or regulatory restrictions.

E. Longer required takeoff run: An overweight aircraft will require a longer takeoff run to achieve a safe takeoff speed and lift off the ground.

F. Greater required takeoff speed: An overweight aircraft will require a higher takeoff speed to achieve the necessary lift to become airborne.

B. Higher service ceiling: An overweight aircraft is not likely to have a higher service ceiling. In fact, the opposite is usually true, as the increased weight will limit the altitude the aircraft can safely fly at, and may require a lower cruising altitude for safe operation. Therefore, this statement is incorrect.

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