pls help me… I missed the whole lesson and I have no clue what to do-

Answer:

yes its okay i think so it is the correct answer

I think the answer is d.right?

Answer:

a. as the distances increases , the gravitational force decreases

Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.

the answer I need points for my math test

Density of liquid try thank you so much

Answer:

  a) measure the change in volume when the object is immersed; compute from range data

  b) find the ratio of mass to volume for a measured mass and volume

Explanation:

a) The volume of a small enough irregular body can be found by measuring the difference in volume of the (semi-)fluid in which it is immersed, before and after immersion.

For irregular bodies for which that approach does not work, various 3D scanners are available for measuring volume and surface area. They may rely on optical (laser or camera), sonic, or radar measurements, and generally involve computation from distances to various points.

__

b) Density is the ratio of mass to volume. So, measurements of mass and volume of a liquid sample are sufficient to provide the basis for determining density.

Other methods include measuring buoyancy forces, and/or the depth of submersion of something that floats in the liquid. For specific liquids, hydrometers are available for measuring their density relative to that of water.

Answer:

Explanation:

Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.

Each half will have a mass of 770/2 = 385 kg

Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s

Now add back the reference frame velocity to get velocity relative to earth.

Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s

Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s

In the moving reference frame, each half will have kinetic energy which could only come from the explosion

KE = ½(385)(1.3 x 10³)² = 325,325,000 J

2(325,325,000) = 650,650,000 J released in the explosion.

Rounding to the three significant figures of the problem numerals

E = 6.50 x 10⁸ J or 650 MJ released

what's your question?

Answer:

Explanation:

acceleration a = F/m = 44000/33000 = 1.3m/s²

F = ma = 70(1.3) = 93 N

We have that for the Question, it can be said that

From the question we are told

From,

[tex]tan\theta = \frac{h}{2}[/tex]

differentiate with respect to h

[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]

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Answer:

Explanation:

Answer:

14.625 years

Explanation:

Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:

First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:

0.8t=6.5

solve for t by dividing by 0.8

t=6.5/0.8 or 8.125

so, she will arrive in 8.125 years.

Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:

1.0t=6.5 which is just t=6.5

so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.

The time elapsed in earth will be 14.625 years.

To understand this we need to find how much time she took to travel 6.5 light years and then how much time the signal took to reach the earth.

 0.8t = 6.5 ( because 1 light year = distance travelled by light in a year)

      t = 6.5/0.8

      t = 8.125 years

So, she travelled for 8.125 year.

1.0 t = 6.5 ( because radio signal travells with the same speed of light)

     t  = 6.5/1.0

     t  = 6.5 years

So, 8.125 + 6.5 = 14.625 years

So, the time elapsed on earth between the launch and the arrivel of the first radio signal is 14.625 years.

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[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]

[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]

[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]

[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]

[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]

[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]

The wavelength of the light is 550 nm

For a double slit interference pattern with slit spacing, d we have

dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.

Thus sinθ = mλ/d

Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m

Since θ is small, sinθ ≅ tanθ

So, mλ/d = L/D

Making λ subject of the formula, we have

λ = dL/mD

Substituting the values of the variables into the equation, we have

λ = dL/mD

λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)

λ = 0.22 × 10⁻⁵ m²/4.0 m

λ = 0.055 × 10⁻⁵ m

λ = 0.55 × 10⁻⁶ m

λ = 550 × 10⁻⁹ m

λ = 550 nm

So, the wavelength of the light is 550 nm

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Answer:

you have patience the distance.

Explanation:

the train leaves at 6.30.

[tex]\mu = 0.306[/tex]

Explanation:

Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as

[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]

[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]

where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as

[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]

Combining these two equations together, we get

[tex]mg\sin17° = \mu(mg\cos17°)[/tex]

Solving for [tex]\mu, [/tex]

[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]

Answer:

ω = 3.1 rad/s

θ = 36° from vertical

Explanation:

I will ASSUME that the bob and string is acting as a pendulum.

Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.

At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.

F = mg + mω²R

F/m = g + ω²R

F/m - g = ω²R

ω = √((F/m - g)/R)

ω = √((3/0.220 - 9.8)/0.40)

ω = 3.09691...

ω = 3.1 rad/s

Potential energy will convert to kinetic energy

       mgh = ½mv²

             h = v²/2g

R - Rcosθ = v²/2g

R(1 - cosθ) = v²/2g

   1 - cosθ = v²/2gR

        cosθ = 1 - v²/2gR

        cosθ = 1 - (Rω)²/2gR

        cosθ = 1 - Rω²/2g

        cosθ = 1 - 0.40(3.1²)/(2(9.8))

        cosθ = 0.804267

              θ = 36.46045...

              θ = 36°

Answer:

I actually don't understand what you are asking for....but if you denote second then...

Explanation:

Just divide 120/6=20

Hence, it's traveling 20m per second.

Answer:

20 m/s

Explanation:

Speed = Distance/Time

Speed = 120m/6s

Speed = 20 m/s

Answer:

bonjour

Explanation:

bye

Answer:

83.8851 mi/h              

Answer: Moses

Explanation: The Israelites are angry at Aaron and Moses, because they lead them into this desert where there is no food, or water. It was told that after the Israelites left Egypt they would wander the desert for forty years. The Israelites complain that they would have had plenty of food if they had just stayed enslaved in Egypt. They are not grateful that God has liberated them from slavery.

The projectile launch equations allow to find the results for the questions about the movement of the ball are:

Given parameters

To find

Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.

The range is the distance traveled for the same departure height, see attached.

.

          R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]  

         [tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]  

Let's calculate.

          v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)

          [tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46

          v₀ = 17.7 m / s

At the point of maximum height the vertical speed is zero.

          v² = v₀² - 2 g y

          0 = v₀² - 2g y

          y = [tex]\frac{v_o^2}{2g}[/tex]  

Let's calculate.

         y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]  

         y = 16 m

In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:

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For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.

The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:

For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.

The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.

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Answer:

Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.

Answer:

yes

Explanation:

because is pure ohmic risistance

Answer:

it would remain the same speed

Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace

Answer:

hi there is that OK for the weekend of the following week as well

Explanation:

6th of March is fine for me

The minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.

To determine the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop, we need to consider the maximum acceleration the pilot can withstand without losing consciousness.

Given:

Maximum acceleration without losing consciousness = 5.00g

Acceleration with g-suits to avoid blackout = 9.00g

First, we need to convert the speed of the F-15 from km/h to m/s:

Speed = 729 km/h = (729 * 1000) m/3600 s ≈ 202.5 m/s

Next, we'll calculate the acceleration experienced by the pilot in the circular loop. In a horizontal circular motion, the centripetal acceleration is given by:

Acceleration = ([tex]\rm Velocity^2[/tex]) / Radius

We can rearrange the equation to solve for the radius:

Radius = ([tex]\rm Velocity^2[/tex]) / Acceleration

Using the maximum acceleration of 5.00g, we convert it to [tex]\rm m/s^2[/tex]:

Maximum acceleration = 5.00g ≈ (5.00 * 9.8) [tex]\rm m/s^2[/tex] = 49 m/s^2

Now, we can calculate the minimum safe radius of curvature:

Radius = ([tex]\rm 202.5^2[/tex]) / 49 ≈ 838.1 meters

Therefore, the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.

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Answer: Varies

Explanation: It became more relied upon, technology is responsible for this.

Answer:

1m² = 10000cm²

1m² = 1000000mm²

1m² = 1 × 10^-6

Explanation:

When coverting from square meter to square centimetre, multiply the area value by 10.

When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )

When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)