Please help if u can thanksss

[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 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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Answer:

you have patience the distance.

Explanation:

the train leaves at 6.30.

Answer:

RNA polymerase reads a template strand of DNA 5' to 3'

Explanation:

RNA polymerase is the enzyme accountable for the transcription of the DNA. It shows one coast of the DNA and transcribed within the RNA. It combines the nucleotide to the 3' conclusion of the expanding series of the RNA. It commences the method by connecting to the promoter area of the gene. The enzyme increases nucleotide, not ribonucleotide.

Therefore, A. RNA polymerase reads a template strand of DNA 5' to 3' is false.

Answer:

RNA polymerase reads a template strand of DNA 5' to 3'.

Explanation:

RNA synthesis is less processive during elongation than during initiation. Synthesis begins with the polymerase binding two rNTP molecules. The polymerase frequently releases the nascent transcript before it reaches 8-10 nucleotides in length.

Answer:

Explanation:

Given,

Let Alex be = x years

Then June will be = (7 + x) years

We know that,

Their total combined age is 29 years

Therefore,

By the problem,

=> x + (7 + x) = 29

=> 2x = 29 - 7

=> 2x = 22

=> x = 22 ÷ 2

=> x = 11

So,

Required age of June is = (7 + x) years

= (7 + 11) years

= 18 years (Ans)

Answer:

917 N

Explanation:

917 N, this is your answer!!

Glad to help.

I think the answer is d.right?

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°

The period of oscillation of the system nearest to three decimal places

= 1.092 seconds

The period of an oscillation occurring in a system is the time taken to complete one cycle.

The formula that is used to calculate the period of oscillation (T) is

                = 2π√[tex]\frac{l}{g}[/tex]

But,

π = 3.14159 (constant)

g= 10m/s² (acceleration due to gravity)

l = 0.302 m

Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]

                    = 6.28318 x √0.0302

                    = 6.28318 x 0.17378

                    = 1.09189s

                    = 1.092 seconds ( to the nearest three decimal places)

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reduction

deflation

these are examples of the opposite of enlarged to make something smaller is really the key thing here

Answer:

Antonie van Leeuwenhoek

Explanation:

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

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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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:

No, it is closer to 1.30

Explanation:

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:

is 6

Explanation:

h

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:

The average atomic mass of an element is the sum of masses of it's isotopes

Each are multiplied by it's natural abundance

Explanation:

The average atomic mass of an element is the sum of masses of it's isotopes

Each are multiplied by it's natural abundance

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:

deformation : elastic deformation is reversed when the force is removed. inelastic deformation is not fully reversed when the force is removed – there is a permanent change in shape.

Explanation:

tysm

Explanation:

let v1 = 4.3 m/s

v2 = 12.9 m/s

t = 2 seconds

v2 = v1 + at

12.9 = 4.3 + a×2

2a = 12.9 - 4.3 = 8.6

a = 8.6/2

a = 4.3 m/s^2

Answer:

Its Friction

Explanation:

the pins are not floating and they are not a magnet and not involved with air

The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.

Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.

Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.

To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.

The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.

Thus, the correct option is A.

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Answer:

Acceleration due to gravity

Answer:

Attracting means pulling toward you and repelling means pushing away

Explanation:

Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.

Explanation: Hope this helps!

Answer:

Explanation:

Power is the rate of doing work

P = 72 / 1.2 = 60 Watts

Answer:

Give me some hint please

The conservation of momentum and energy allows to shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

Kinematics studies the movement of bodies, looking for relationships between the position, velocity and acceleration of bodies.

The momentum is defined by the product of mass and the velocity, when a system is isolated the momentum is conserved.

The mechanical energy is the sum of the kinetic energy plus the potential energies, when there is no friction in the system the mechanical energy is conserved.

Let's solve this exercise in parts:

a) Let's use kinematics to find the speed of the girl before she grabs the box

              v² = v₀² - 2 g y₁

              v² = 8² - 2 9.8 2.00

              v = R 24.8 = 4.98 m / s

b) Let's use momentum conservation for when the speed of the girl and the box together. Let's write the moment in two moments.

Initial instant. Just before you grab the box.

              p₀ = M v + 0

Final moment. Right after taking the box

             [tex]p_f[/tex]  = (m + M) [tex]v_f[/tex]

In system this form by the girl and the box therefore it is an isolated system and the momentum is conserved.

           [tex]p_o = p_f[/tex]  

           mv = (m + M) [tex]v_f[/tex]  

           [tex]v_f = \frac{m}{m+M} \ v[/tex]

Let's calculate

           [tex]v_f = \frac{15}{15+ 60} \ 4.98[/tex]  

           [tex]v_f[/tex]  = 0.996 m / s

c) Now we use conservation of energy after the girl has the box.

Starting point. When the girl has the box

           Em₀ = K + U

           Em₀ = ½ (m + M) v² + (m + M) g y₁

Final point. At the highest point of the trajectory

          [tex]Em_f[/tex]  = U

         [tex]Em_f[/tex] = (m + M) g y₁

As there is no friction, the energy is conserved.

           [tex]Em_o = Em_f[/tex]  

          ½ (m + M) v² + (m + M) g y₁ = (m + M) g y

          y = [tex]\frac{v^2}{2g} + y_1[/tex]  

Let's calculate

           y = [tex]\frac{0.996^2}{2 \ 9.8} + 2.0[/tex]

           y = 2.05 m

In conclusion using the conservation of momentum and energy we can shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

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Answer:

Vbefore = 4.98 m/s

Vafter = 3.98 m/s

Hmax = 2.81 m