A and E are wrong i think please solve this thank you.

A And E Are Wrong I Think Please Solve This Thank You.

Answers

Answer 1

Answer:

I think number 3.

Explanation:


Related Questions

What diameter telescope is needed to resolve the separation between an Earth-like planet and its star at 550 nm if the linear separation between them is 1 AU and the star system is 3 pc from Earth

Answers

The Rayleigh criterion allows finding the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

The diameter of the telescope is D = 0.415 m

The Rayleigh criterion is used to find the separation of two points, it is based on the fact that the diffraction maximuum pattern of the first object coincides with the first minimum of the second object.

By entering in the diffraction ratio for slits you will find.

           sin θ  = [tex]\frac{\lambda}{a}[/tex]  

In general in diffraction experiments the angles are very small,

           [tex]tan \theta = \frac{y}{x} = \frac{sin \theta}{cos \theta} \\sin \theta = \frac{y}{x}[/tex]

 

For the case of circular apertures, when solving in polar coordinates, a constant appears.

 

        [tex]\frac{y}{x} = 1.22 \frac{\lambda}{D}[/tex]

       [tex]D = 1.22 \frac{\lambda \ x}{y}[/tex]

Where λ is the wavelength of light and D is the diameter of the aperture.

They indicate that the separation between the star and the planet is 1 AU and the distance from the system to the Earth is 3 parce.

Let's reduce the parce to astronomical units

       x = 3 pc (  [tex]\frac{206264 AU}{1 pc}[/tex] )

       x = 6.18 10⁵ AU

Let's calculate

          D = [tex]D = 1.22 \ \frac{550 \ 10^{-9 } \ 6.18 \ 10^5 }{1}[/tex]  

          D = 0.415  m

In conclusion, using the Rayleigh criterion we can find the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

 The diameter of the telescope is D = 0.415 m

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a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?

Answers

Answer:

Explanation:

F = ma

3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a

a = 1.10566...

a = 1.1 m/s²

the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.

A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the
average power produced by the soccer player?

Answers

The  average power produced by the soccer player is  710 Watts.

Given the data in the question;

Mass of the soccer player; [tex]m = 67kg[/tex]Energy used by the soccer player; [tex]E = 5100KJ = 5100000J[/tex]Time; [tex]t = 2.0h = 7200s[/tex]

Power; [tex]P =\ ?[/tex]

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

[tex]Power = \frac{Energy\ converted }{time}[/tex]

We substitute our given values into the equation

[tex]Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W[/tex]

Therefore, the  average power produced by the soccer player is  710 Watts.

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5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.

Answers

Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

of the following which is the largest body?
a. the moon
b. Pluto
c. Mercury
d. Ganymede

Answers

Answer:

Ganymede is the largest body

Explanation:

it is the satellite of jupiter

Ganymede is the largest body

Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?

Answers

Answer:

Ammonium perchlorate NH4ClO4

Ammonium Nitrate

Calcium Cyanamide

When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.

Hopefully this helps :)

what is the specific latent heat of fusion of water?

Answers

Answer:

334

Explanation:

In an oscillating LC circuit, when 81.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor

Answers

Answer:

21

Explanation:

9+10=21

what do i do for #17

Answers

Answer:

whats the question...............

Explanation:

A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?

Answers

Answer:

30.3 meters, 172 degrees

Explanation:

To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.

Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West

Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North

The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.

Applying the Pythagorean theorem leads to the magnitude of the resultant (R).

R2 = (30.0 m)2 + (4.0 m)2 = 916 m2

R = Sqrt(916 m2)

R = 30.3 meters

The angle theta in the diagram above can be found using the tangent function.

tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)

tangent(theta) = 0.1333

theta = invtan(0.1333)

theta = 7.59 degrees

This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.

Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points

Answers

The potential difference between the initial and final point is 3.0 V.

The given parameters:

distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5m

The potential difference between the initial and final point is calculated as follows;

[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]

where:

[tex]d_2[/tex] is the distance of the electron between the positive and negative plate

[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]

[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]

Thus, the potential difference between the initial and final point is 3.0 V.

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HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena​

Answers

A, C, D, but there’s a possible point where b and e could be up there

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship

Answers

The length of the ship in terms of Emily's equal steps is 84.

The given parameters;

equal steps forward = 210equal steps backward = 42

Defined parameters:

Let the constant velocity of the ship = VLet the velocity of Emily = VsLet the length of the ship = dLet the time of motion, = t

The velocity of Emily when moving forward in the direction of the ship:

[tex]V_s_1 = \frac{210}{t}[/tex]

The velocity of Emily when moving in opposite direction to the ship:

[tex]V_s_2 = \frac{42}{t}[/tex]

The constant velocity of the ship:

[tex]V = \frac{d}{t}[/tex]

Apply relative velocity formula to determine the length of the ship:

For forward (same direction) motion:

[tex](V_s_1 - V)t = d[/tex]

For backward (opposite direction) motion:

[tex](V_s_2 + V)t = d[/tex]

[tex](V_s_1 - V)t = (V_s_2 + V)t\\\\V_s_1 - V = V_s_2 + V\\\\\frac{210}{t} - \frac{d}{t} = \frac{42}{t} + \frac{d}{t} \\\\\frac{210}{t} - \frac{42}{t} = \frac{d}{t} + \frac{d}{t} \\\\\frac{210 - 42}{t} = \frac{d+ d}{t} \\\\\frac{168}{t} = \frac{2d}{t} \\\\2d = 168\\\\d = \frac{168}{2} \\\\d = 84[/tex]

Thus, we can conclude that the length of the ship in terms of Emily's equal steps is 84.

"Your question is not complete, it seems to be missing the following information;"

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship

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A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 55.0 s.

Answers

Hi there!

(A)

A grinding wheel is the same as a disk, having moment of inertia of:

[tex]I = \frac{1}{2}MR^2[/tex]

Plug in the given mass and radius (REMEMBER TO CONVERT) to find the moment of inertia:

[tex]I = \frac{1}{2}(0.380)(0.085)^2 = 0.00137 kgm^2[/tex]

(B)

We can use the rotational equivalent of Newton's Second Law to calculate the needed torque:

Στ = Iα = τ₁ - τ₂

Begin by solving for the angular acceleration. Convert rpm to rad/sec:

[tex]\frac{1750r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 183.26 rad/sec[/tex]

Now, we can use the following equation:

ωf = wi + αt (wi = 0 rad/sec, from rest)

183.26/5 = α = 36.65 rad/sec²

τ = Iα = 0.0503 Nm

Since there is a counter-acting torque on the system, we must begin by finding that acceleration:

[tex]\frac{1500r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 157.08 rad/sec[/tex]

ωf = wi + αt

-157.08/55 = α = -2.856 rad/sec²

τ₂ = Iα = 0.0039 Nm

Now, calculate the appropriate torque using the above equation:

[tex]\Sigma\tau = \tau_1 - \tau_2[/tex]

[tex]\Sigma\tau + \tau_2 = \tau_1[/tex]

[tex]0.0503 + 0.0039 = \large\boxed{0.054 Nm}[/tex]

A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?

Answers

Answer:

that is all i know

Explanation:

radius= 25.0cm

height= 8m

inner radius= 0.582cm

density= 1.00 × 103kgf= m× a

Ignoring any effects of dc resistance, what is the total reactance of a 250 mH coil in series with a 4.7 microfarad capacitor at a signal frequency of 60 Hz

Answers

The total reactance of the inductor and the capacitor is 470.1 ohms.

The given parameters;

inductance of the coil, L = 250 mHcapacitance, C = 4.7 μfrequency of the circuit, f = 60 Hz

The inductive reactance of the coil is calculated as follows;

[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]

The capacitive reactance of the capacitor is calculated as follows;

[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]

Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.

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Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.

Answers

Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.

Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:

[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]

Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:

[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]

Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:

[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]

Finalmente, reemplazamos los valores para obtener:

[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]

Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.

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A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipes,
it is observed that the number of beats per second first diminishes to zero and then increases again to 4.
By how much has the temperature of the air in the pipe been altered?

Answers

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]

These combinations of frequency produce 4 beats per sound.

i.e.

[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]

[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]

[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]

When it is altered, the beats first diminish and increase again by 4.

i.e.

[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]

[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]

If we equate both equations (1) and (2) together, we have:

[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

when the temperature of the pipe  = unknown ???the temperature of the open orang pipe = 15

[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]

By squaring both sides, we have:

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]

[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]

[tex]\implies \mathbf{273 +T =306.726912 }[/tex]

T = 306.726912 - 273

T ≅ 33.73 ° C

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

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What is the effect of erosion?

A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.

Answers

The answer is A new land forms at the mouth of a river

please help meee please please

Answers

Answer:

I think it's c but don't know for sure

There are 270 students and teachers going on a field trip to a science center. If each school bus holds 54 people, how many buses are needed?

Answers

5 buses!
270/54 is five:)


I will mark brainlist

A wave is disturbance that transfers energy and matter.


true
false

Answers

Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

Answer:

I'm pretty sure it true sorry if I'm wrong

which statement is true about heat and light energy

Answers

Answer:

statements?

Explanation:

PLEASE HELP I NEED TO PASS THIS CLASS ​

Answers

Can you please tell me what is the problem that we need to answer?

Answer:

I think that the answer is 4.318 as a=f/p

i need help with these please

Answers

Answer:

                             

Explanation:

Check all true statements about nonmetals. Group of answer choices
All nonmetals are gaseous at room temperature.
Nonmetals have low melting and boiling points. ✅
Nonmetals are brittle and have a relatively low density. ✅
Nonmetals are good at thermal and electrical conduction.

Using the colored periodic table below, match the group or area to its color.
alkali metals red alkaline ✅
earth metals orange ✅
transition metals white ✅
halogens purple ✅
Nobel gasses teal✅

How many of each subatomic particle are in a neutral atom of Potassium-39? What is its mass number? Potassium is in group 1 and has the symbol "K".
protons 19 ✅
neutrons 20 ✅
electrons 19 ✅
mass number 39✅

Match the ion with its charge.
11 protons, 12 neutrons, 10 electrons = +1 ✅
Groups 17, period 2 = -1 ✅
31 protons, 39 neutrons, 28 electrons = +3✅
Group 2, period 5 = +2

Which best describes a metal such as Silver (Ag)?
Answer: Lustrous, malleable, and forms cations.

Determine if the "quoted" word(s) makes each statement True or False.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". False ✅
If matter has a higher temperature, that means there will be "more" molecular motion. True✅
If you put a balloon in a freezer, its volume would "increase". False ✅
If you put the balloon into a chamber where there is half the pressure, its volume will "double". True ✅
If you cool down a propane tank, there will be "less" pressure in it. True ✅
The average atomic mass of Aluminum (Al) is "16.00 amu".
False✅
Oxygen's atomic number is "8" therefore it has "8" protons in its nucleus. True✅
An "electron" has the same mass as a neutron. False✅
The nucleus contains "protons and neutrons", virtually all the mass of an atom.✅

Answers

Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.

Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.

Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.

Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.

The following is an accurate matching of the ions;

11 protons, 12 neutrons, 10 electrons = +1 - Na^+

31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+

Group 2, period 5 = +2 - Sr^2+

Silver is a lustrous metal.

Elements that share the most characteristics are found on the periodic table in the same "horizontal period". FalseIf matter has a higher temperature, that means there will be "more" molecular motion. TrueIf you cool down a propane tank, there will be "less" pressure in it. TrueIf you put a balloon in a freezer, its volume would "increase". FalseIf you put the balloon into a chamber where there is half the pressure, its volume will "double". TrueThe average atomic mass of Aluminum (Al) is "16.00 amu".  FalseOxygen's atomic number is "8" therefore it has "8" protons in its nucleus. TrueAn "electron" has the same mass as a neutron. FalseThe nucleus contains "protons and neutrons", virtually all the mass of an atom - True

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Answer:

If your answers are the green check marks all of them are correct

Explanation: I took the test

what were your preparetion before going the different physical fitness test?​

Answers

Answer:

Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened

A boat is headed north at a velocity of 8 km h-1. A strong wind is blowing whose pressure on the boat’s superstructure causes it to move sideways to the west at a velocity of 2 km h-1. There is also wave present that flows in a direction 30° south of east at a velocity of 5 km h-1. What is the velocity of the boat?​

Answers

The velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.

The given parameters;

velocity of boat northwards, = 8 km/hvelocity of the boat westwards, = 2 km/hvelocity of the wind, = 5 km/h 30° south of east

The resultant vertical velocity of the boat is calculated as;

[tex]v_y = 8 \ - \ 5 sin(30)\\\\v_y = 8 - 2.5\\\\v_y = 5.5 \ km/h[/tex]

The resultant horizontal velocity of the boat is calculated as;

[tex]v_x = -2 \ + 5cos(30)\\\\v_x = 2.33 \ km/h[/tex]

The resultant boat velocity is calculated as follows;

[tex]v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.33^2 + 5.5^2} \\\\v = 5.97 \ km/h[/tex]

The direction of the velocity;

[tex]\theta = tan^{-1} (\frac{v_y}{v_x} )\\\\\theta = tan^{-1}(\frac{5.5}{2.33} )\\\\\theta = 67 \ ^0[/tex]

Thus, the velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.

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the push up is dynamic or static​

Answers

Answer:

Dynamic exercises

Explanation:

A Projectile of mass 0.277 kf is shot from a cannon. The end of the cannon's barrel is at heiht 6.8 m. The initial velocity of the projectile has a horizontal component of 6.1 m/s. The projectile rises to a maximum height above the end of the cannon's barrel and strikes th ground a horizontal distance past the end of the cannon;s barrel. What is the time it takes for the projectile to reach its maximm height? The acceleration of gravity is 9.8 m/s^2.

Answers

Ms2 is the answer to this Q

The time taken by  the projectile to reach its maximum height is 0.62 second.

Given parameters:

Horizontal component of initial velocity of the projectile: [tex]u_x[/tex] = 6.1 m/s.

To reach maximum height, the projectile should be throwed in 45 degree. So, vertical component of initial velocity of the projectile: [tex]u_y[/tex] = 6.1 m/s.

Given,  The acceleration of gravity is 9.8 m/s².

At maximum height the vertical component of velocity of the projectile becomes zero due to acceleration due to gravity acts downwards.

So, from formula [tex]v_y[/tex] = [tex]u_y[/tex] - gt; we can write:

0 = 6.1 - 9.8t

⇒ t = 6.1/9.8 = 0.62 s.

Hence, the time taken by  the projectile to reach its maximum height is 0.62 second.

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