The launch speed of a projectile is three times the speed it has at its maximum height.what is the elevation angle at launch?

Answers

Answer 1

Answer:

Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.


Related Questions

A very light rod 40cm long is pivoted at the centre. A weight of 50N is placed at one end. Where is the place to put a weight of 200N in order that the rod is in equilibrium?

Answers

Hi there!

We can go about this problem using a summation of torques.

In order to ensure the rod is in equilibrium, we must satisfy the condition:

Στ = 0

Since the rod is "very light", we can disregard its mass.

The equation for torque is:

τ = rFsinθ

In this instance, the torques are the weights of the objects and their distance from the pivot.

As the rod is 40 cm, the pivot is at 20 cm. Also, the torques must sum up to 0, so:

0 = rF1 - rF2

r1F1 = r2F2

0.20(50) = r2(200)

Solve:

10 = r2(200)

r2 = 0.05 m = 5 cm

The 200N weight must be put at a distance of 5 cm from the OTHER SIDE of the pivot in order to balance the rod.

how have astronomers interpreted the unexpectedly fast rotation of galaxies

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Answer:

There must be a lot of dark matter that can be felt but not seen

The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.

What is a galaxy?

Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.

A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.

The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.

Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.

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What’s the mass of the continent?

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Answer:

Assume that the north american continent can be respanterd by a slab of rock 4000 km on a side and 30 km deep and that the rock

iam help for youthankscommentflowing

A particular roller coaster has a mass of 3500 kg, a height of 4.0, and a velocity of 12m/s. What is the potential energy? If needed, use g=10.m/s^2

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[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Gravitational Potential Energy of an object is calculated by formula ~

[tex] \large\boxed{\sf P = mgh}[/tex]

where,

m = mass of the object = 3500 kg

g = Acceleration due to gravity = 12 m/s²

h = height attained by the object = 4 m

Now, let's calculate its potential energy ~

[tex]3500 \times 10 \times 4[/tex]

[tex]140000 \: \: joules[/tex]

[tex]140 \: \: kj[/tex]

Answer:

Potential Energy of an object is calculated by formula:

Potential Energy (P.E)=m×g×h

Where,

m=mass of bodyg=acceleration due to gravityh=height from the earth surface

Now, let's solve the question.

Given,

mass(m)=3500 kgheight (h)=4mvelocity (v)=10m/s²

Now,

We know that,

Potential Energy (P.E)=mgh

[tex] = 3500 \times 10 \times 4[/tex]

[tex] = 3500 0\times 4[/tex]

[tex] =140000 joules [/tex]

[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]

Compare the amplitude of these waves.

help?

Answers

Answer:

The greater the amplitude of a wave then the more energy it is carrying.

Explanation:

E is carrying the most energy which means that it has the highest amplitude

A has the least energy because it has the lowest amplitude

please help me solve all of them ( a, b, c and d ) thankiew !!
I’m also kind of in a rush

Answers

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

What is an electron cloud 5th grade short answer.

Answers

Answer:

the system of electrons surrounding the nucleus of an atom

If you double the pressure of a constant amount of gas at a constant temperature, what happens to the volume

Answers

Answer: if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.

Explanation:

Alejandro Kirk is the catcher for the Blue Jays’ baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m.
Determine is the acceleration of the ball.
Determine the force applied by Alejandro.

Answers

The kinematics and Newton's second law allow to find the results for the questions about the motion of the ball are:

The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N

Given parameters

Mass of the ball m = 0.145 kg Ball starts speed vo = 38.2 m / s Setback distance x = 0.135 m

To find

Acceleration The force

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

         v² = v₀² - 2 ax

Wher vo es la initial veloicity, a the acceleration y x is the distance,

When the ball stops the velocity is zero.

         0 = v₀² - 2ax

         a = [tex]\frac{v_o^2}{2x}[/tex]  

Let's calculate

         a = [tex]\frac{38.2^2}{2 \ 0.135}[/tex]  

         a = 5.4 10³ m / s²

Newton's second law establishes a relationship between force, mass, and acceleration of the body.

             F = ma

             F = 0.145 5.4 10³

             F = 7.84 10² N

In conclusion using kinematics and Newton's second law we can find the results for the questions about the motion of the ball are:

The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N

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A car traveling at 35.6m/s crashes into a concert barrier and comes to a stop in 0.35 seconds. Calculate the average force applied to the 75kg driver.A 3.2kg steel ball traveling at 4.1m/s strikes a second ball of a mass 2.3kg Initially at rest. Calculate the velocity of the second ball when the first one continues traveling in the same direction with a speed of 1.5m/s2 balls of putty are shot towards one another. Ball 1 has a mass of 4.3kg and is moving at 18.6m/s . Ball 2 has a mass of 5.8kg and is moving at 9.5m/s. They collide and stick together. Calculate their final combine velocity.I really appreciate those attempting the problems. I do know the answers but I’m unaware of the steps to get there. Please include all formulas in your response and steps so I can learn and understand.Check your answer:7629N3.6m/s2.46m/sThank you all!

Answers

The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.

We have to find the acceleration using;

v = u - at

v = final velocity = 0 m/s

u = initial velocity = 35.6m/s

a = acceleration = ?

t = time = 0.35 s

u = at

a = u/t = 35.6m/s / 0.35 s

a = 101.7 ms-2

The force on the driver =  75kg ×  101.7 ms-2 = 7629 N

Using the principle of conservation of momentum;

Momentum before collision = momentum after collision

m1u1 +m2u2 = m1v1 + m2v2

Hence

(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2

13.12 = 4.8 + 2.3v2

13.12 - 4.8 = 2.3v2

v2 = 13.12 - 4.8/2.3

v2 = 3.6 m/s

Using the principle of conservation of linear momentum;

m1u1 + m2u2 = m1v1 + m2v2

(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v

v = 79.98 + 55.1/10.1

v = 13.37 m/s

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which wave has a higher frequency and why?

Answers

Explanation:

the figure in the left side has higher frequency.

because it has more nos. of wave in 1sec.

I need help in question 7, a and b.

Answers

Answer:

The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)

Explanation:

PLS HELP ASAP it's just 2 questions and they are multiple choice!

Answers

It’s the first option heat factories (A)

Virtual image formed by concave mirror

Answers

Answer:

yes

Explanation:

because it is a diverging mirror

The _____________ experiences a temperature increase primarily due to the absorption of gamma rays and X-rays.

Answers

Answer:

The thermosphere experiences a temperature increase primarily due to the absorption of gamma-rays and X-rays.

a bus increase its velocity from 30m/s to 45m/s and covers 300m distance. Find the time​

Answers

Answer:

3.5 m/s²

Hope this helps, have a nice day/night! :D

Which of the following is NOT true of a hotspot?

Answers

Given what we know about hotspots and their characteristics, we can confirm that the option that is false regarding hotspots is "A. Hotspots are constantly moving forming chains of volcanoes".

In regards to the study of geology, we refer to hotspots as spots in an oceanic mantle where magma located below causes them to become intensely hot. This heat forces magma and therefore the mantle itself upwards as a tectonic plate moves across the spot, and therefore results in the formation of volcanic chains. Some very popular examples of volcanoes and volcanic islands formed by this method include:

Yellowstone National ParkHawaiiThe country of Iceland

Despite the other options being true, option A is false given that hotspots are stationary elements of the oceanic mantle. Although some scientists believe that there is evidence of the movement of hotspots, this movement is slow enough to not be taken into account.

This question was answered in regards to the complete question found online which states:

Which of the following is NOT TRUE about hotspots? a. Hotspots are constantly moving forming chains of volcanoes.

b. Hotspots are stationary.

c. Hotspots form volcanic island arcs

d. Hotspots lie along an oceanic plate

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easy one - giving brainly if correct.​

Answers

Gas.

HOPE YOU GET 100!

a gas increases in pressure from 2.00 atm to 6.00 atm at a constant volume of 1.00 m3 and then expands at constant pressure to a volume of 3.00 m3 before returning to its initial state as shown in the figure below. how much work is done in one cycle?

Answers

The work done on the given gas in one cycle is -405.3 kJ.

The given parameters;

initial pressure of the gas, P₁ = 2 atmfinal pressure of the gas, P₂ = 6 atminitial volume of the gas, V₁ = 1 m³final volume of the gas, V₂ = 3 m³

Convert the pressure to Pascal (N/m²);

1 atm = 101325

The work done in one cycle is the area of the triangle and it is calculated as follows

[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]

the net work done on the gas = - 405.3 kJ

Thus, the work done on the given gas in one cycle is -405.3 kJ.

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a block weighing (Fg) 500 N is resting on a steel table ( us=0.74) the minimum force start this block moving is?

Answers

The minimum force required to start this block moving is 370 N.

The given parameters;

weight of the block, W = 500 Ncoefficient of static friction, [tex]\mu_s[/tex] = 0.74

The minimum force required to start this block moving is calculated as follows;

[tex]F= \mu_s F_n[/tex]

where;

[tex]F_n[/tex] is the normal force on the block which is equal to the weight of the block

[tex]F= \mu_s F_n \\\\F= \mu_s W\\\\F = 0.74 \times 500 \\\\F= 370 \ N[/tex]

Thus, the minimum force required to start this block moving is 370 N.

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The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?

Answers

The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The  variation  of the volume is: ΔV = 7.67 cm³

Pressure is defined by the relationship between force and area.

       P = F / A

The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.

          PV = nR T

Where P is pressure, V is volume, and T is temperature.

Let's write this equation for two points assuming that the temperature has not changed.

          P₀ V₀ = P₁ V₁

          V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex]                 (1)

The subscript "o" is used for the start point and the subscript "1" for the end point.

The pressure in a barometer is:

         P = ρ g y

They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.

 

The volume of the cylinder is

         V = π r² y

Let's calculate the initial volume.

         V₀ = π 1 9

         V₀ = 28.27 cm³

We substitute in equation 1.

         V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]  

         V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]  

Let's calculate.

        V₁ = [tex]\frac{75}{59} \ 27.27[/tex]  

        V₁ = 35.94 cm³

The volume to be incremented is

         ΔV = V₁ - V₀

         ΔV = 35.94 - 28.27

         ΔV = 7.67 cm³

Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The change of the volume is: ΔV = 7.67 cm³

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can someone sent some gravity questions

Answers

What do you mean ???

Answer:

can gravity from waves

Explanation:

yes gravity can froms way

Analyze the effects on a passenger riding in a car traveling at
50 km/h that collides head-on with a solid object.

Answers

Without:

The passenger will get seriously injuredColliding with the windshield at 50km/hCould result in death

With safety:

The seatbelt holds tge passengers body back so they dont collide with anythingTheir head is cushioned by the airbagThe passenger might get minor injuries but nothing serious will happen

What happens if you drop a chicken wing from the top of the Eiffel Tower

Answers

Answer:

I think it will brake or squashed

The wave height is equal to twice the _____.wave period wave height wave amplitude wave velocity wavelength

Answers

The wave height is equal to twice the amplitude of the wave.

The wave height of a wave of given wave with amplitude, period and wavelength is equal to twice the amplitude of the wave.

The amplitude of a wave is the maximum displacement of the wave, starting from the zero position of the wave. The wave height measures twice the maximum displacement of the wave.

Thus, we can conclude that the wave height is equal to twice the amplitude of the wave.

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Calculate torque using angular momentum

Answers

Answer:

The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.

Which of toby's answers is a correct description of what happens when a 1-kg cart traveling at 1 m/s collides inelastically with another 1-kg cart at rest?.

Answers

Answer:

hi I don't know sorry sorry forgive me

Explanation:

sorry

two objects are sitting 6m apart, one object has a mass of 100kg and the other has a mass of 200kg. what is the gravitational attraction between them

Answers

F=G(m1m2/r^2)

F= 6.67x10^-11(100 x 200/ 36)

F= 3.7 x 10^-8

what is the molecular formula for the hydrocarbon in the mass spectrum above?

Answers

What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following $1 \mathrm{H}$ NMR spectrum? 7.25 $\delta(5 \mathrm{H}, \text { broad singlet); } 2.90 \delta(1 \mathrm{H}, \text { septet, } J=7 \mathrm{Hz}) ; 1.22 \delta(6 \mathrm{H},\text { doublet, }$ $J=7 \mathrm{Hz})$

What symbols are these?

Answers

Answer:

the bottom one is wollsiegel

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