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

Answer:

Answer:

the system of electrons surrounding the nucleus of an atom

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:

Answer:

can gravity from waves

Explanation:

yes gravity can froms way

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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Answer:

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

Explanation:

Answer:

the bottom one is wollsiegel

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.

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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#SPJ12

Answer:

3.5 m/s²

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

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

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

The given parameters;

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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Answer:

yes

Explanation:

because it is a diverging mirror

Without:

With safety:

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

Given parameters

To find

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:

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Answer:

hi I don't know sorry sorry forgive me

Explanation:

sorry

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

The given parameters;

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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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:

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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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:

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:

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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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Explanation:

the figure in the left side has higher frequency.

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

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

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Gravitational Potential Energy of an object is calculated by formula ~

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

where,

Now, let's calculate its potential energy ~

Answer:

Potential Energy of an object is calculated by formula:

Where,

Now, let's solve the question.

Given,

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]

Answer:

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

Answer:

I think it will brake or squashed

Answer:

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

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.

Gas.

HOPE YOU GET 100!