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

Answer:

441 Newtons equals 44.97 kilograms.

The maximum displacement angle of the bob is 13⁰.

The given parameters;

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]

The maximum displacement angle is calculated as follows;

[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]

Thus, the maximum displacement angle of the bob is 13⁰.

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Answer:

Atmospheric pressure is caused by the gravitational attraction of the planet on the atmospheric gases above the surface and is a function of the mass of the planet, the radius of the surface, and the amount and composition of the gases and their vertical distribution in the atmosphere

Explanation:

Answer:

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{42}{1.8} = 23.3333... \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

[tex]\boxed {\boxed {\sf C. \ 23.33 \ kg}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F=ma[/tex]

The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.

Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².

Substitute the values into the formula.

[tex]42 \ kg*m/s^1 = m * 1.8 \ m/s^2[/tex]

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}[/tex]

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m[/tex]

The units of meters per second squared cancel.

[tex]\frac {42 \ kg}{1.8 } =m[/tex]

[tex]23.3333333 \ kg=m[/tex]

Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.

[tex]23.33 \ kg \approx m[/tex]

The mass of the ball is approximately 23.33 kilograms.

Answer:

3.5 m/s²

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

Answer:

74.4 mph

Explanation:

Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.

speed = mi/hr

speed = 93/1.25

speed = 74.4 mph

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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Answer:

The second deal.

Explanation:

I had that before, if u pick the second one you will get twice as much in a year as the first one.

Answer:

?

Explanation:

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

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:

5m/s ez

Explanation:

plz make me brainliest

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

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Answer:

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

The distance in which the spring is compressed is 1.065 m.

The given parameters;

Apply the principle of conservation of energy to determine the distance in which the spring is compressed.

[tex]K.E = W + U\\\\\frac{1}{2} mv^2 = \mu Fx \ + \ \frac{1}{2} kx^2\\\\\frac{1}{2}(4.2)(3.5)^2 = 0.25(4.2 \times 9.8)(x) \ + \frac{1}{2} (26)(x^2)\\\\25.725 = 10.29x \ + \ 13x^2\\\\13x^2 + 10.29x - 25.725 = 0\\\\[/tex]

solve the quadratic equation using formula method;

a = 13, b = 10.29,  c = -25.725

[tex]x = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(10.29) \ \ +/- \ \sqrt{(10.29)^2 \ - \ 4(13\times 25.725)} }{2(13)} \\\\x = 1.065 \ m, \ \ or \ -1.86 \ m[/tex]

choose positive distance, x = 1.065 m.

Thus, the distance in which the spring is compressed is 1.065 m.

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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

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]

To get the required work done, we will divide the mass by the speed of one knot to have:

[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

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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:

Gas.

HOPE YOU GET 100!

9) a) formula = h(ρ)g by default g=9.8m/s²

Pressure = 2×1000×9.8 = 19600Pa

b)

Levelized production cost is a big challenge in the production of wind energy.

The main challenge in the production of wind energy is the high production cost. It has to minimize the Levelized Production Cost (LPC) which is the ratio between energy production cost and economic lifetime. This problem can be solved by doing modification in the structure of the turbine as well as we must ensure minimal energy loss.

The modification in the structure can produce more electrical energy and reduce the energy loss also provides more profit to the company so we can conclude that levelized production cost is a big challenge in the production of wind energy.

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Answer:

can gravity from waves

Explanation:

yes gravity can froms way

Explanation:

the figure in the left side has higher frequency.

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

Answer:

90 Km/h

Explanation:

60 mins is 1 hr

so in an hour bus covers 60 Km

so new speed:

d/t

(60km/40mins)*60mins/h

90Km/h

Answer:

I think it will brake or squashed

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:

55.84 look at periodic table

so 55.84*3= 167.52gFe

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 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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