What would be the weight of an astronaut of the gravitational attraction has been reduced by 1/10 of earth's gravitational attraction?

Answer:

[tex]\textsf{a)} \quad x=\dfrac{25(7-\sqrt{13})}{3}=28.2870727\; \sf mm[/tex]

[tex]\textsf{b)} \quad V=379037.8082\; \sf mm^3[/tex]

Explanation:

[tex]\boxed{\begin{minipage}{5 cm}\underline{Volume of a rectangular prism}\\\\$V=w\:l\:h$\\\\where:\\ \phantom{ww}$\bullet$ $w$ is the width of the base. \\ \phantom{ww}$\bullet$ $l$ is the length of the base. \\ \phantom{ww}$\bullet$ $h$ is the height.\\\end{minipage}}[/tex]

Given:

The dimensions of the rectangular prism in terms of x are:

Substitute these values into the formula for volume to create an equation in terms of x:

[tex]\begin{aligned}\implies V&=(150-2x)(200-2x)x\\&=(30000-700x+4x^2)x\\&=4x^3-700x^2+30000x\end{aligned}[/tex]

To find the value of x that will give the maximum volume, differentiate the equation for volume.

[tex]\begin{aligned}\implies \dfrac{\text{d}V}{\text{d}x}&=3 \cdot 4x^{3-1}-2 \cdot 700x^{2-1}+30000x^{1-1}\\&=12x^2-1400x+30000\end{aligned}[/tex]

Set the derivative to zero and solve for x using the quadratic formula:

[tex]\implies x=\dfrac{-(-1400) \pm \sqrt{(-1400)^2-4(12)(30000)}}{2(12)}[/tex]

[tex]\implies x=\dfrac{1400 \pm \sqrt{520000}}{24}[/tex]

[tex]\implies x=\dfrac{1400 \pm \sqrt{40000 \cdot 13}}{24}[/tex]

[tex]\implies x=\dfrac{1400 \pm \sqrt{40000}\sqrt{13}}{24}[/tex]

[tex]\implies x=\dfrac{1400 \pm 200\sqrt{13}}{24}[/tex]

[tex]\implies x=\dfrac{175\pm 25\sqrt{13}}{3}[/tex]

[tex]\implies x=\dfrac{25(7\pm\sqrt{13})}{3}[/tex]

To determine which value of x will give the maximum volume, differentiate again:

[tex]\begin{aligned}\implies \dfrac{\text{d}^2V}{\text{d}x^2}&=2 \cdot12x^{2-1}-1400x^{1-1}+0\\&=24x-1400\end{aligned}[/tex]

Substitute both values of x into the second derivative:

[tex]x=\dfrac{25(7+\sqrt{13})}{3} \implies \dfrac{\text{d}^2V}{\text{d}x^2}=721.1102551 > 0\implies \sf minimum[/tex]

[tex]x=\dfrac{25(7-\sqrt{13})}{3} \implies \dfrac{\text{d}^2V}{\text{d}x^2}=-721.1102551 < 0\implies \sf maximum[/tex]

Therefore, the value of x which will give the maximum volume is:

[tex]x=\dfrac{25(7-\sqrt{13})}{3}=28.2870727\; \sf mm[/tex]

To find the maximum volume of the box, substitute the found value of x into the equation for volume:

[tex]\begin{aligned}\implies V_{\sf max}&=4\left(\dfrac{25(7-\sqrt{13})}{3}\right)^3-700 \left(\dfrac{25(7-\sqrt{13})}{3}\right)^2+30000 \left (\dfrac{25(7-\sqrt{13})}{3}\right)\\\\&=379037.8082\; \sf mm^3\end{aligned}[/tex]

The dimensions of the box with maximum possible volume will be:

More force is exerted by the table against her finger than she exerts against the surface.

Active power is the force behind motion or change that is exerted or brought to bear. the elements of nature

The term "force" has a clear definition in science. It is quite acceptable to refer to a force of this level as a push or a pull. An object does not "have in it" or "contain" a force. One thing is subject to a force from another. There is no distinction between living and non-living things in the concept of a force.

The newton  is the SI unit of force. As they relate to force, the base units are: The symbol for the length measurement known as the meter is m. The symbol for the kilogram, a unit of mass the sign s represents the second, a unit of time.

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Mv is equal to m - 2.5 log(d/10)2. There is a minus sign in the expression because stars distant than 10 pc have Mv that is more negative than m. Make careful you provide the star's given distance in parsecs (1 pc = 3.26 ly = 206265 AU) if you use this formula.

The moon can appear to be nearly -13 magnitude bright, the moon can appear to be -6 magnitude bright, and stars can get as bright as nearly -1.5 magnitude.

The star's actual magnitude is determined by subtracting its apparent magnitude by five times the trigonometric parallax of its distance from the star.

First-magnitude stars have apparent magnitudes less than +1.50, making them the brightest stars in the night sky. In the first century B.C., Hipparchus invented the magnitude scale. He assigned first magnitude to the 20 brightest stars and sixth magnitude to the stars that are faint enough to be seen with the unaided eye.

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The stick figure that is walking at a velocity of 1 m / s to the west is moving quickly relative to the ground than the stick figure that is walking at a velocity of 1 m / s to the east.

Vsg = Vsw + Vwg

Vsg = Relative velocity of stick figure wrt ground

Vsw = Relative velocity of stick figure wrt walkway

Vwg = Relative velocity of walkway wrt ground

Let the stick figure that is walking to the west be A and the stick figure that is walking to the east be B.

For A,

Vsw = - 1 m / s

Vwg = - 0.5 m / s

Vsg = - 1 - 0.5

Vsg = - 1.5 m / s

For B,

Vsw = 1 m / s

Vwg = - 0.5 m / s

Vsg = 1 - 0.5

Vsg = 0.5 m / s

The negative symbol indicates direction. So Vsg for A > Vsg for B

Therefore, the stick figure that is walking at a velocity of 1 m / s to the west is moving quickly relative to the ground than the stick figure that is walking at a velocity of 1 m / s to the east.

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The total resistance of the circuit is 25 ohms.

In the same way, if three or more resistors of the same value are connected in parallel, the equivalent resistance will be equal to R/n, where R is the resistor's value and n is the total number of resistors in the arrangement.

The total current flowing from the source is the sum of the currents through each of the paths. Using the following formula, you can get the overall resistance in a parallel circuit: 1/Rt = 1/R1 + 1/R2 + 1/R3 +... Even if one of the parallel channels breaks, current will still flow.

[tex]R_{total}[/tex] = 100/4 = 25ohms

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The two measurements, which are needed in order to calculate the speed of an object in motion, be: amount of distance travelled at a particular time interval and the time interval.

Speed is characterised as the rate at which an object's position changes in any direction.

The distance travelled in relation to the time it took to travel that distance is how speed is defined. Since speed simply has a direction and no magnitude, it is a scalar quantity. SI unit of speed be meter/second.

The equation provides the speed formula: s = d/t

Where,

the speed in m.s^-1 is s.

The distance in m is represented by d.

The time interval in s is t.

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The initial speed of the projectile launched at an angle of 30 above the horizontal from the top of a 10m high cliff is 34.2 m / s

According to equations of motion,

s = u t + 1 / 2 at²

s = Displacement

u = Initial velocity

t = Time

a = Acceleration

H = 10 m

t = 4 s

θ = 30°

g = 9.8 m / s²

Here the object is launched art an angle from a height. So,

s = - H = - 10 m

u = uy

While considering vertical motion,

a = - g = - 9.8 m / s²

- H = uy t - 1 / 2 g t²

- 10 = 4 uy - 1 / 2 * 9.8 * 4²

4 uy = 68.4

uy = 17.1 m / s

sin θ = uy / u

u = 17.1 / sin 30°

u = 34.2 m / s

Therefore, the initial speed of the projectile is 34.2 m / s

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The floor normally presses down on the carton with a force of 1049.67N.

The box normally applies 667.08 N of force to the person.

A perpendicular force that acts on two surfaces that are touching. It stands in for the pressure pressing the two surfaces together. With the size of the usual response force, the value of limiting friction grows. If the weight is the only vertical force acting on an object that is lying or moving on a horizontal surface, While acting in the opposite direction from the weight, the normal response force is of identical size. Raising the weight results in increased friction.

Given,

The crate's mass, mp=68kg, is maintained on a horizontal floor.

mc=39kg is the mass of the guy standing above the crate.

(a) determine the normal force the floor applies to the crate:

N =(mc+mp)g

N =(39kg+68kg)×9.81m/s²

N = 1049.67kg⋅m/s² [tex]\frac{1N}{1Kg.m/s^{2} }[/tex]

N = 1049.67N

∴ The floor normally presses down on the carton with a force of 1049.67N.

(b) determine the typical force the crate would apply to the person:

N = mp g

N = 68kg×9.81m/s²

N = 667.08m/s² [tex]\frac{1N}{1Kg.m/s^{2} }[/tex]

N = 667.08 N

∴ The box normally exerts 667.08 N of force on the person.

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Yes, a spectrophotometer can be used to measure the absorption of a particular wavelength of light that is not visible to the human eye.

A spectrophotometer is an instrument used to measure the intensity of electromagnetic radiation at different wavelengths.

Spectrophotometry is a standard biological technique used to measure light absorption or the amount of chemicals in a solution. This technique uses a light beam which passes through the sample, and each compound in the solution absorbs or transmits light over a certain wavelength.

Spectrophotometry is the quantitative analysis of electromagnetic spectra by use of a spectrophotometer; especially in order to determine the structure or quantity of a substance.

Using the spectrophotometer, absorption of light can be measured, even if we cannot observe it unaided.

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The wool becomes positively charged because it gives up electrons to the rubber rod whereas wool is rubbed against a Nylon strip so the wool will steal electrons from the Nylon and form a negative charge.

When the woolen cloth is stroke against the plastic, both wood and plastic get electrified due to friction built while rubbing. So, plastic has greater electron affinity (ability to attract electrons) in a contrast to wool, therefore, plastic acquires a negative charge and wool obtains a positive charge. When you rub a balloon with a cloth made of wool, the balloon captures electrons from the wool, leaving the balloon with a negative charge and the wool with a positive charge. If both balloons are pat with wool, they will both have the same charge (negative) and therefore will repel each other.

So we can conclude that Wool is a conductive material, which means it readily gives away its electrons. accordingly, when you rub a balloon on wool.

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The average acceleration of the dragster during this time interval is 44.57 m/s² and the average net force on the dragster during this time is 38915N.

The initial speed U of the dragster is zero while the final speed V of the dragster is 26.3m/s. The director runs for a total of time t of 0.59 seconds.

According to the equation of motion,

V = U+at

Putting values,

26.3 = 0.59a

a = 44.57 m/s².

So, the acceleration of the dragster is 44.57 m/s².

The force on the director is given by,

F = ma

m mass of the dragster and a is the acceleration of the dragster.

F = 873×44.57

F = 38915N.

So, the net average force on the dragster during this time is 38915 Newton.

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The average acceleration vector of the ball is - 9.78 m/s².

When you throw a ball upward with an initial speed, it will go up until reaching its highest peak then it will go down. So the ball doing non-uniform motion. The formula for non-uniform motion including initial speed, and final speed with time is

v = u+at

v-u = at

[tex]a = \frac{v - u}{t}[/tex]

[tex]a = \frac{- 4.5 - 4.5}{0.92}[/tex]

[tex]a = \frac{- 9.0}{0.92}[/tex]

a = 9.78 m/s²

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The greater amount of energy is transferred during the first case and the difference in energies is: 84 × 10³ J

Internal energy refers to all energy within a particular system, including the kinetic energy of molecules and the energy stored in chemical bonds between molecules. The interaction of heat, work, and internal energy involves the transfer and conversion of energy whenever a change is made to the system.

The amount of heat (Q) gained or lost from the sample can be calculated using the formula Q = mcΔT. where m is the mass, c is the specific heat, and ΔT is the temperature change.

For the first case:

Q = mcΔT

Q = mc(T₁ - T₂)

m = 1 kg or 1000 g

c =  4.2 J g⁻¹/°C

T₁ = 10 °C

T₂ = -15 °C

Q₁ = 1000 × 4.2 × (10 -(-15))

Q₁ = 105 × 10³ J

For the second case:

Q = mcΔT

Q = mc(T₂ - T₁)

m = 1 kg or 1000 g

c =  4.2 J g⁻¹/°C

T₁ = 60 °C

T₂ = 65 °C

Q₂ = 1000 × 4.2 × (65 - 60)

Q₂ = 21 × 10³ J

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Explanation:

Major activity injuries include:

Ansver:

All of the above

The heated (brown) air is separated from the cold (blue) air bubble on the left. A(n) is the term used to describe

"Fronts" are the borders that are formed when air masses converge. Cold front in three dimensions. Temperature changes caused by a front's motion can be used to identify it. A cold front causes a warmer air mass to be replaced by a colder one. Another name for it is a "dryline" or "dry front." a line or zone between two air masses that have different densities and, as a result, are typically at different temperatures. A moving front is given a name based on the air mass that is advancing; for example, a cold front if colder air is going forward. The front is the line separating two air masses. A front typically has overcast, stormy weather. Four separate fronts exist: the Cold, Warm, Stationary, and Occluded fronts. A cold front arises when two cold air masses collide with warm air masses.

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Answer:

Explanation:

Given:

V₀ = 1 000 km/h = 1  000 000 m / 3 600 s ≈ 280 m/s

V = 0 m/s - the Blue Flame car stopped

D = 2.0 km = 2 000 m  - distance

________________________________

a - ?

D = (V₀² - V²) / (2·a)

Acceleration:

a = (V₀² - V²) / (2·D)

a = (280² - 0²) / (2·2000) ≈ 20 m/s²

Other than 120 and 240 volts, what is another common voltage used inside a building is 208,277and 480volts.

When the same signal is applied to the IN(+) and IN(-) terminals of an op-amp, it must operate satisfactorily within a range of input voltages known as the common-mode input voltage (CMVIN), which is described in the datasheet of the device. You may think of it as never applying common-mode signals to an op-amp.

The voltage shared by an electrical device's two input terminals is known as a common-mode signal. The common-mode signal on a transmission line in telecommunications is also referred to as longitudinal voltage. The signal is typically transmitted between two conductors using a differential voltage in electrical circuits.

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Answer:

Explanation:

Refraction of a light ray may be defined as:

D. changing direction while crossing a boundary ​

To find the collision rate, known the magnitude.

What is magnitude?

Magnitude typically refers to size or distance. By comparing the size and motion speed of the thing, we can relate magnitude to movement. The object's or quantity's size determines its magnitude.

What is mean by collision rate?

The frequency of collisions between a single particle of radius m and particles with masses ranging from (m1, m1 + m1) to (m, m1) is proportional to their concentration, n(m1) m1, where (m, m1) is the rate constant for collisions between the two particles of masses m and m1 and depends on the process. 

u=140kmph

t=7.oms

v=0

E=M(V-U)/t

⇒m(0-140)/0.007×5/18m/s

⇒5555.55 (m) N

E=(5555.55) (mass) N

Therefore, the collision lasts is 5555.55(m) N.

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The law that explains after energy conversions, the total amount of energy has the same amount as the original amount of energy is the Law of conservation of energy.

The state law of conservation of energy is that energy cannot be created, the energy cannot be destroyed, and the energy can be converted from one form of energy to another form of energy. Examples like mechanical energy. Mechanical energy is the sum of potential energy and kinetic energy.

If an apple fell from a branch of a tree

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Work done by the system is calculated to  - 456.863 J, when pressure and volume change is given.

For a gas, work is the product of the pressure and the volume during a volume change. Relationship between pressure and volume can be found out by Boyle's Law. As the pressure on a gas increases then the volume of the gas decreases because the gas particles are forced closer together.

W= -p ΔV

p is pressure and ΔV is volume change

Given that volume changes from 0.160 liters to 0.570 liters

ΔV = 0.570 - 0.160

= 0.41 L

Given p = 11 atm

Now, W = -11 * 0.41

As 1 atm L = 101.3 J

Hence W=  -11 * 0.41 * 101.3

Work = - 456.863 J (work done by system)

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The second glider will stop after it travels the distance of 8 times the distance of the first glider.

Let subscript 1 and 2 represent parameters related to glider 1 and glider 2.

According to the Newton's law:

F = m . a

Where:

F = force

m = mass

a = acceleration

Since both gliders received the same forces,

F₁ = F₂

m₁ . a₁ = m₂ . a₂

m₁ . a₁ = 1/2m₁ . a₂

a₂ = 2a₁

Hence, the acceleration of the 2nd glider is twice of that of the 1st glider.

Velocity formula for decelerated move

v² = u² - 2.a.s

Where:

v = final velocity

u = initial velocity

s = distance.

Since the gliders stop, v = 0

Hence,

u₁² = 2.a₁.s₁

u₂² = 2.a₂ . s₂

Substitute u₂ = 4u₁,  a₂ = 2a₁

(4u₁)² = 2.(2a₁) . s₂

16 u₁² = 4 a₁ . s₂

16. (2.a₁.s₁) = 4 a₁ . s₂

s₂ = 8 s₁

Hence, the distance of glider 2 is 8 times the distance of glider 1

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The speed of this child from this surface would be 14.69 m / s

We have the formula for energy conservation as:

E - potential = E - kinetic

We have the following details with which we are to solve the problem

weight of child w = 350 N

length of rope L = 7.2m

angle θ = 12 degrees

we would have the solution as

m x 9.8m/s² x (10.2 + 1.79 - 7.2 x cos 12) = 1/2 mv²

we would have to solve out the expression that we have above. Such that :

9.8m * (11.99 - 0.9781) = 1/2 mv²

107.9 m = 1/2 mv²

cancel out m

107.9 = 0.5v²

divide through by 0.5

107.9 / 0.5 = v²

215.8 = v²

v = √215.8

v = 14.69

Hence the speed of the child at the surface of the water would be given as 14.69 m / s

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Answer:

Radius 20 m.

Explanation:

Given:

aₙ = 1.4 m/s²

V = 5.3 m/s

___________

R - ?

Centripetal acceleration:

aₙ = V² / R

Radius:

R = V² / aₙ

R = 5.3² / 1.4 ≈ 20 m

The magnitude of its acceleration is a= g ([tex]W_{B}[/tex])/( 2[tex]W_{A}[/tex] +[tex]W_{B}[/tex] )

The length of the vector is all there is to an acceleration's magnitude. To put it another way, writing the magnitude of acceleration indicates how quickly velocity varies. In the standard international (SI) system, the acceleration is measured in units of meters per second square.

Let T be the rope's tension, assuming it has no mass.

T =( 2[tex]W_{A}[/tex])a/g ------- 1

([tex]W_{B}[/tex]) - T = ([tex]W_{B}\\[/tex])a/g ------- 2

Add 1 and 2

[tex]W_{B}[/tex] =(a/g)( 2[tex]W_{A}[/tex]+ [tex]W_{B}[/tex])

a= g ([tex]W_{B}[/tex])/( 2[tex]W_{A}[/tex] +[tex]W_{B}[/tex] ).

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Answer:

See below

Explanation:

1800 j/(kg C)   =     x J  / (.20 kg * 50 C)

solve for x J = 180 000 J  

The amount of energy needed to raise the temperature of 0.20 kg of cooking oil by 50°C is 18000 J.

The amount of energy (Q) needed to raise the temperature of a substance can be calculated using the formula:

Q = mcΔT

where Q is the amount of energy, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

In this case, the mass of the cooking oil is 0.20 kg, the specific heat capacity of the oil is 1800 J/(kg·°C), and the temperature needs to be raised by 50°C. Substituting these values into the formula, we get:

[tex]Q = (0.20 kg) x (1800 J/(kg·°C)) x (50°C)[/tex]

Q = 18000 J

Therefore, the amount of energy needed to raise the temperature of 0.20 kg of cooking oil by 50°C is 18000 J.

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The maximum height travelled by the ball before returning to the ground is 8.62 m.

The maximum height travelled by the ball before returning to the ground is calculated by applying the following kinematic equation as shown below.

H = (u²sin²θ) / 2g

where;

H = (26² (sin 30)²) / (2 x 9.8)

H = 8.62 m

Thus, the maximum height travelled by a projectile depends on the initial velocity and angle of projection.

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