two people are in a boat that is capable of a max speed of 5 km per hour in still water, and wish to cross a river 1 km wide to a point directly across starting point. if the speed of the water in the river is 5 km per hour, how much time is required for the crossing

The kinetic coefficient of friction is 0.40

Kinetic friction, also known as sliding friction or gliding friction, is the slowing force between two objects moving relative to each other. This is important in understanding the dynamic friction coefficient. The dynamic coefficient of friction is the pulling force required to keep an object in motion divided by the pulling force on the two sliding surfaces. This factor is specific to the materials involved in the contact surfaces. The rougher the surface, the higher the coefficient.

Calculation:

given

mass =10kg

force(Ff)=40N

so

Ff=μ F=μmg

   40=μ*10*9.8

   μ=0.40

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Answer:

power

power is the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the SI system, the unit of power is the joule per second (J/s).

Since potential energy is inversely proportional to height times mass, an automobile that is twice as large as another will have twice as much potential energy as that vehicle for a given height.

A doubling in height will lead to a doubling in gravitational potential energy since an object's gravitational potential energy is directly proportionate to its height above the zero point. The gravitational potential energy will triple with a tripling of height.

The car's position at the top of the hill's potential energy equals kinetic energy is being created. where "g" is the gravitational constant and equals 9.8 m/s2.

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(a) Equation for the radius of the ball, r, treating it as a sphere, in terms of its surface area will be  [tex]r & =\sqrt{\frac{A}{4 \pi}}[/tex]

(b) The density of the ball is 4.9307 g/cm^3.

(c) The density of the ball is 4903.7 Kg/m^3 or 4904 Kg/m^3

Surface area of a spherical ball, A=45 cm^2

Part a

The surface area of a spherical ball is given as

[tex]$$\begin{aligned}A & =4 \pi r^2 \\\therefore r & =\sqrt{\frac{A}{4 \pi}}\end{aligned}$$[/tex]

So, this is the expression for the radius of the ball in terms of surface area.

Part b

Mass of the ball, M=140 grams

The density of the ball is given as [tex]$$\rho=\frac{M}{V}$$[/tex]

Where, V is the volume of the spherical ball.

And [tex]$V=\frac{4}{3} \pi r^3$[/tex]

So, the value of the density will be

[tex]$$\begin{gathered}\rho=\frac{M}{\frac{4}{3} \pi r^3} \\\rho=\frac{3 M}{4 \pi r^3}\end{gathered}$$[/tex]

Now, putting the value of "r" in the above expression,

We will get

[tex]$$\begin{aligned}& \rho=\frac{3 M}{4 \pi\left(\sqrt{\frac{A}{4 \pi}}\right)^3} \\& \therefore \rho=\frac{3 M}{4 \pi\left(\frac{A}{4 \pi}\right)^{\frac{3}{2}}}\end{aligned}$$[/tex]

Now putting the value of M and A in above,

We will get,

[tex]$$\begin{gathered}\therefore \rho=\frac{3 \times 140}{4 \pi\left(\frac{45}{4 \pi}\right)^{\frac{3}{2}}} \\\therefore \rho=\frac{420}{4 \times 3.14(3.583)^{\frac{3}{2}}} \\\therefore \rho=\frac{420}{4 \times 3.14 \times 6.782} \\\therefore \rho=\frac{420}{85.18} \mathrm{~g} / \mathrm{cm}^3 \\\therefore \rho=4.9307 \mathrm{~g} / \mathrm{cm}^3\end{gathered}$$[/tex]

Part c

The density in Kg/m^3 is to be calculated.

[tex]$1 \mathrm{~g} / \mathrm{cm}^3=1000 \mathrm{~kg} / \mathrm{m}^3$[/tex]

So,

[tex]$$\begin{gathered}\therefore 4.9307 \mathrm{~g} / \mathrm{cm}^3=1000 \times 4.9307 \mathrm{~kg} / \mathrm{m}^3 \\\therefore \rho=4903.7 \mathrm{~kg} / \mathrm{m}^3\end{gathered}$$[/tex]

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Joe throws his ball vertically upward with an initial speed twice as high as the Bill. The maximum height of Joe's ball is 4 times of the maximum height of Bill's ball.

The equation of motion for the upward vertical motion is:

v² = u² - 2.gh

Where

v = final velocity

u = initial velocity

h = height

g = acceleration due to gravity

At the maximum height, the final speed is zero, hence:

0 = u² - 2.gh

u² = 2.gh

Therefore, the maximum height is directly proportional to the square of initial speed.

In the problem, u_joe =  2 x u_bill

Hence,

(u_joe)² : (u_bill)² = h_joe : h_bill

(2xu_bill)² : (u_bill)² = h_joe : h_bill

h_joe : h_bill = 4 : 1

Thus, the maximum height of Joe's ball is 4 times of the maximum height of Bill's ball.

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When traveling at the specified speed and wavelength, a wave has a frequency of 13Hz.

The distance that the shapes of waves repeat is known as the wavelength. It is a periodic wave's spatial period.

Expression derived from the relationship between wavelength, frequency, and speed; = v /f

Where v is speed or velocity and f is frequency, is the wavelength.

Wavelength of wave λ = 1m

Speed of wave v =13m/s

Frequency f = ?

We substitute our values into above expression.

λ = v ÷ f

1m = 13m/s ÷ f

f = 13m/s ÷  1

f = 13Hz

As a result, the wave's frequency in hertz is 13Hz when it travels at the given speed and wavelength.

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Based on the circled elements in the periodic table’s location, we could infer that most of the circled elements are gases at room temperature. The correct answer is B.

There are 11 elements that are in gaseous state at room temperature. They are Hydrogen (H), Helium (He), Neon (N), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), Fluorine (F), Chlorine (Cl), Nitrogen (N) and Oxygen (O).

Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and also are found as diatomic molecules (H2, N2, O2, F2, Cl2).

Although part of your question is missing, you might be referring to this full question: Consider the circled elements in the periodic table as seen on attached image. Based on their location, we could infer that most of them are:

A) very reactive gases.

B) gases at room temperature.

C) do not react readily with metals.

D) nonreactive solids at room temperature.

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The amount of heat that the water absorbs in cooling system of the automobile is 4688320 Joules.

The water holding capacity of the cooling system of the automobile s 16L.

The density of water is 1000Kg/m³. The specific heat of water is 4.186  J/KgC°.

The temperature of the water rises to 87°C from 17°C.

Now, the amount of heat that the water absorbs is,

Q = mcT

Where,

Q is the heat absorbed,

m is the mass of the water,

c is the specific heat of water,

T is the change in temperature of water.

Putting values,

Q = 1000 x 16 x 4.186 x 70

Q = 4688320 Joules.

So, the amount of heat absorbed is 4688320 Joules.

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If the coil is then rotated through an angle of 30o in a time of 0.10 s. The average induced emf is 0.94 V.

Given :

A = 0.020 m²

N = 50

B = 0.18 T

t = 0.10 sec

Ф = 30° = ( π / 180 ) × 30 rad

Ф = 0.5236 rad

Now, angular velocity (ω) = Ф / t = 0.5236 / 0.10

ω = 5.236 rad/s

Average induced emf is given by

ε = N × A × B × ω

ε = 50 × 0.02 × 0.18 × 5.236

ε = 0.94 V

So, The average induced emf is 0.94 V.

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Every 557 years, the dwarf planet Eris revolves around the Sun. The average separation from the Sun is 67.7 a.u. (semimajor axis).

With an orbital period of 557 years, Eris is nearly at its closest point to the Sun, 96.6 AUs, as of 2011. Its perihelion occurred between 1698 and 1699, its aphelion occurred around 1977, and its perihelion will occur again between 2256 and 2258. At the aphelion position (approximately 152 million kilometers), the Sun-Earth distance is at its greatest, while at the perihelion position, it is at its shortest. The largest asteroid belt object between Mars and Jupiter is Ceres, the smallest of the five dwarf planets. However, Pluto is typically 39 AU, or 3.7 billion miles (5.9 billion kilometers), from the Sun. Pluto's closest approach to the Sun, or perihelion, occurred between 1979 and 1999.

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Answer:

0.59 seconds

Explanation:

To find the time it takes for a tennis ball to reach the ground and bounce back to a height of 7 meters, we need to use the formula for the time it takes an object to fall to the ground.

The formula for the time it takes an object to fall to the ground is given by:

t = sqrt(2 * d / g)

where t is the time it takes the object to fall, d is the distance the object falls, and g is the acceleration due to gravity.

In this case, we are given that the height from which the tennis ball is dropped is 3 meters, the height to which the ball bounces is 7 meters, and the acceleration due to gravity is 9.8 m/s^2. We can use these values to solve for the time it takes for the ball to reach the ground and bounce back to a height of 7 meters.

First, we need to calculate the distance the ball falls. The distance the ball falls is equal to the difference between the height from which the ball is dropped and the height to which the ball bounces:

d = H1 - H2

d = 3 - 7

d = -4

Next, we can use the formula for the time it takes an object to fall to the ground to solve for the time it takes for the ball to reach the ground and bounce back to a height of 7 meters:

t = sqrt(2 * d / g)

t = sqrt(2 * -4 / 9.8)

t = 0.59 seconds

Therefore, it takes 0.59 seconds for the tennis ball to reach the ground and bounce back to a height of 7 meters.

Answer:

It slows down.

Explanation:

When light travels from one medium to another, its speed changes according to the properties of the two media. In the case of light traveling from air to water, the speed of light will decrease. This is because water is denser than air, so the light waves encounter more resistance as they pass through the water. This causes the light to slow down and its path to bend, a phenomenon known as refraction.

Answer: It slows down.

Explanation:

they are said to be “denser” than air. what happens is that light slows down when it passes from the less dense air into the denser glass of water. this slowing down the ray of light also causes the ray of light to change directions. it is the change in the speed of the light that causes refraction.

The index of refraction of water (1.33) is greater than that of air (1.00).

With that said the index of refraction is related to the velocity of light by:

[tex]n=\frac{c}{v}[/tex]

where n is the index of refraction, c is the speed of light in a vacuum and v is the velocity of light in the medium.

As c is constant we can see that the speed of light in a medium is inversely proportional to the index of refraction

[tex]n = \frac{1}{v} \\[/tex]

therefore the medium with the lower index of refraction will have the greatest velocity of light (Air).

The new (resultant) vector could be written as 0 m, since the vectors cancel each other out. The vector going 5 m West is cancelled out by the vector going 5 m East, and the vector going 11 m West is cancelled out by the vector going 11 m East. This means that the final result is a vector with a magnitude of 0 and no direction.

The moment about AB using the position vector AC is 55.9 i + 342.3 j + 0 k N.m and the position vector from A to C, calculate the moment about segment AB due to force F is -95.4 i + 191.07j + 0 k N.m.

Given that:

x₁ = 1.4m

y₁ = 1.7m

z₁ = 1.5m

Applied force at point C = F = | - 165 i  + 100 j + 140 k | N

The moment about AB using the position vector AC :

Using the equation: M(AB) = u(AB) . [ r(AC) X F ]        (equation 1)

We have:

OA = 0

OB = 1.4 i + 1.7 j

OC = 1.4 i + 1.7 j + 1.5 z

Now,

vector AB = vector OB - vector OA

vector AB = 1.4 i + 1.7 j

and, u(AB) = vector AB | vector AB     (equation 2)

= |vector AB| =

= AB  

= √ ( 1.4² + 1.7² )    (inserting this in equation 2)

= u(AB)  = (1.4 i + 1.7 j ) / √ ( 1.4² + 1.7² )

= u(AB) = 0.636 i + 0.772 j

distance r(AC) =  vector OC - Vector OA

= r(AC) = 1.4 i + 1.7 j + 1.5 z

Now, inserting values of u(AB), r(AB) and F in equation 1,

= M(AB) = 0.636 [1.7 X 140 - 1.5 X 100] + 0.772[1.4 X 140 + 165 X 1.5] + 0

= M(AB) = 0.636 (238 - 150) + 0.772(196 + 247.5) + 0

= M(AB) = 55.9 i + 342.3 j + 0 k N.m

Now,

r(BC) = vector OC - vector OB

r(BC) = (1.4 i + 1.7 j + 1.5 z) - (1.4 i + 1.7 j )

r(BC) = 1.5 z

Now, inserting values of u(AB), r(BC) and F in equation 1,

= M(AB) = 0.636 [0 X 140 - 1.5 X 100] + 0.772[ 0 X 140 + 165 X 1.5] + 0

= M(AB) = 0.636 (0- 150) + 0.772(0+ 247.5) + 0

= M(AB) = -95.4 i + 191.07j + 0 k N.m

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The complete question:

Calculating the moment about AB using the position vector AC

Using the position vector from A to C, calculate the moment about segment AB due to force F .

Express the individual components to three significant figures, if necessary, separated by commas.

The dimensions are: x₁ = 1.4m, y₁ = 1.7m, z₁ = 1.5m and the force applied at point C = [-165 i + 100 j + 140 k] N.

From the given information, we get these harmonics as 8th and 9th harmonics.

The length of nth harmonic is given as 0.54 m.

This means, (n* 0.54)/2 = n*0.27

0.27n = l ----(1)

The length of (n+1)th harmonic is given as 0.48 m.

This means, (n+1)* 0.48/2 = (n+1)* 0.24

0.24(n+1) = l ----(2)

From (1) ÷ (2), we can write

n/(n + 1) * 9/8 = 1

n/(n + 1) = 8/9

9n = 8(n + 1)

9n - 8n = 8

n = 8

n + 1 = 9

Thus, these harmonics are said to be 8th and 9th harmonics.

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Answer:

The answer is: It cannot be zero.

Explanation:

Displacement is the change in position of an object. Displacement can be 0 because the object just wouldn't move and its initial position and final position remain the same.

If the receiver has a totally absorbing panel measuring 20.0 cm by 35.0 cm oriented with its plane perpendicular to the direction the waves travel, the average force these waves exert on the pane is 1.44X10-18N

Given that length of panel (l)  = 35cm = 0.35m

Breadth of panel (b) = 20cm = 0.2m

Force exerted on the plane (F) =Prad x area

where Prad = pressure of radiation wave and a = area of the pane

To find pressure we know that Prad = I/c where I = intensity of radiation and c = speed of light = 3x10^8m/s

Given that power of wave (P) = 25KW

Diameter of satellite (d) = 575km

I = Power/ 4πd^2 = 25x10^3/4x3.14x(575x10^3)^2 = 6.2x10-9

Prad = 6.2x10-9 /3x10^8 = 2.06 x 10-17

Force (F) = 2.06x10-17x 0.2x0.35=1.44x10-18N

Hence the force exerted on the pane by the waves  = 1.44x10-18N

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For complete question: A satellite 575 km above the earth's surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 kW. Find If the receiver has a totally absorbing panel measuring 15.0 cm by 40.0 cm oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

The pooled estimated value of p is 0.435

Given values

n1 = 100

n2 = 100

x1 = 42  and  x2 = 45

H0: p1 = p2  

Calculation:

p1 = [tex]\frac{x_1}{n_1}[/tex]

=> 42/100

=>0.42

so p1=0.42

p2= [tex]\frac{x_2}{n_2}[/tex]

=> 45/100

=>0.45

so shown that p1 ≠ p2 so the claim that p1 = p2 is false and proved above using two tail test.

p = [tex]\frac{x_1+x_2}{n_1+n_2}[/tex]

=>(42+45)/(100+100)

=>87/200

=>0.435

so p = 0.435

The pooled estimate p. is 0.435

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Complete question :

Assume that you plan to use a significance level of α = 0.05 to test the claim that wheather p1 = p2. Use the sample sizes mentioned  below  and numbers of successes to find the pooled estimate p. Round your answer to appropriate decimal.

Given values:

n1 = 100, n2 = 100 ,x1 = 42 ,x2 = 45

The pressure inside the tank would be 213.65 atm.

According to  Gay-Lussac's law, for a particular gas, the pressure of the gas is directly proportional to the temperature at a constant volume.

p ∝ T: where p is the gas's pressure and T is the gas's Kelvin temperature.

Given that:

initial temperature: T₁ = 23° C =273 + 23 K = 296 K.

initial pressure: p₁ = 204 atm.

final temperature: T₂ = 37° = 273 + 37 K = 310 K.

final pressure: p₂ = ?

Now: according to Gay-Lussac's law:

p₂/T₂ = p₁/T₁

p₂ = p₁ (T₂/T₁)

= 204 ( 310/296) atm.

= 213.65 atm.

Hence, the pressure inside the tank would be 213.65 atm.

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Each photon's momentum in the light beam is p=h=632. 8106. The first gas laser was the helium-neon laser.

Although the first laser transitions were in the near infrared, the red laser, with a wavelength of 632.8 nm, is now the most widely employed transition. This laser is offered in lengths ranging from around 10 cm to more than 100. For longer wavelength lines, helium-neon lasers with a 1- to 10-m W output at 543 nm (green), 595 nm (yellow), or 633 nm provide an option to krypton ion lasers (red). A helium-neon laser can produce light with a power output of 1 mW at a wavelength of 632.8 nm.

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The distance between the second maximum of laser 1 and the third minimum of laser 2, is 0.36 m.

x1 = nλL / d

n=second maxima of laser 1

  = 2 * (d/20) *5.40 /d

  = 0.54 m

X2= (n+1/2)λL/d

n= 2= third minima of laser 2

  = 2.5 *(d/15)* 5.40 /d

  = 0.9m

distance del(y)max-min between second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum =

0.9 -0.54 = 0.36 m

A laser is a device that produces light by optically amplifying electromagnetic energy through stimulated emission. The term "laser" stands for "light amplification by stimulated emission of radiation" (abbreviated "laser"). The first laser was created in 1960 by Theodore H. Maiman at Hughes Research Laboratories, based on theoretical research by Charles Hard Townes and Arthur Leonard Schawlow.

Because it emits coherent light, a laser is unique among other light sources. Laser cutting and lithography are made possible by spatial coherence, which makes it possible to focus a laser on a small area.

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The wavelength of light is 432 nm in a vacuum.

A plate is a particular type of dish. A dish is usually something that meal is eaten or served from. A plate has a tendency to refer in particular to a flat dish that is suitable for holding meals that don't have a high liquid content material. the opposite common sort of dish is a bowl.

There are 3 forms of plate tectonic limitations: divergent, convergent, and transform plate limitations. This picture suggests the three predominant sorts of plate obstacles divergent, convergent, and transform.

A plate is an extensive, especially flat vessel on which food may be served. A plate also can be used for ceremonial or decorative functions. maximum plates are round, but they may be in any form.

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The terrestrial planets have small size, high density, and are closest to the Sun because the extremely high temperatures associated with formation allowed only silicate minerals and metals to form solids.

The substance's mass per unit of volume is its density (also known as specific mass or volumetric mass density). The sym Density and mass concentration have the same numerical value for a pure substance. Density can be important to packaging, purity, and buoyancy because different materials typically have varied densities. At conventional settings for temperature and pressure, osmium and iridium are the densest elements currently known. The Greek letter rho, written in lower case, is the bol most frequently used to denote density, while the Latin letter D is also acceptable. According to mathematics, density is determined by dividing the mass by the volume: where is the density, m is the mass, and V is the volume. In some circumstances, such as in the oil and gas business in the United States, density is merely defined as something's weight.

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213.65 atm is the pressure inside the tank if left outside in the sun in the temperature is 37°C.

Gay-Lussac's law states that " the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas".

According to the law, we have,

[tex]\frac{p_2}{T_2} = \frac{p_1}{T_1}[/tex]

Where,

[tex]p_2[/tex] is final pressure

[tex]p_1[/tex] is initial pressure

[tex]T_2[/tex] is final temperature

[tex]T_1[/tex] is initial temperature

Initial temperature: T₁ = 23° C =273 + 23 K = 296 K.

Initial pressure: p₁ = 204 atm.

Final temperature: T₂ = 37°C = 273 + 37 K = 310 K.

Final pressure: p₂ = ?

Substituting the values we get,

[tex]\frac{p_2}{T_2} = \frac{p_1}{T_1}[/tex]

[tex]p_2[/tex] = [tex]\frac{(p_1)(T_2)}{T_1}[/tex]

[tex]p_2[/tex] = [tex]\frac{(204)(310)}{296}[/tex]

[tex]p_2[/tex] = [tex]\frac{63,240}{296}[/tex]

[tex]p_2[/tex] = 213.648 ≈ 213.65atm.

Therefore, the pressure inside the tank if left outside in the sun is 213.65atm.

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The temperature of a black body that would radiate most intensely at 560 nanometers (nm) would be approximately 5,379 Kelvin.

Wavelength is the distance between two consecutive peaks (or troughs) of a wave. It is most commonly associated with light and sound waves, but can also refer to electromagnetic radiation, like gamma rays, x-rays, and radio waves. Wavelength is usually measured in meters or nanometers. The shorter the wavelength, the higher the frequency of the wave, and the higher the energy of the wave.

This is because black bodies emit radiation with a peak intensity that is determined by their temperature. The hotter they are, the shorter the wavelength of the peak radiation. The equation used to calculate this is Wien's displacement law, which states that the wavelength of peak radiation is inversely proportional to the temperature of the black body.

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Nuclear fission is used by all nuclear power plants, and uranium atoms are typically used in nuclear power plants.In nuclear fission, an neutron splits an atom of uranium by slamming against it; two neutrons are created in the process.

A U-235 nucleus immediately splits in half after taking in an additional neutron.Fission is the term for this action (see diagram below).A split U-235 nucleus releases two to three neutrons each time it divides.As a result, starting a chain reaction is possible.

Continuous fission was achieved as more neutrons are successively emitted.Fission-generated neutrons have tremendous energy and move very swiftly.The procedure of moderation is used in most reactors to slow down these so-called rapid neutrons because they do not promote fission as effectively as slower-moving ones do.

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The collision is perfectly elastic.

A collision is considered elastic if it does not result in a net decrease in the system's kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved.

Kinetic Energy(K.E.)=[tex]\frac{1}{2} mv^{2}[/tex]

Let [tex]K.E_i[/tex] and [tex]K.E_f[/tex] be the initial and final kinetic energy.

From the figure, we have

[tex]K.E_i[/tex] = [tex]\frac{1}{2}[/tex]×4×[tex](1.8)^2[/tex] +  [tex]\frac{1}{2}[/tex]×6×[tex](0.2)^2[/tex]

[tex]K.E_i[/tex] = 6.48+0.12

[tex]K.E_i[/tex] =6.6

and

[tex]K.E_f[/tex] = [tex]\frac{1}{2}[/tex]×4×[tex](0.6)^2[/tex] +  [tex]\frac{1}{2}[/tex]×6×[tex](1.4)^2[/tex]

[tex]K.E_f[/tex] = 0.72+5.88

[tex]K.E_f[/tex] =6.6

Since the initial and final kinetic energy is equal there is perfectly elastic.

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Complete Question:

In the figure, determine the character of the collision. the masses of the blocks, and the velocities before and after are given. the collision is

The relationship between time of flight t₁ and t₂ is t₂² = 20t₁ +  4.9t₁² = t₂²

The time it takes for the two cannon balls to land or the time of flight is determined using the equation below:

h = ut + ¹/₂gt²

where;

For ball 1:

h = 20t₁ + ¹/₂ * 9.8 * t₁²

For ball 2:

h = 0 * (t₂) + ¹/₂* 9.8 * t₂²

Equating the two values of h:

20t₁ + ¹/₂ * 9.8 * t₁² = 0 * (t₂) + ¹/₂ * 9.8 * t₂²

20t₁ +  4.9t₁² = t₂²

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The accelerating expansion of the universe is thought to be driven by thermal energy.

Acceleration is the charge at which velocity modifications with time, in terms of each pace and path. A factor or an object moving in an instant line is extended if it quickens or slows down. Movement on a circle is expanded even though the speed is steady because the path is always converting.

Mainly, Uniform acceleration, non-uniform acceleration, and average acceleration are the three styles of elevated motions.

Acceleration (a) is the alternate in speed (Δv) over the trade in time (Δt), represented through the equation a = Δv/Δt. This lets you measure how rapidly the pace changes in meters in step with second squared (m/s^2).

Learn more about Acceleration here:-brainly.com/question/605631

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