Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 1.0 meters above the ground and is trying to spray Ferdinand,who is standing 10.0 meters away. I know so far that she cannotspray Ferdinand at the current position and with the curreentspeed of spray. I got stuck inthe following question:To increase the range of the water, Isabellaplaces her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the cross-sectional area of the hose hole does shehave to cover to be able to spray her friend?

Assume that the cross section of the hoseopening is circular with a radius of 1.5 centimeters.

Answers

Answer 1

Answer:

Answer:

Explanation:

According to the formula below, with constant flow rate, the less cross-sectional area there is, the faster water would flow, and vice-versa

$\dot{V} = A*v$

where $\dot{V} m^3/s$ is the constant flow rate,

A m2 is the cross-sectional area

v m/s is the water speed.

So if the flow rate is constant, when A decreases, v must increase proportionally.

Since this problem is missing the water speed, here are the steps to solve it

Step 1: find the new spray speed that could reach Ferdinand

Step 2: find the ratio of this new spray speed to the old one, this will also be the ratio of the old cross-sectional area to the new one.

Step 3: find the fraction f of the cross-sectional area of the hose hole

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Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?

Answers

Answer:

The electric force increases by a factor of 4.

Explanation:

The electric force between two charges $q_1$ and $q_2$ separated a distance d can be calculated using Coulomb's Law:

$F=(kq_1q_2)/(d^2)$

where $k=9*10^9Nm^2/C^2$ is the Coulomb constant.

If the value of each charge is doubled, then we will have a force between them which is:

$F'=(k(2q_1)(2q_2))/(d^2)=4(kq_1q_2)/(d^2)=4F$

So the new force is 4 times larger than the original force.

Final answer:

Doubling the charge on each particle increases the electric force between them by a factor of 4.

Explanation:

The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.

Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.

So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.

Learn more about Coulomb's Law here:

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List Five examples from daily life in which you see periodic motion caused by a pendulum(Marking Brainliest)

Answers

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.

Answers

Answer:

change in length is 3.397 cm

Explanation:

Given data

long = 91 m = 9100 cm

coefficient for concrete (a) = 1.2 × 10−5 ( ◦C)−1

temperature = 56 F = (56× 5/9) ◦C

to find out

how much spacing is needed to allow

solution

we know allow space is given by this formula

change in length = coefficient for concrete × given length × temperature .............1

put all value in equation 1

change in length = 1.2 × 10−5 × 9100 × (56× 5/9)

change in length = 3.397 cm

so change in length is 3.397 cm

Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.

Answers

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

$E = (13.6Z^(2) )/(n^(2) ) eV$ .....(1)

Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for Z and 1 for n in the equation (1).

$E = (13.6*(6)^(2) )/(1^(2) )$

E = 489.6 eV

The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

$(1)/(\lambda) =RZ^(2)[(1)/(n_(1) ^(2))-(1)/(n_(2) ^(2) )]$ ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

$(1)/(\lambda) =1.09*10^(7) *6^(2)[(1)/(2 ^(2))-(1)/(3 ^(2) )]$

$(1)/(\lambda)$ = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

Make the following conversion.56.32 kL = _____ L

0.056320
0.56320
5,632
56,320

Answers

Answer:

The answer would be D 56,320

Explanation:

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

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