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3 years ago

11

If water is leaking from a certain tank at a constant rate, is it leaking at a rate that is greater than 12 liters per hour?(1 l

iter = 1,000 milliliters)

1 answer:

Margarita [4]3 years ago

5 0

Answer:

Yes it is possible because 3.33 is greater than 2.

Explanation:

Given that,

1 liter = 1000 milliliters

Suppose water is leaking from the tank at a rate that is greater than 2 milliliters per second.

We need to change the unit in ml/s

$12 lit/hr=\dfrac{12000}{3600}$

$12 lit/hr=3.33\ ml/s$

Hence, Yes it is possible because 3.33 is greater than 2.

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The sun will eventually become a

nadezda [96]

<span>The sun will eventually become a

>white dwarf 5 billion years from now then 2 billion years after that the core will crystallize, </span><span>leaving a giant diamond in the center of our solar system</span><span />

5 0

3 years ago

A proton moves from left to right across the screen at speed v0. Then a uniform magnetic field of strength 0.80 T is directed in

Anton [14]

Answer: The force is directed upward

Explanation: Considering the Lorentz force, given by:

F= qv×B

using the right hand rule and considering the direction of electron velocity and the magnetic field from the figure, the vectorial product gives a force vector upwards .

8 0

4 years ago

A 706-N man stands in the middle of a frozen pond of radius 8.0 m. He is unable to get to the other side because of a lack of fr

mariarad [96]

Answer:

t=62 s

Explanation:

Applying the conservation of linear momentum formula:

$(m1+m2)*v_{o1}=m1*v_{f1}+m2*v_{f2}$

the initial velocity is zero, we can calculate the man's mass using the gravitational force formula:

$F_g=m.g\\\\m=\frac{706N}{9.81}\\\\m=72.0kg$

now:

$m*v_{f1}=-m_b*v_b\\\\V_{f1}=-\frac{m_b*v_b}{m}\\\\V_{f1}=-\frac{1.2kg*8.0m/s}{72.0kg}\\\\V_f=-0.13m/s$

That is 0.13m/s due south.

because there is no friction, the man will maintain a constant velocity, so:

$d=v*t\\t=\frac{d}{v}\\t=\frac{8m}{0.13m/s}\\\\t=62s$

8 0

3 years ago

Davisson and Germer performed their experiment with a nickel target for several electron bombarding energies. At what angles wou

pochemuha

Answer:

The angle of diffraction are 67.75 deg and 53.57 deg.

Explanation:

Given:

Davisson and Germer experiment with nickel target for electrons bombarding.

Voltages : $38\ eV$ and $54\ eV$

We have to find the angles that is $\phi_3_8$ and $\phi_5_4$ .

Concept:

Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
When electrons get reflected from the surface of a metal target with an atomic spacing of $D$ , they form diffraction patterns.
The positions of diffraction maxima are given by $Dsin(\phi) = n\lambda$ .
An atomic spacing is $D = 0.215\ nm$ , when the principal maximum corresponds to n=1
The wavelength is $\lambda$ , and $\lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm$ .

Solution:

Finding the wavelength at $V_o=38\ eV$ .

⇒ $\lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm$

⇒ $\lambda_3_8 =0.199$ nm

Plugging the values of wavelength.

⇒ $sin(\phi)=\frac{\lambda}{D}$

⇒ $\phi_3_8=sin^-1(\frac{0.199}{0.215} )$

⇒ $\phi_3_8 =67.75$ degrees.

Now

For for the electrons with energy $54\ eV$ , $V_0=54V$ the wavelength is.

⇒ $\lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173$ nm

And

⇒ $\phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57$ degrees.

So,

The angles of diffraction maxima are 67.75 deg and 53.57 deg.

6 0

3 years ago

A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65

Sedaia [141]

The thermal energy was produced is 116J

<h3>What is the thermal energy produced?</h3>

Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;

Ff= μmgcosθ

Ff = 0.65 * 5.0 kg * 9.8 m/s^2 * cos 25

Ff = 29 N

Hence, the thermal energy is;

29 N * 4.0 m = 116J

Learn more about frictional force:brainly.com/question/1714663

#SPJ1

4 0

2 years ago