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Rasek [7]
3 years ago
5

There are 3.78 liters in one gallon, therefore 3.78L = 1 Gallon. If I need 58 half-gallon containers for my boat, determine the

buoyant force that would be created.
Physics
1 answer:
Digiron [165]3 years ago
8 0

Answer:

F(b) =2142.106N

Explanation:

Given the following data

Volume is 3.78L which is 1 gallon

Therefore, 58gallons = 3.78×58 = 219.24L

Buoyant force F(b) = V(s) × d × g

Where V(s) = submerged volume = 219.24L = 0.21924m³

d = density of water = 997kg/m³

g = acceleration due to gravity = 9.8m/s²

Therefore F(b) = 0.21924m³× 997kg/m×9.8m/s²

F(b) =2142.106N

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What happens when an electron moves from an excited state to the ground state?
Darina [25.2K]
<span>When an electron moves from an excited state to the ground state, "Energy releases"

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6 0
3 years ago
A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th
lana66690 [7]

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

or

v₂ = 14.2 m/s

8 0
3 years ago
What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz
deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

7 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
3 years ago
What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

Explanation:

When a particle with charge q travels across a potential difference \Delta V, then its change in electric potential energy is

\Delta U = q \Delta V

In this problem, we know that:

The particle is an electron, so its charge is

q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

V_+=0V

While the negative terminal is at potential

V_-=-12 V

Therefore, the potential difference (final minus initial) is

\Delta V = -12-0 = -12 V

So, the change in potential energy of the electron is

\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J

This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

3 0
3 years ago
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