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inessss
1 day ago
9

A tall tube is evacuated, and its stopcock closed. The open end of the tube is immersed into a container of water (density 103 k

g/m3) that is open to the atmosphere (pressure 105 N/m2). When the stopcock is opened, how far up the tube will the water rise?
Physics
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A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Softa [3030]

Answer:

 Baseball mass:  m_b=3.992*10^{14}kg  

Explanation:

 Circumference of a baseball is calculated using 2πr = 23 cm

 Thus, the radius comes out to be 3.66 cm, which equals 3.66*10^{-2} m

 The mass density of the baseball matches that of a neutron or proton.

 Proton mass = 10^{-27} kg  

 Proton diameter = 10^{-15} m

 Proton radius =  5*10^{-16} m

 Volume of the baseball is \frac{4}{3} \pi r^3

 Now by substituting all values into the mass per unit volume equation for the baseball, we get:  

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       Therefore, the baseball mass amounts to m_b=3.992*10^{14}kg              

5 0
2 months ago
Read 2 more answers
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Applying the natural logarithm to both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

5 0
1 month ago
Marcia is given an incomplete chemical equation that includes the number of nitrogen atoms present in the products of the reacti
ValentinkaMS [3465]
If the products have three nitrogen atoms, the reactants must have had the same quantity, as mass is conserved in a chemical reaction.
8 0
2 months ago
Read 2 more answers
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
serg [3582]

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

6 0
3 months ago
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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame
kicyunya [3294]
For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

                         = 2 * 2π  radians for each second

                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

                                                         ≈13.817  m/s²

. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

                                   = 2.418 N
5 0
2 months ago
Read 2 more answers
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