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2 years ago

Baseball Digest is but one of several publications on the sport of baseball. True or False?

Physics

2 answers:

Gnoma [55]2 years ago

3 0

Answertrue

Explanation:

Dafna11 [192]2 years ago

3 0

Baseball Digest is one of the publications on the sport of baseball is a True statement.

<u>Explanation: </u>

The baseball digest is the publication in regard to the baseball sport. It is published by the Grand Stand Publishing LLC in the city of Gurnee, state of Illinois, in United States. It was first published in the year of 1942 in August.

It is being published in every 2 months in a year thereby, a total of 6 publications in a year. There are other publications as well for baseball namely, Baseball America, Baseball Hobby News, Baseball Register.

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Which characteristics is common in mature rivers river

amm1812

It channels erode wider fed by many tributaries and has more discharge and is less steep

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2 years ago

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Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

Lana71 [14]

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the given Normal on block 2 = N

and $N = m_2 g \cos \theta$

and the tension in the given string is said to be $T = m_2 g \sin \theta$

When the acceleration $a=\frac{F}{m_1}$

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

$F = m_1g-T.$

It is known that if the said

$\theta=\frac{\pi}{2}$ ,

then Tension $T= m_2g$ $[since \sin(\pi/2) = 1]$ ,

Now making $"F = m_1g - m_2g"$

So If we are to make Force equal to zero

$F=0 => m_1g = m_2g \ or \ m_1 = m_2$

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2 years ago

What is the differece between force and motion?

Jobisdone [24]

Answer:

---

Explanation:

Force is like the amount of push

Motion is the result of Push

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2 years ago

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El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d

Doss [256]

Answer:

R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

ΔR = 1000 (-0,0005) (120-0)

ΔR = -60 Ω

ΔR = R (120) + R (0) = -60

R (120) = -60 + R (0)

R (120) = -60 + 1000

R (120) = 940Ω

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3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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2 years ago