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vivado [14]
3 years ago
6

Use the crisscross method to find the chemical formula for the ionic compound formed by strontium (Sr) and bromine (Br).​

Physics
2 answers:
nignag [31]3 years ago
8 0

Answer:it’s c SrBr2

Explanation:

Vikentia [17]3 years ago
3 0

Answer above is correct

Explanation:

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The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
dimaraw [331]

The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



3 0
3 years ago
The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?
Aleks04 [339]
The formula for the period of wave is: wave period is equals to 1 over the frequency.waveperiod=\frac{1}{frequency}
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s. 
6 0
3 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?
andrey2020 [161]

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19×\frac{1}{2\pi} *60 revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

4 0
3 years ago
A street light is on top of a 8 foot pole. Joe,
zubka84 [21]
8/4 = y/y-x

8y - 8x  = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps
5 0
3 years ago
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