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3 years ago

An elements atomic number is 85 how many protons would an atom of this element have?

Physics

2 answers:

garik1379 [7]3 years ago

5 0

85 protons are in astatine have.

ArbitrLikvidat [17]3 years ago

3 0

The atomic number and number of protons is the same thing :)

So here, your atom would have 85 protons.

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The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

3 years ago

Explain how Pascal's principle can be used to design a fluid power system and describe how a fluid power system works.

zhenek [66]

Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.

4 0

3 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

4 0

3 years ago

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