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4 years ago

12

A 61.5 KG student sits at a desk 1.2 5M away from a 70.0 KG student. What is the magnitude of the gravitational force between th

e two students?

1 answer:

muminat4 years ago

6 0

mass of two students are

$m_1 = 61.5 kg$

$m_2 = 70 kg$

distance between them is given as

$r = 1.25 m$

now gravitational force between them is given as

$F = \frac{Gm_1m_2}{r^2}$

now plug in all values

$F = \frac{6.67\times 10^{-11} \times 61.5 \times 70}{1.25^2}$

$F = 1.84 \times 10^{-7} N$

so above is the force of gravitation between them

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Air at 38oC and 97% relative humidity is to be cooled to 14oC and fed into a plant area at a rate of 510 m3/min. Calculate the r

bekas [8.4K]

Answer:

mass flow rate at water condenses is 36.72 kg/min

Explanation:

given data

temperature t1 = 38°C

temperature t2 = 14°C

humidity ∅= 97 % = 0.97

rate v = 510 m³/min

to find out

mass flow rate at water condenses

solution

by gas equation we find here mass flow rate that is

pv = mRT

put here value and p is 0.066626 bar at 38°C and find m

m = 0.06626 × $10^{5}$ × 510 / 287×311

m = 37.85 kg/min

so at water condenses mass flow rate is express as

∅ = M / m

Mass flow rate M = ∅ × m

M = 0.97 × 37.85

mass flow rate = 36.72 kg/min

so mass flow rate at water condenses is 36.72 kg/min

8 0

3 years ago

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

$h=ut+\frac{1}{2}at^2$

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

3 years ago

A uniform electric ﬁeld exists in the region between two oppositely charged plane-parallel plates. An electron is released from

Zigmanuir [339]

Answer:

Explanation:

given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s

q = 1.6 x 10_19C

using S = at^2/2

acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

From F = ma

F = qE

ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric ﬁeld. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0

3 years ago

Evolutionary psychology is a relatively new approach to psychology that has been especially influenced by

Andrews [41]

the answer is David Buss

5 0

3 years ago

Read 2 more answers

Fill in the blanks for the following:

storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0

3 years ago