If

Grain is pouring into a silo to be stored for later use. Due to the friction between pieces of grain as they rub against each ot

jolli1 [7]

Answer:

The electric force between them if the pieces of grain are 2 cm apart is $3.15\times 10^{-11}\textrm{ N}$ .

Explanation:

Given:

Charge on one grain, $Q_{1}=6\times 10^{-10}\textrm{ C}$

Charge on another grain, $Q_{2}=2.3\times 10^{-15}\textrm{ C}$

Separation between them, $d=2\textrm{ cm}=0.02\textrm{ m}$

Electric force acting between two charges $Q_{1}\textrm{ and }Q_{2}$ separated by a distance $d$ is given as:

$F_{elec}=\frac{kQ_{1}Q_{2}}{d^2}$

Where, $k$ is Coulomb's constant equal to $9\times 10^9\textrm{ }Nm^2/C^2$ .

Now, plug in all the values and solve for $F{elec}$ .

$F{elec}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}$

Therefore, the electric force between them if the pieces of grain are 2 cm apart is $3.15\times 10^{-11}\textrm{ N}$ .

3 0

3 years ago

A fission reaction is one in which the weaker electrostatic repulsion force wins out over the ___________ nuclear force and, as

natulia [17]

The answer is b. Strong

6 0

4 years ago

Read 2 more answers

An angle has a value of 2.2 radians. Find its value in degrees.

Irina18 [472]

Answer:

The value of 2.2 radians in degree is 126.11°.

Explanation:

Given that,

Angle = 2.2 radians

We need to calculate the value in degrees

We know that,

$1\ radians = \dfrac{180}{\pi}$

$2.2\ radians = \dfrac{180}{3.14}\times2.2$

$2.2\ radians = 126.11^{\circ}$

Hence, The value of 2.2 radians in degree is 126.11°.

8 0

3 years ago

A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie

Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.

But the work done can be written as

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel

each other.

e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

= Q_C1(T_C/T')

(T1 is the intermediate temperature)

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 = -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by

e_12=1-Tc/Th

This is same as the original Carnot engine.

7 0

3 years ago

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance It takes the pot 0.420 s to pa

eduard

Answer:

the distance of the top of the window below the windowsill from which the flowerpot fell is 0.31 m

Explanation:

given information:

time, t = 0.420 s

height, h = 1.9 m

the the vertical motion equation:

h = $v_{0}$ t + $\frac{1}{2} gt^{2}$

where

h = height (m)

$v_{0}$ = initial velocity (m/s)

t = time (s)

g = gravitational force (9.8 $m/s^{2}$ )

first we calculate the velocity of the flowers pot when it reaches top window

h = $v_{0}$ t + $\frac{1}{2} gt^{2}$

1.9 = $v_{0}$ (0.42) + ( $\frac{1}{2} 9.8 0.42^{2}$ )

$v_{0}$ = (1.9 - ( $\frac{1}{2} 9.8 0.42^{2}$ ))/0.42

= 2.47 m/s

now we can find the distance of the windowsil

$v^{2} =v_{0} ^{2}$ + 2gh

v = 0, thus

2gh = - $v_{0}^{2}$

h = - $v_{0}^{2}$ /2g

= $(- 2.47^{2} )$ / (2 x 9.8)

= 0.31 m

6 0

3 years ago