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3 years ago

9

A 3.5 kg book is sitting at rest on a table. what is the value of the normal force being exerted by the table on the book?

1 answer:

Ira Lisetskai [31]3 years ago

6 0

This is 3.5 * 9.8 (where 9.8 ms-2 is the acceleration due to gravity).

= 34.3 Newtons ( answer).

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Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measuremen

lorasvet [3.4K]

Answer:

(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area $A=(1.3\times1.3\times10^{-6})^2$

Potential difference = 92.2 mV

Resistivity of the material $\rho=1.30\times10^{7}\ \ohm$

We need to calculate the resistance

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.30\times10^{7}\times7.50\times10^{-9}}{(1.3\times1.3\times10^{-6})^2}$

$R=3.413\times10^{10}\ \Omega$

(a). We need to calculate the amount of current that flows through this portion of the membrane

Using Ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{92.2\times10^{-3}}{3.413\times10^{10}}$

$I=2.70\times10^{-12}\ A$

The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). If the side dimensions of the membrane portion is halved

We need to calculate the new resistance

Using formula of resistivity

$R'=\dfrac{\rho \dfrac{l}{2}}{A}$

Put the value into the formula

$R'=\dfrac{1.30\times10^{7}\times\dfrac{7.50\times10^{-9}}{2}}{(1.3\times1.3\times10^{-6})^2}$

$R'=1.706\times10^{10}\ \Omega$

We need to calculate the new current

Using Ohm's law

$V=I'R'$

$I'=\dfrac{V}{R'}$

Put the value into the formula

$I'=\dfrac{92.2\times10^{-3}}{1.706\times10^{10}}$

$I'=5.404\times10^{-12}\ A$

We need to calculate the factor

$\dfrac{I'}{I}=\dfrac{5.404\times10^{-12}}{2.70\times10^{-12}}$

$I'=2I$

The factor of the current is increase by factor of 2.

Hence,(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

6 0

3 years ago

How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

$\text { 1 miles }=\frac{1}{0.621} \mathrm{km}$

So, 2.50 miles will be equal to

$2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}$

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,

$1 \mathrm{km}=1000 \mathrm{m}$

Thereby,

$4.026 \mathrm{km}=4026 \mathrm{m}$

7 0

3 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

3 years ago

Which statement about cellulose is true?

zlopas [31]

Is there any answers? Or is it asking you to choose?

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3 years ago

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10)A car is moving from rest and attained a velocity of 80 m/s. Calculate the

Mashcka [7]

Explanation:

the velocity graph of a ball mass 20mg moving along a straight line

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3 years ago

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