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3241004551 [841]

4 years ago

6

A cylindrical can holds 1 liters (1000 cm^3) of liquid. How should the height and radius be chosen to minimize the amount of mat

erial needed to manufacture the can?

1 answer:

ArbitrLikvidat [17]4 years ago

4 0

I was struck by this many years ago while in graduate school. ... People with mental illness tend to die young , but so do their families. ... there is no medical assessment that can diagnose the absence of illness.

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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

A 3-kg object is attached to a spring and moving in simple harmonic motion. Its angular frequency is 20 rads/sec. When the mass-

Trava [24]

Answer:19.5 J

Explanation:

Given

mass of block=3 kg

angular frequency=20 rad/sec

spring constant $k=\omega _n^2m=1200 N/m$

we know total energy remain conserved

$E_T at x=0.1 m$

$E_T=E_P+E_K$

Where $E_K$ =kinetic energy

$E_P$ =potential Energy

$E_P=\frac{1}{2}kx^2$

$E_P=600\times 0.01=6 J$

$E_K=\frac{1}{2}mv^2$

$E_K=\frac{1}{2}\times 3\times 3^2=13.5 J$

$E_T=13.5+6=19.5 J$

When mass reaches amplitude its velocity becomes zero

there is only potential energy which is equal to Total energy

$E_T=E_P=19.5 J$

7 0

3 years ago

If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t

Liono4ka [1.6K]

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²

Take square root of each side:

√ (ratio of their masses) = ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

5 0

3 years ago

I need help ASAP please :)

rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0

2 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago