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sergij07 [2.7K]

3 years ago

15

Bacteria vary in size, but a diameter of 2.0 µm is not unusual. What are the volume (in cubic centimeters) and surface area (in

square millimeters) of a spherical bacterium of that size?

1 answer:

telo118 [61]3 years ago

5 0

A = 4\pi r^2

A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2

A = 1.33*!0^{-5}MM^2

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John pushes forward on a car with a force of 125n while bob pushes backward on the car with a force of 225n. what is the net for

Pie

Answer:

100N

Explanation:

because 225-125= 100

5 0

3 years ago

Read 2 more answers

A uniform disk with radius 0.650 m

VashaNatasha [74]

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

$a = \sqrt{a_{t}^{2}+a_{n}^{2}}$

$a = 13.758\,\frac{m}{s^{2}}$

3 0

3 years ago

The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

sesenic [268]

See attachment for answer:

4 0

3 years ago

Velocities of two bodies A and B are given in vectors notation as va =i+2j-3k and Vb=3i+2j-k what will be the relative velocity

ArbitrLikvidat [17]

Answer:

$V_{B/A}=2i+2k$

Explanation:

The relative velocity can be calculated by means of the difference between vector B minus vector A.

$V_{A}=i+2j-3k\\V_{B}=3i+2j-k\\V_{B}-V_{A}=(3-1)i + (2-2)j+(-1-(-3))k\\V_{B/A}=2i+2k$

5 0

3 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago