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Dmitry_Shevchenko [17]
2 years ago
5

A tissue cross-section viewed on a microscope slide shows a cell of almost rectangular shape with dimensions 6.9 × 10-4 cm by 1.

1 × 10-3 cm. Calculate the cross-sectional area of the cell. Express the area in powers of ten notation and use the appropriate number of significant digits for the number multiplying the power of 10.
Physics
1 answer:
zimovet [89]2 years ago
4 0

Answer:

The  cross-sectional  area is A = 7.59 *10^{-11} \  m^2

Explanation:

From the question we are told that

   The dimension of the cell is    6.9 × 10-4 cm by 1.1 × 10-3 cm = 6.9*10^{-6} \ m \ by\  1.1*10^{-5} \  m

   Generally the cross-sectional area is mathematically represented  as

    A  =  6.9*10^{-6}   *  1.1*10^{-5}

=>   A = 7.59 *10^{-11} \  m^2

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Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.
Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

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3 years ago
Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely
m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

  • Space long= 84 ft
  • Space wide= 42 ft
  • Space deep= 9 ft
  • Cost by yard3 = $39/yd3
  • Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0
1 year ago
What type of heat does not require matter?
Lana71 [14]
It would be Thermal Radiation
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2 years ago
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HEY CAN YALL PLS HELP ME IN DIS!!!!!!!!!!!!!
Serga [27]

Answer:

Hey!!

Your answer is: 0.72

Explanation:

if 760=1  then...

550=x

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