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3 years ago

What are the units of measurement for length

Physics

2 answers:

Sonbull [250]3 years ago

8 0

Inches, feet, yards, meters

lord [1]3 years ago

4 0

Inches, cm, m, mm, miles, km

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8. Where are the ribosomes usually located in plant and animal cells?

professor190 [17]

Answer:

Ribosomes are found 'free' in the cytoplasm or bound to the endoplasmic reticulum (ER) to form rough ER.

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3 years ago

Determine the speed of a falling object 7.8 sec after it dropped.

butalik [34]

Answer:

76.4m/s

Explanation:

Given parameter:

Time taken = 7.8sec

Unknown:

Speed after it dropped = ?

Solution:

To solve this problem, we use one of the kinematics equation:

V = U + gt

V is the final speed

U is the initial speed = 0m/s

g is the acceleration due to gravity

t is the time taken

V = 0 + 9.8 x 7.8 = 76.4m/s

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2 years ago

The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det

seraphim [82]

(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s

orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg.

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3 years ago

Use the component form of newton's second law to write an expression for the x component of the net force, σfx

nordsb [41]

Draw a free body diagram to show which forces act in the x and y directions. The x component equation is σfx = 0. The σfx being all the forces acting in the x direction.

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3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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3 years ago