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neonofarm [45]
2 years ago
5

A physical (beam) is used to measure?​

Physics
2 answers:
murzikaleks [220]2 years ago
7 0

Answer:

the weight of something, that is

Explanation:

to compare weights

Kisachek [45]2 years ago
6 0

Answer:

mass of a object is the answer

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One baked potato provides an average of 30 mg of vitamin C. If 70 potatoes weigh 20 lb., how many milligrams of vitamin C are pr
JulsSmile [24]

Answer:

105 mg

Explanation:

Given that:

1 baked potato provides 30 mg of vitamin C.

So,

70 baked potatoes provide 30\times 70 mg of vitamin C

Also,

70 potatoes = 20 lb

So,

20 lb potatoes provide 30\times 70 mg of vitamin C

Thus,

1 lb potatoes provide \frac {30\times 70}{20} mg of vitamin C

<u>Thus, 105 mg of Vitamin C are provided per pound of the potatoes.</u>

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3 years ago
When Jennifer is out for a
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The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.

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2 years ago
What is the Acceleration of a car traveling 25 m/s and in 5 seconds<br> traveling 65 m/s? *
WARRIOR [948]

Answer:

8m/s²

Explanation:

a = change in v/t

a = (65-25)/5

a = 8

7 0
2 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
2 years ago
You toss a coin into a wishing well full of a liquid denser than the coin. Which of the following could be true?
maks197457 [2]
If one of the answers are the coin will float that one would be correct.
8 0
3 years ago
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