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lions [1.4K]
3 years ago
7

What will sunshine help produce in the body?

Physics
1 answer:
andreev551 [17]3 years ago
4 0
Sunshine will help prode vitamin C
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The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0
3 years ago
Define oxidation number
Delicious77 [7]

<span>a number assigned to an element in a chemical combo that represents the number of electrons lost or gained by atom of the element in the compound.</span>

8 0
3 years ago
A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe
marysya [2.9K]

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration.  So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

7 0
3 years ago
Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35
tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
3 0
3 years ago
A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down
Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

wsin\theta-\mu F_n=ma That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

6 0
3 years ago
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