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BabaBlast [244]
2 years ago
5

Describe 5 steps you would take when trouble shooting a test kit

Physics
1 answer:
Alecsey [184]2 years ago
5 0

Answer:

1. Information Gathering

2. Analysis and Planning.

3. Implementation of a solution.

4. Assessment of the effectiveness of the solution.

5. Documentation of the incident.

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A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls
anastassius [24]

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

                     radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

KE_r = \dfrac{1}{2}I\omega^2

Moment of inertia of coin

I = \dfrac{1}{2}MR^2

so,

KE_r = \dfrac{1}{4}MR^2\omega^2

Transnational Kinetic energy

KE_t = \dfrac{1}{2}Mv^2

v = r ω

KE_t = \dfrac{1}{2}MR^2\omega^2

now,

using conservation energy

Kinetic energy of the coin is converted into the potential energy  

KE_r + KE_t = PE

\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh

\dfrac{3}{4}R^2\omega^2=gh

\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h

h = 0.0259 m

Vertical height gain by the coin is equal to 0.0259 m

7 0
4 years ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the
Law Incorporation [45]

Answer:

y and length is directly relation

Explanation:

Given data

A single-slit diffraction pattern is formed on a distant scree

angles involved = small

to find out

what factor will the width of the central bright spot on the screen change

solution

we know that  for single slit screen formula is

mass ƛ /area = sin θ and y/L = sinθ

so we can say mass ƛ /area =  y/L

and y = mass length  ƛ / area       .................1

in equation 1 here we can see y and length is directly relation so we can say from equation 1 that  the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled

8 0
3 years ago
Someone help me pls
Alexeev081 [22]

Answer:

a = heart and brain          b = lungs

Explanation:

7 0
4 years ago
In science, a hypothesis must be true.
Anon25 [30]
I would believe that this is false.
6 0
4 years ago
Suppose that 3.00 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the Earth's
blondinia [14]

Answer:

Explanation:

Atomic mass of hydrogen is 1 . so 1 g of hydrogen will have 6.02 x 10²³ atoms

3 g of hydrogen will have 3 x 6.02 x 10²³ atoms . This will give 3 x 6.02 x 10²³  protons and same number of electrons

So number of protons at north pole = 3 x 6.02 x 10²³

charge on these protons = 1.6 x 10⁻¹⁹ x  3 x 6.02 x 10²³ C

= 28.9 x 10⁴ C  

similarly total charge on electrons at south pole = 28.9 x 10⁴ C

distance between them = diameter of the earth = 2 x 6356 x 10³ m

= 12712 x 10³ m

Attractive force between these charges

= k q₁q₂ / r² , q₁ ,q₂ are charges and r is distance between charges.

= 9 x 10⁹ x (28.9 x 10⁴)² / (12712 x 10³ )²

= 4.6517 x 10⁻⁵ x 10¹¹

= 4.6517 x 10⁶ N

8 0
4 years ago
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