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3 years ago

Which type of friction keeps a mound of rocks from falling away from each other

2 answers:

6 0

Hello Lunaerica62ovuaaf, Which type of friction keeps a mound of rocks from falling away from each other,
<span>B) static friction.</span>

nirvana33 [79]3 years ago

6 0

b) static friction

Static means "not moving." So friction that is static is friction that keeps things from moving.

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

5 0

4 years ago

A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes

Lerok [7]

Answer:

R = 715.4 N

L = 166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L = 166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

0 = 0 checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0

3 years ago

A jogger runs 20 mi West and then 6.0 mi North. Find the magnitude and direction of the resultant displacement.

Black_prince [1.1K]

Answer:

The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west

Explanation:

Hi there!

Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:

The vector for the first displacement is:

First displacement = (20 mi, 0)

The second displacement:

Second displacement = (0, 6.0 mi)

The resultant displacement will be:

R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)

The magnitude of this vector will be:

$|R| = \sqrt{(20 mi)^{2} + (6.0 mi)^{2}} = 21 mi$

The magnitude of the vector displacement is 21 mi.

To find the direction of the vector R, we have to apply trigonometry:

In a right triangle the following trigonometric rule applies:

cos θ = adjacent side to the angle/ hypotenuse

In this case:

cos θ = 20 mi / magnitude of R

θ = 16.7°

The direction of the vector is 16.7° north of west.

4 0

3 years ago

What is the normal force in the image below?

saveliy_v [14]

Answer:

980.7N

Explanation:

F = m * g

F = 100 * 9.807m/ $s^{2}$

F = 980.7N

3 0

3 years ago

Answer this question

kow [346]

Answer:

15) C

16) A

17) D

18) D

19) A

22) B

23) D

I might be wrong though

6 0

4 years ago