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Reptile [31]
3 years ago
10

Answers for 13 are

Physics
1 answer:
Veseljchak [2.6K]3 years ago
5 0
A. Andesite

Members of the diorite family<span> have an </span>intermediate<span> composition that is neither felsic nor mafic but has characteristics of both. ... Diorite, a coarse-grained rock, has less quartz than </span>granite<span> and less plagioclase feldspar than gabbro. Andesite is a fine-grained member of the diorite </span>family<span>.</span>

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A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Feliz [49]

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

7 0
3 years ago
Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high.
Over [174]
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity. 

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

√(2*g*h)=v, we plug in the numbers and get:

v=9.9 m/s. 

So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
8 0
3 years ago
Read 2 more answers
What is hypothesis testing
Stells [14]
Hypothesis testing is basically testing the results of a experiment to see weather your results are valid or not.
5 0
3 years ago
a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
8 0
3 years ago
The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.​
Galina-37 [17]

Answer:

Explanation:

weight on moon = 1/6* weight on earth

69.3=1/6*weight on earth

weight on earth = 69.3*6

weight on earth = 415.8 N

8 0
3 years ago
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