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Mandarinka [93]
3 years ago
5

The density for potassium is 0.856 g/cm3. What would be the mass of a 40 cm3 piece of potassium?

Physics
2 answers:
SCORPION-xisa [38]3 years ago
3 0

Answer:

38.52g

Explanation:

il63 [147K]3 years ago
3 0

Answer:

34.24g

Explanation:

for study island said I got it correct.

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52.38 m/s

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To move the Center of Gravity (CG) forward, you could do which of the following?
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I think it's no. 3

Decrease the volume of the horizontal and vertical stabilizer material.

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A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move
Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity  = 21.7m/s

Time  = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

  Acceleration  = \frac{v - u}{t}

  v is final velocity

   u is initial velocity

   t is the time

     Acceleration  = \frac{21.7 - 10.5}{4.34}   = 2.58m/s²

Distance traveled;

     V² = U² + 2aS

    21.7² = 10.5² + 2 x 2.58 x S

   360.64 = 2 x 2.58 x S

     S = 69.9m

3 0
3 years ago
A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne
alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E =  U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

3 0
1 year ago
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