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ser-zykov [4K]
3 years ago
7

A 0.83-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one

, and the amount that the spring stretches from its unstretched length triples. What is the mass of the second block?
Physics
1 answer:
vodomira [7]3 years ago
6 0

Answer:

1.66 kg

Explanation:

Given that a 0.83-kg block is hung from and stretches a spring that is attached to the ceiling.

From Hook's law

F = Ke

But F = mg

Substitute mg for force in the Hook's law

Mg = ke

0.83 × 9.8 = ke

Make K the subject of formula

8.134 = Ke

K = 8.134 /e

Given that a second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples.

That is

(0.83 + M) × 9.8 = K (3e)

Substitutes K into the above equation

(0.83 + M) × 9.8 = 8.134 / e (3e)

The e will cancel out

(0.83 + M) × 9.8 = 24.402

0.83 + M = 24.402/9.8

0.83 + M = 2.49

M = 2.49 - 0.83

M = 1.66 kg

Therefore, the mass of the second block is 1.66kg

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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
4 years ago
A ball is thro.d hit the grouvelocity)?
kati45 [8]

Answer:

The velocity of the ball is 3.52 m/s.

Explanation:

A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.

Given that the ball was thrown vertically upward on the top of a skyscraper of  height 61.9 m. So that the velocity can be determined by;

      u = \sqrt{\frac{2H}{g} }

Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/s^{2}, then;

 u = \sqrt{\frac{2*61.9}{10} }

   = \sqrt{\frac{123.8}{10} }

u  = 3.5185

The velocity of the ball is 3.52 m/s.

3 0
4 years ago
A drowsy cat spots a flower pot that sails first up and then down past an open window. The pot was in view for a total of 0.56 s
liberstina [14]

Answer:

h = 0.028 m

Explanation:

As we know that

d = \frac{v_2 + v_1}{2} t

here we have

1.95 = \frac{v_2 + v_1}{2}(0.56)

v_2 + v_1 = 6.96

also we know

v_2 - v_1 = at

v_2 - v_1 = (9.81)(0.56)

v_2 - v_1 = 5.49

so we have

v_2 = 6.23 m/s

v_1 = 0.74 m/s

so the height above window is given as

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0.74^2 - 0 = 2(9.81)h

h = 0.028 m

4 0
4 years ago
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