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Vera_Pavlovna [14]
3 years ago
12

REALLY NEED HELP, 3 QUESTIONS 20 POINTS. ONLY ANSWER IF YOU KNOW THE CORRECT ANSWER. DO NOT ANSWER WITH "I DON'T KNOW, SORRY" JU

ST FOR THE BRAINLY POINTS. Thanks very much, have a great day. ( I Will report you if you answer with "I don't know, sorry"
1) A spring with a natural height of 60 mm is compressed by a 300 g mass to a new height of 54 mm.
 Find the spring constant in SI units.
 Find the height of the spring if the 300 g mass were replaced by a 400 g mass

2) A spring is stretched by 2m with a force of 100 N. Calculate its spring constant.

3) You have just bought an innerspring mattress that contains coil springs in a rectangular array 20 coils wide and 40 coils long. You estimate that when you lie on the mattress, your weight is supported by about 200 springs (about one-fourth of the total number of springs in the mattress). You observe that the springs compress
about 2.0 cm when you lie on the mattress. Assuming that your weight of 600 N is supported equally by 200 springs, find the force constant of each spring.
Physics
1 answer:
Lostsunrise [7]3 years ago
6 0

Answer:

1. a. -490N/m

  b. -653.333N/m

2. -50N/m

3.  (i don't know this one but i hope the rest is helpful)

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A man is jogging at 3 m/s. If his mass is 85 kg, what is his kinetic energy?
Lina20 [59]

Answer:

This is the complete answer and step by step explanation.

Given:

a = 3m/s

m = 85 kg

Unknown:

The Kinetic energy of the man jogging

Formula and solution:

KE = 1/2 mv²

KE = 1/2(85 kg)(3 m/s)²

KE =  (42.5 kg)(9 m²/s²)

Final Answer:

KE = 382.5 kg.m²/s² or simplify as 382.5 J is the kinetic energy of man jogging

8 0
3 years ago
the air in a tire is initially at 380 kpa, 20 C, when the tire volume is 0.120 m^3. as the tire is warmed by the sun, the pressu
castortr0y [4]

Answer:

final temperature is 364.32 K

mass of air is 0.5423 kg

Explanation:

given data

pressure p1 = 380 kPa

volume v1 =  0.120 m³

temperature = 20°C = 20 + 273 = 293 K

pressure p2 = 3450 kPa

volume v2 =  5% increase

to find out

final temperature and the mass

solution

we consider here ideal gas

so equation is

p1v1 / t1 = p2v2/t2

put here all value and find t2

and v2 is = 0.120 ( 1 + 0.05) = 0.126 m³

so

t2 = p2v2/ p1v1 × t1

t2 = (450 × 0.126) /  (380 × 0.120)  ×  293

t2 = 364.32 K

so final temperature is 364.32 K

and

mass = p2v2 / Rt2

here R gas constant is 0.287 kJ/kg.K

so

mass = (450 × 0.126) / (0.287 × 364.32 )

mass = 0.5423

so mass of air is 0.5423 kg

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