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kramer
3 years ago
13

A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 1

03 N/m. By how much is the spring compressed?
Physics
1 answer:
dimaraw [331]3 years ago
6 0

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

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10. Convert the following:<br> a. 37.4 mL into ML<br> b. 689 km/hr into m/s<br> c. 34.5 m² into mm²
Snezhnost [94]

A. When we convert 37.4 mL to ML, the result obtained is 3.74×10¯⁸ ML

B. When we convert 689 km/hr to m/s, the result obtained is 191.39 m/s

C. When we convert 34.5 m² to mm², the result obtained is 3.45×10⁷ mm²

<h3>A. How to convert millimeters (mL) to megaliter (ML)</h3>
  • Volume (mL) = 37.4 mL
  • Volume (ML) =?

1 mL = 1×10¯⁹ ML

Therefore,

37.4 mL = 37.4 × 1×10¯⁹

37.4 mL = 3.74×10¯⁸ ML

Thus, 37.4 mLis equivalent to 3.74×10¯⁸ ML

<h3>B. How to convert 689 km/hr to m/s</h3>

Conversion scale

3.6 Km/hr = 1 m/s

Therefore,

689 km/hr = 689 / 3.6

689 km/hr = 191.39 m/s

Thus, 689 km/hr is equivalent to 191.39 m/s

<h3>C. How to convert 34.5 m² to mm²</h3>

Conversion scale

1 m² = 1×10⁶ mm²

Therefore,

34.5 m² = 34.5 × 1×10⁶

34.5 m² = 3.45×10⁷ mm²

Thus, 34.5 m² is equivalent to 3.45×10⁷ mm²

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

6 0
2 years ago
A chemist must dilute 58.0 ML of 69.6 M aqueous silver nitrate (AgNO3) solution until the concentration falls to 5.00 M. He'll d
sdas [7]
1929292929292929393939339
3 0
3 years ago
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What i
NISA [10]

Answer:

C_{eq}=1.97\ \mu F

Explanation:

Given that,

Capacitance 1, C_1=0.5\ \mu F

Capacitance 2, C_2=11\ \mu F

Capacitance 3, C_3=1.5\ \mu F

C₁ and C₂ are connected in series. Their equivalent is given by :

\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}

\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}

C'=0.47\ \mu F

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

C_{eq}=C'+C_3

C_{eq}=0.47+1.5

C_{eq}=1.97\ \mu F

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

3 0
3 years ago
A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car
anygoal [31]

Answer:h=19.4 m

Explanation:

Given

mass of automobile m=750\ kg

Initial height of automobile h_o=5\ m

Velocity at this instant v=16.8\ m/s

If the car stops somewhere at a height h

Thus conserving total energy we get

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh

\frac{v^2}{2g}+h_o=h

h=5+\frac{16.8^}{2\times 9.8}

h=5+14.4

h=19.4\ m

6 0
3 years ago
Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a
mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0
3 years ago
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