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Gnoma [55]
4 years ago
15

A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on

the box but the box is observed to be at rest. What is the value of the static friction force
Physics
1 answer:
Ira Lisetskai [31]4 years ago
4 0

Answer:23 N

Explanation:

Given

Weight of box W=60\ N

Coefficient of static friction is \mu _s=0.5

Applied force F=23\ N

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction F_s=applied force

F_s=23\ N

Although it maximum value can go up to (F_s)_{max}=\mu _sN

F_s=0.5\times 60

F_s=30\ N

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As sea floor spreading occurs, the oceanic plate _______.
Ira Lisetskai [31]

Becomes older

Explanation:

As sea floor spreading occurs at divergent margins, the oceanic plate becomes older. Younger plate margin are the closest to the margin whereas the older plates bushes backward away from the spreading centers.

  • The idea that the sea floor spreads was postulated by Harry Hess shortly after the second world war around the 1960's.
  • At divergent margins new crust materials from the mantle are brought to the surface.
  • They crystallize and settle at the flanks of plate margins.
  • Older ones are pushed backward away from the margin into far away subduction zones.

Learn more:

Sea floor spreading brainly.com/question/9912731

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8 0
4 years ago
If the period of a wave is 0.05 s, then what is its frequency?
yKpoI14uk [10]
Frequency = 1/time period = 1/0.05 = 20s^-1. 
8 0
3 years ago
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Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
4 years ago
It is MOST accurate to say the the body mass index (BMI) provides information about ____________. A. Body composition. B. An ind
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It is most accurate to say that body mass index (BMI) provides information about an individual's height-weight ratio. The correct answer is B.
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3 years ago
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