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pishuonlain [190]
3 years ago
10

How many pounds must a sandbag weigh to test if a safety net can absorb the proper amount of force

Physics
2 answers:
inessss [21]3 years ago
3 0
I believe its 400lbs
777dan777 [17]3 years ago
3 0

Answer;

400 pounds

400 pounds is the number of pounds that a sand bag must weigh to test if a safety net can absorb the proper amount of force.

Explanation;

The drop test consists of a 400 lb. sandbag dropped from the highest height from which a potential fall might occur, and from no fewer than 42 inches above that level.

Safety nets must be drop-tested on the job site at the following times:

  • After installation and prior to first use
  • If the net is relocated
  • After a major repair
  • After six months, if the net is left in place
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Using the given values

F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N

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2 years ago
A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the
nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c)  1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

b)

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

c)

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

d)

Ke = 1/2mv²

    = (1.67×10⁻²⁷×)(1.4106)²/2

 =  1.661×10⁻²⁷KJ

5 0
3 years ago
A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio
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in the attachment

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3 years ago
A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
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Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
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