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3 years ago

What can you do to mitigate these risks?

Physics

1 answer:

andrey2020 [161]3 years ago

7 0

Answer: your question is incomplete, however I want you to know that there are four ways to mitigate a risk, these are:

avoid, accept, reduce/control, or transfer. Whatever the risk in your question is, it will definitely fit into any of the aforemened risk strategies.

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Which of the following is NOT a characteristic of noble gases?

Rudik [331]

Solid at room temperature

8 0

2 years ago

Read 2 more answers

A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient

e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

$Fr=\mu N$ [1]

Where $\mu$ is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for $\mu$ :

$\displaystyle \mu=\frac{Fr}{N}$

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

$\displaystyle \mu=\frac{80}{196}$

$\mu=0.41$

The coefficient of static friction between the crate and the floor is 0.41

7 0

2 years ago

When measuring speed, you will measure the

qaws [65]

Answer:

Speed is the rate at which an object's position changes, measured in meters per second. The equation for speed is simple: distance divided by time

Explanation:

5 0

3 years ago

A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica

-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 81 mN when sepa

Gekata [30.6K]

Answer:

Explanation:

Check the attachment for solution

8 0

3 years ago