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3 years ago

13

changes to OSHA's regulations in 2013 require chemical information be provided via the _______ format. a) safety data sheets b)

classification and labeling chemicals c) globally safety data sheets

1 answer:

Talja [164]3 years ago

8 0

Don't listen to the other guy I just took the test and got it wrong because of him..

I re-took it and the correct answer is

A) Safety Data Sheets (SDS)

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If two asteroids moved closer together, what would be the result on the gravitational force each asteroid exerts on the other?

spayn [35]

Gravitational force depends on inverse square law. That is, gravitational force is inversely proportional to square of distance between asteroids.
As distance between them decreases, gravitational force increases. Hence A is correct.

7 0

3 years ago

Read 2 more answers

Orbital Motion<br> Project: Career Multimedia Presentation

Yuliya22 [10]

Answer:

can you post the full question plz

Explanation:

4 0

2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago

A large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the

exis [7]

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

7 0

3 years ago

Read 2 more answers

Someone pls help me with this!

Lina20 [59]

Answer:

<em>F=8.87 N</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge in coulomb

d= The distance between the objects in meters

Object 1 has a charge of

$q_1=-2.3\cdot 10^{-6}\ c$

Object 2 has a charge of

$q_2=-4.2\cdot 10^{-6}\ c$

They are separated by a distance of

d = 0.099 m

Calculate the force:

$\displaystyle F=9\cdot 10^9\frac{2.3\cdot 10^{-6}*4.2\cdot 10^{-6}}{0.099^2}$

F=8.87 N

5 0

2 years ago