All categories

3 years ago

Which of these hazmat products are allowed in your FC?

Physics

1 answer:

aliina [53]3 years ago

7 0

Answer: Hazmat products are allowed in your FC are:

A GPS unit (lithium batteries)
A subwoofer (magnetized materials)

Explanation:

Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.

Hazardous material allowed in FC are as follows.

Magnetized material products like as speakers.
Non-spillable battery products like toy cars.
Lithium-ion battery containing products like laptops, mobile phones etc.
Non-flammable aerosol.

So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).

Thus, we can conclude that hazmat products are allowed in your FC are:

A GPS unit (lithium batteries)
A subwoofer (magnetized materials)

You might be interested in

1. Stops: Using the information learned in this course, explain three things you will not do when driving.

vazorg [7]

Don't text while driving
don't get your eyes off the road
don't get distracted

4 0

3 years ago

A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.

Alex Ar [27]

Good. You can do some very interesting experiments with that equipment.

3 0

3 years ago

An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (

vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg

angular speed of the disk = 4 rev/s

arm outstretched = 1 m

centripetal force of the disk in the circular path

F = m ω² r

ω = 4 x 2 x π

ω = 25.13 rad/s

F = m ω² r

F = 2 x 25.13² x 1

F = 1263.03 N

hence, centripetal force equal to the F = 1263.03 N

6 0

3 years ago

In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

jarptica [38.1K]

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$ :
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$

3 0

3 years ago

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0

3 years ago