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Volgvan
3 years ago
9

The results from Rutherford's alpha particle scattering experiment led to

Physics
1 answer:
Brut [27]3 years ago
7 0

Answer:

Explanation:

Can you help me in mathematics

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Which of these hazmat products are allowed in your FC?
aliina [53]

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)
7 0
3 years ago
An organism that cannot make its own food called
Rzqust [24]

Answer:

heterotroph

Explanation:

8 0
3 years ago
Read 2 more answers
Assuming 70% of Earth's surface is covered in water at an average depth of 2.5 mi, estimate the mass of the water on Earth in Ki
saw5 [17]
This is an excellent question that i do not have the answer to.
7 0
3 years ago
A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.
shutvik [7]
<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>
7 0
4 years ago
A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can
Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

\mu is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

v=85 mph \cdot \frac{1609}{3600}=38.0 m/s is the speed

And solving for \mu, we find the coefficient of friction required to keep the car in circular motion:

\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196

Learn more about circular motion:

brainly.com/question/2562955  

brainly.com/question/6372960  

#LearnwithBrainly

8 0
3 years ago
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