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Evgesh-ka [11]
3 years ago
12

Why is gum bad for you

Physics
2 answers:
valina [46]3 years ago
7 0
Actually gum is not all that bad yes you could swallow it but you got to be that smart not to do that or choke on it and sugar sweetened gum is bad for your teeth the American dental association says chewing sugarless gum stimulates saliva flow, which reduces plaque and helps prevent cavities and there is also a saying that if you chew gum while studying it might help you more on the test because you will remember the question that you studied if you chew the same brand of gum but i don't know if that is true
viva [34]3 years ago
6 0
There are a lot of reasons why gum is bad for you but a main reason because if you make a bubble and swallow it you could choke. Another reason is because it breaks down your gums.
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A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram
kogti [31]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

7 0
3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
Electricity does not commonly occur in nature except in the form of ( hydrocarbons, photosyntheis, lightning, thunder)
horrorfan [7]
Hydrocarbons are compounds of hydrogen and carbon.
Photosynthesis is a process of nutrition in plants.
Thunder is a sound.
Lightning and electric eels are electrical phenomena in nature.
7 0
3 years ago
Read 2 more answers
A portion of the mesosphere and thermosphere known for its ability to "bounce" radio signals is the _____________
Lerok [7]

It is B: ionosphere. This has the ability to bounce radio signals
8 0
3 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
2 years ago
Read 2 more answers
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