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Natasha2012 [34]
3 years ago
7

What happens to the boiling point of water when the pressure is increased?

Physics
1 answer:
Artemon [7]3 years ago
4 0
You can drink it nothing will happen you caN BOIL ME SOME EGGS IF YOU WANT TOO
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Gymnasts often practice on foam floors, which increase the collision time when a gymnast falls. What effect does this have on co
krok68 [10]

Gymnasts often practice on foam floors, which increase the collision time when a gymnast falls. What effect does this have on collisions

The effect of the gymnast on the collision will increase. An elastic collision is when two bodies collide and separates after collision conserving the total kinetic energy before and after collision.

8 0
2 years ago
a block initially at rest has a mass m and sits on a plane incline at angle. it slides a distance d before hitting a spring and
KiRa [710]

Answer:

k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}

Explanation:

Let assume that spring reaches its maximum compression at a height of zero. The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}=U_{k,B} + W_{f}

m\cdot g \cdot (d + x_{f})\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x_{f}^{2}+\mu_{k}\cdot m \cdot g \cdot (d+x_{f})\cdot \cos \theta

m\cdot g \cdot (d + x_{f})\cdot (\sin \theta-\mu_{k}\cdot \cos \theta) = \frac{1}{2}\cdot k \cdot x_{f}^{2}

The spring constant is cleared in the expression described above:

k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}

6 0
2 years ago
I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
3. Suppose you take a pendulum with length L and mass m having a period T to a
natta225 [31]

(C)

Explanation:

t = 2\pi \sqrt{ \frac{l}{g} }

If g is only 1/6 on another planet, then

t = 2\pi \sqrt{ \frac{l}{ \frac{g}{6} } } =  2\pi \sqrt{ \frac{6l}{g} }

=  \sqrt{6} \: (2\pi \sqrt{ \frac{l}{g} } ) = 2.4  \times t(on \: earth)

6 0
2 years ago
Two thermometers are calibrated, one in de
zlopas [31]
-40 c = -40 f but k would be 233.15
6 0
3 years ago
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