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lara31 [8.8K]
3 years ago
6

Which does a reference point provide

Physics
1 answer:
mr_godi [17]3 years ago
5 0
Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"
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What is fluorine sidekicks
Talja [164]

Fluorine is the most reactive and the most electronegative of all the elements. Fluorine is a pale yellow, diatomic, highly corrosive, flammable gas, with a pungent odor. It is the lightest halogen.

6 0
3 years ago
Explain why water boiling in a container stops<br>boiling momentarily when the lid is removed​
Dmitriy789 [7]

Answer:

It's because it tips over the threshold from nucleate boiling, which we can see, to convection boiling, which we can't. ... Even if the steam stayed in the pot, it would still stop boiling when you removed the heat. The steam and water in a liquid/vapour mixture are at the same temperature (100ºC).

Explanation:

8 0
3 years ago
A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0
viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
KE = \frac{1}{2}mv^2 \\\\PE = mgh

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

E_f = \frac{1}{2}mv_f^2

And:
W_A = E_i - E_f

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) -  \frac{1}{2}mv_f^2

Solving for the work done by air resistance:
W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) -  \frac{1}{2}(.450)(19.89^2)

W_A = \boxed{42.552 J}

8 0
2 years ago
A piece of metal has a mass of 10g and a mass of 2cm
qwelly [4]

Answer:

so whats the  questain?!

Explanation:

idgi

3 0
3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

8 0
3 years ago
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