Which of the following activities belongs on the top of the physical activity for pyramid

Which feature of a heating curve indicates a change of state

ziro4ka [17]

Answer:

The diagonal or the inclined lines shows the changes in terms of temperature, and the horizontal lines shows the changing of phases.

Explanation:

hope it is useful

3 0

3 years ago

Two particles have a mass of 8kg and 12kg, respectively. if they are 800mm apart, determine the force of gravity acting between

natima [27]

The magnitude of the force of gravity acting between the particles is:
$F=G\frac{m_1.m_2}{d^2}$
The weight of each particle is:
$P=mg$
Now let's plug in the numbers knowing that $G=6.67\times10^{-11}$ , $g=9.81$ , $d=0.8$ and m1 and m2 are already given in kilograms. We get then:
$P_1=m_1.g=78.48$ N
$P_2=m_2.g=117.72$ N
$F=G\frac{m_1.m_2}{d^2}=1.00\times10^{-8}$ N

This results shows us why we don't often see objects being attracted to each other, their mass is too small compared to the earth gravitational pull.

8 0

3 years ago

∁=f

Sindrei [870]

Answer:

so you have a question

Explanation:

either way have a nice day

5 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma

Gre4nikov [31]

Answer:

correct answer is b

Explanation:

The frequency of a wave depends on the properties of medium density and the elasticity properties change the amplitude depends on the energy carried by the wave, that is, the amplitude is proportional to the height of the wave (oscillation).

Consequently the amplitude of independent of the frequency because it depends on different factors.

Therefore when changing the amplitude the wavelength remains constant

the correct answer is b

8 0

3 years ago