All categories

2 years ago

Find the total resistance ....

Physics

1 answer:

Marianna [84]2 years ago

4 0

The Equivalent resistance is :

$\qquad \tt \dashrightarrow \: \dfrac{14}{9} \: \: ohms$

The solution is in attachment for solution ~

You might be interested in

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

Crazy boy [7]

Answer:

$v=1.93m/s$

Explanation:

From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

$v=\sqrt{2gh}$

Since we know gravity and its hight

$v=\sqrt{2*(9.81m/s^{2})(0.19m) }=1.93m/s$

5 0

2 years ago

How to hope with challeges of today,&how to veaw our current

mel-nik [20]

Answer:

hey how are you

are these points for free or something

5 0

2 years ago

List the location in order from lear to greatest pressure

kramer

Lear vv
F
B
C
D
A
E
G
greatest pressure ^^

I’m really sorry if I’m wrong

7 0

3 years ago

Today, scientists believe that the main reason the Earth has had periodic ice ages is:____________.

horsena [70]

Answer:

Today, scientists believe that the main reason the earth has had periodic ages is that the climate of earth is controlled by the difference in heating of it's surface by the sun. As for example, the equatorial regions are the warmest as the sun is vertically overhead of those areas and the polar regions where the sun is at the extreme angels are the coldest ones.

Explanation:

6 0

3 years ago

What can you say about the magnitudes of the forces that the balloons exert on each other?

maxonik [38]

Answer:

$F_G=G. \frac{m_1.m_2}{R^2}$ gravitational force

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$ electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

$F_G=G. \frac{m_1.m_2}{R^2}$

where:

$G=$ gravitational constant $=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$