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Natali [406]
4 years ago
6

I will give Brainliest to WHoever answers truthfully!!!!!T/F net force charge and net electric force are the same thing

Physics
2 answers:
MakcuM [25]4 years ago
5 0

Answer:

it's true I'm pretty sure

algol134 years ago
4 0

Answer:

the answer is true.I'm sure

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A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration
umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

6 0
3 years ago
Prentice Hall. All
iVinArrow [24]

Answer:

Decreases/Reduces

Explanation:

Fill in the blank:

Consider the equation Work = Force X Distance.

<em>If a machine  increases the distance over which a force is exerted, the force </em>

<em>required to do a given amount of work</em> .........

If the work is a constant value, then by isolating force from the equation, we get:

Force = Work / Distance

By increasing the value of the Distance, then the quotient Work. Distance diminishes, and therefore the required force decreases (diminishes, reduces)

Answer: Decreases/Reduces

7 0
3 years ago
Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu
ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0
3 years ago
Two black holes (the remains of exploded stars), separated by a distance of
jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

  • <em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
  • <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

3 0
2 years ago
Two slits in an opaque barrier each have a width of 0.020 mm and are separated by 0.050 mm. When coherent monochromatic light pa
Semenov [28]

Answer:

The answer is 5

Explanation:

The maximum interference is:

m * λ = d * sinθi

Where m = 0,1,2,3,...

The first minimum diffraction is:

λ = a * sinθd

|sinθi| < sinθd

Where

(|m| * λ)/d < λ/a

|m| < d/a = 2.5

|m|max = 2

It can be concluded that coherent monochromatic light passes through the slits, therefore the maximum number of interference is 5.

5 0
4 years ago
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