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3 years ago

Your friend asks you to describe pseudoscience. You reply pseudoscience is

Physics

2 answers:

exis [7]3 years ago

6 0

your answer is 'D' something that resembles science but cannot be tested or proven using the scientific method

oksian1 [2.3K]3 years ago

5 0

The answer is most likely D. hope that helped

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2. Can the frictional force in this experiment be ignored? Explain.

neonofarm [45]

Answer:

i dont know

Explanation:

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2 years ago

Convert 10 kilowatt to watt

Nataly [62]

Answer:

0.01 kilowatts cubed

Explanation:

just ask alexa lol

3 0

3 years ago

Read 2 more answers

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Irina18 [472]

$P = w \times delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

$P = 35 \div 70 \\ P = 0.5 \: watts$

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.

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3 years ago

A proton traveling along the x-axis enters a region at x = 0 where the x-component of the electric field is given by E = Ao/x1/2

storchak [24]

.Answer:

The value of the work done is $\bf{ 5.29 qA_{0}}$ .

Explanation:

When a charged particle having charge $q$ is moving through an electric field $E$ , the net force ( $F$ ) on the charge is

$F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

and the work done ( $W$ ) by the particle is

$W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Given, $E = \dfrac{A_{0}}{x^{1/2}}$ .

Substitute the value of electric field in equation (1) and then substitute the result in equation (2).

$W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}$

7 0

3 years ago

A 9 kg box is placed on a 6 m tall shelf. What will be the kinetic energy of the box right before it hits the ground? (use g = 1

m_a_m_a [10]

Answer:

540 J

Explanation:

U at top equals to K at bottom if it's an isolated system therefore U at top is equals to 540 J so we can assume that at 0m U=0 (mgh) therefore the box has gained some velocity due to the acceleration due to g and we can calculate it using 1/2mv²

5 0

3 years ago