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3 years ago

I need help with 1-3 and 6-9 its do today need help with all of the questions plzzzzz

Physics

2 answers:

solniwko [45]3 years ago

6 0

Answer:

1.C

2.B

3.A

i need a better picture to help with 6-9

Elodia [21]3 years ago

5 0

Answer:

56767475457345621251461

Explanation:

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Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field

kirza4 [7]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

= (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ 0_∫^11 r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

3 0

4 years ago

When the light on the moon is increasing, what is it called

nataly862011 [7]

Answer:

Waxing Gibbous phase occurs when the Moon is mostly lit and the illuminated portion is egg-shaped (gibbous) with the eastern edge shaded. The amount of illuminated area visible is increasing from one day to the next which is what is meant by "waxing"

hope this helps

have a good day :)

Explanation:

6 0

3 years ago

Read 2 more answers

A 2.0 kg tumor is being irradiated by a radioactive source.The

navik [9.2K]

Answer:

426136363636.36365 Bq

Explanation:

Grays = 12

Mass = 2 kg

Time = 880

Energy = 0.4 MeV

Activity is given by

$Activity=\dfrac{Grays\times Mass}{Time\times 0.4}\\\Rightarrow Activity=\dfrac{12\times 2}{880\times 0.4\times 1.6\times 10^{-13}}\\\Rightarrow Activity=426136363636.36365\ Bq$

The activity N//t of the radioactive source is 426136363636.36365 Bq

6 0

4 years ago

What is each end of magnet A called?

leonid [27]

Answer: The ends of a magnet are called the poles. Each magnet has a north pole and south pole.

$\textit{\textbf{Spymore}}$

4 0

3 years ago

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Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

4 years ago

Read 2 more answers