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3 years ago

7

In the diagram below astronauts on different planets are dropping apples.

2 answers:

mixer [17]3 years ago

4 0

Answer:

jupiter

Explanation

it says in the description that is 2 1/2 times earth's gravity thingy. you gotta read the question

Keith_Richards [23]3 years ago

3 0

The apple would weigh the most on jupiter since it has 2 and 1/2 the amount of gravity as earth

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Look at the variables for this lab. Which variable would we have to change in order to change the amount of current flowing thro

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Where are the variables from the lab?

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2 years ago

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

Without sea otters, sea urchins would overgraze on kelp beds, dramatically changing the marine community. true/false

Bumek [7]

True is the anwser to your question
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5 0

3 years ago

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A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

Llana [10]

Answer:

a) $C = 40.138\,pF$ , b) $q = 16.056\,nC$ , c) $U = 3.212\,\mu J$

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

$C = K\cdot \epsilon_{o}\cdot \frac{A}{d}$

Where:

$K$ - Dielectric constant.

$\epsilon_{o}$ - Vaccum permitivity.

$A$ - Plate area.

$d$ - Distance between plates.

Hence, the capacitance of the system is:

$C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)$

$C = 4.014\cdot 10^{-11}\,F$

$C = 40.138\,pF$

b) The charge can be found by using the definition of capacitance:

$q = C\cdot V_{batt}$

$q = (4.014\times 10^{-11}\,F)\cdot (400\,V)$

$q = 1.606\times 10^{-8}\,C$

$q = 16.056\,nC$

c) The energy stored in the charged capacitor is:

$U=\frac{1}{2}\cdot Q\cdot V_{batt}$

$U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)$

$U = 3.212\times 10^{-6}\,J$

$U = 3.212\,\mu J$

3 0

3 years ago

Read 2 more answers