How much work is done on 10.0C of charge to move it through a potential difference of 9V in 10s?

Enny and Anne wanted to see who could throw a ball the hardest. They decided to each throw a ball against a wall as hard as they

mihalych1998 [28]

Answer:

wha t kind of variables

independent variables

dependent variables

controled variables

4 0

2 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s

BabaBlast [244]

Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is $g_m=1.81\ m/s^2$ .

Weight of watermelon in earth , $W=44\ N$ .

Acceleration due to gravity at the surface of earth is $g=9.81\ m/s^2$ .

We know , weight is given by :

$W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg$

Therefore , mass at the surface of Jupiter's moon Io is :

$W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N$

Hence , this is the required solution .

6 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

4 years ago

How far does a boat travel in 5 hours at 32 miles per hour? 162 mi 160 mi 210 mi

sleet_krkn [62]

We know that:
d=vt
d=32mph*5h
d=160mi

4 0

3 years ago

Read 2 more answers

3.00 m^3 of water is at 20.0°C.

krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume v1= 3m^3

volume v2= ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

6 0

3 years ago