PLEASE SOMEONE HELP ME PLEASE

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

3 years ago

Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual

Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0

3 years ago

A 1000 kg car accelerates from rest at a rate of 10 m/s² for 3 seconds. A) what is the final velocity of the car?

Lorico [155]

Answer:

Refer to the attachment!~

5 0

3 years ago

I don’t have a question because I posted a photoz

Scorpion4ik [409]

Part a:

$Q_{1}$ = 56

$Q_{2}$ = 60

$Q_{3}$ = 63

The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The $Q_{2}$ is the overall medium, $Q_{1}$ is the medium of the first half, and $Q_{3}$ is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between $Q_{1}$ and $Q_{3}$ . In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + $\frac{7}{2}$ ) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0

2 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

3 years ago