I really don’t know the answer for this

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$

7 0

3 years ago

. Maria walked 1.5 miles south to her house in 0.5 hours. What is her speed in miles per hour?

Simora [160]

1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

$v=\frac{d}{t}$

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

$v=\frac{1.5}{0.5}=3 mi/h$

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

$d=1.5 mi \cdot 1609 m/mi = 2414 m$ is the displacement in meters

$t=0.5 h \cdot 3600 s/h =1800 s$ is the time taken in seconds

Substituting,

$v=\frac{2414 m}{1800 s}=1.34 m/s$

And the direction of the velocity is the same as the displacement, so it is south.

6 0

4 years ago

A 100-n force has a horizontal component of 80 n and a vertical component of 60 n. the force is applied to a box which rests on

iragen [17]

<span>160 Joules

For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance: 80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.

So the cart has a final kinetic energy of 160 Joules.</span>

6 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago

Can i get help asap im taking a test

Ainat [17]

Answer:

4 seconds per meter

Explanation:

.......................

5 0

4 years ago