Katie traveled southwest a distance of 4000m on a bicycle in 5400s. What is her velocity?

The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh

jeyben [28]

Explanation:

Given that,

Electrostatic force, $F=3.7\times 10^{-9}\ N$

Distance, $r=5\ A=5\times 10^{-10}\ m$

(a) $F=\dfrac{kq^2}{r^2}$ , q is the charge on the ion

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}$

$q=3.2\times 10^{-19}\ C$

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

$n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2

Hence, this is the required solution.

8 0

3 years ago

The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro

irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}$

f = 34.047 cm = 0.3404 m

Power of the lens = P

$P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter$

6 0

2 years ago

How does the matter and (solar) energy move through the sphere?

Tema [17]

When the Sun's energy moves through space, it reaches Earth's atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet's systems in three ways: by radiation, conduction, and convection.

5 0

2 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0

2 years ago

Read 2 more answers

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago