What are some of the forces that are found in skateboarding?

HELP PLEASE BOYLES LAW

kogti [31]

Answer:

3 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

Using the Boyle's law equation, the new volume (i.e final volume) of the Ne gas can be obtained as:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

0.75 × 2 = 0.5 × V₂

1.5 = 0.5 × V₂

Divide both side by 0.5

V₂ = 1.5 / 0.5

V₂ = 3 L

Thus, the new volume of the Ne gas is 3 L

7 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.

givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/ $s^{2}$ , t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

= 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

= 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + $\frac{1}{2}$ a $t^{2}$

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = $\frac{1}{2}$ a $t^{2}$

= $\frac{1}{2}$ x 14 x $(510)^{2}$

= 7 x 260100

= 1820700 m

The distance that the shuttle has traveled during the given time is 1820.7 Km.

5 0

2 years ago

It is easier to lift the load as the load is pushed towards the wheel in a wheelbarrow

Svetllana [295]

It’s easier to lift the wheelbarrow when the load is near the front because there isn’t any pressure near the back where the handles are.

3 0

2 years ago

The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t

hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

$f=\dfrac{1}{P}$

$f=\dfrac{1}{45.4}$

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Here, u = infinity

$\dfrac{1}{v}=\dfrac{1}{2.2}$

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0

3 years ago