3. В колбе объемом 1,2 л содержится 3*1022 молекул. Какова средняя<br> Кинетическая энергия молекул?

How can you find average speed

BabaBlast [244]

Average speed =

(total distance covered)

divided by

(total time spent covering the distance)

4 0

4 years ago

Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m

Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0

3 years ago

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

choli [55]

Answer:

a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

I = Δp

F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

F = m v₀ / t

F = 63.5 7.89 / 3.99 10⁻³

F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

F = 63.5 7.89 / 0.205

F = 2.44 10³ N

c) The net force is

F_net = Sum F

F_net = F - W

F_net = F - mg

let's calculate

F_net = 2.44 10³ - 63.5 9.8

F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

6 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles

slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

(c) The kinetic energy when its position is 2.50 cm is 0.0195 J

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²

0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²

0.02912 = 1/2 × 0.035 × v²

1/2 × 0.035 × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s

(c) When its position is 2.50 cm = 0.025 m, we have;

E = PE + KE

0.032 = 1/2 × 40 × 0.025² + KE

KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

4 0

3 years ago

3. В колбе объемом 1,2 л содержится 3*1022 молекул. Какова средняя Кинетическая энергия молекул?