Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi

Answer:

Part A) Electric fields at the designated point due to charges q₁ and q₂:

E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) The overall electric field at P (Ep)

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Explanation:

Conceptual analysis

The electric field at point P caused by a point charge is calculated as:

E = k*q/d²

E: Electric field measured in N/C

q: charge magnitude in Newtons (N)

k: electric constant measured in N*m²/C²

d: distance from the charge q to point P in meters (m)

Equivalence:

1 nC = 10⁻⁹ C

1 cm = 10⁻² m

Data:

k = 9 * 10⁹ N*m²/C²

q₁ = -6.00 nC = -6 * 10⁻⁹ C

q₂ = +3.00 nC = +3 * 10⁻⁹ C

d₁ = 4 cm = 4 * 10⁻² m

d₂ = 5 * 10⁻² m

Part A) Calculation for electric fields at point from q₁ and q₂:

Refer to the attached illustration:

E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.

E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from the charge.

E₁ = k*q₁/d₁² = 9 * 10⁹ * 6 * 10⁻⁹ / (4 * 10⁻²)² = 33.75 * 10³ N/C

E₂ = k*q₂/d₂²= 9 * 10⁹ * 3 * 10⁻⁹ / (5 * 10⁻²)² = 10.8 * 10³ N/C

E₁ = 33.75 * 10³ N/C (-j)

E₂x = E₂cosβ = 10.8 * (3/5) = 6.48 * 10³ N/C

E₂y = E₂sinβ = 10.8 * (4/5) = 8.64 * 10³ N/C

E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) Calculation for net electric field at P (Ep)

The electric field at point P from multiple point charges is the vector sum of the individual electric fields.

Ep = Epx (i) + Epy (j)

Epx = E₂x = 6.48 * 10³ N/C (-i)

Epy = E₁y + E₂y = (33.75 * 10³ (-j) + 8.64 * 10³ (+j)) N/C = 25.11 * 10³ (-j) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Answer:

A. Yes, the ball clears the crossbar by 2.83 meters

Explanation:

This situation pertains to projectile motion.

The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s

The vertical velocity component is 26 sin 35 = 14.9 m/s

The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:

= distance / horizontal speed

= 54.9 / 21.3

= 2.577 seconds.

The vertical distance achieved during this time is:

h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.

h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²

= 38.39 - 32.54

= 5.85 m

Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m

Answer:

change in KE = -12.95 Btu

Explanation:

provided data

mass = 3000-lbm

initial velocity of vehicle vi = 10 mph = 14ft/s

final velocity of vehicle vf = 0 mph = 0 ft/s

solution

the crumple zone is designed to absorb kinetic energy upon impact

thus the change in KE is related to the initial and final speeds, expressed as:

change in KE = 0.5 × m × (vf² - vi² ).................1

Substituting in the parameters yields:

change in KE = 0.5 × 3000 × (0² - 14.7² )

change in KE = -324135 × *  

change in KE = -12.95 Btu

Answer:

A flea can attain a maximum elevation of 51 mm.

Explanation:

Hello!

The following equations describe the height and velocity of the flea:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

In free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = flea's height at time t.

h0 = initial height.

v0 = starting velocity.

t = time interval.

a = flea's acceleration while jumping.

v = flea's velocity at that specific time.

g = gravitational acceleration.

Initially, we need to determine the time taken for the flea to attain a height of 0.0005 m. This will help us calculate the flea's velocity during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we assume the ground as the origin, thus h0 = 0. Since the flea starts stationary, v0 = 0. Therefore:

h = 1/2 · a · t²

We need to find the value of t when h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Next, we calculate the velocity achieved during that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

At a height of 0.50 mm, the flea's velocity stands at 1.00 m/s. This initial speed will reduce due to gravity's downward pull. When the speed reaches zero, the flea will have reached its peak height. Using the velocity equation, let's determine the time taken to reach maximum height (v = 0):

v = v0 + g · t

At peak height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, we can compute the height attained by the flea during this time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

A flea reaches a maximum height of 51 mm.

Réponse:

Ve(m) = √(19.2/m)

Ve(0.35) = 7.407 m/s

Explication:

Données:

- La masse de la balle = m

- La vitesse d'entrée de la balle est Vi = Ve

- La vitesse finale de la balle Vf = 0.5*Ve

- La force de frottement constante sur la balle due à la paille est F = 6 N

- L'épaisseur du tas de paille est s = 1.2 m

- On suppose que le lancer se fait horizontalement et qu'on ignore les effets de la gravité

Objectif:

Déterminer une expression pour la vitesse d'entrée typique en fonction de l'inertie de la balle

Quelle est la vitesse d'entrée typique si la balle a une inertie de 0.35 kg?

Solution:

- Pour connaître la vitesse d'entrée comme fonction de l'inertie, nous allons utiliser la troisième équation du mouvement comme suit:

                                    Vf² = Vi² + 2*a*s

où, a est l'accélération de la balle dans la paille. Nous utiliserons la loi du mouvement de Newton pour cela:

                                    F_net = m*a

La seule force agissant sur la balle en traversant la paille est la force de friction F:

                                    - F = m*a

                                    a = -F/m

- Insérer toutes les quantités dans la troisième équation du mouvement:

                                    (0.5Ve)² = Ve² - 2*F*s / m

                                    0.75Ve² = 2*F*s / m

                                    Ve = √(8*F*s/3*m)

- Remplacer les valeurs:

                                    Ve(m) = √(8*6*1.2/3*m)

                                    Ve(m) = √(19.2/m)

- Pour l'inertie de la balle m = 0.35 kg, la vitesse d'entrée est:

                                    Ve(0.35) = √(19.2/0.35)

                                    Ve(0.35) = 7.407 m/s