During a hockey game on a pond, the defenseman passes a 114-g hockey puck over the ice to the center, who fails to catch it. the

Response: 800N

Clarification:

Provided data:

Ball mass = 0.8kg

Contact duration = 0.05 seconds

Final and initial speed = 25m/s

The average force exerted by the ball on the wall can be calculated using the following relationship:

Force (F) = mass (m) * average acceleration (a)

a= (initial velocity (u) + final velocity (v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Thus,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Hence,

The average force magnitude (F)

F=ma

m = ball mass = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

Esta pregunta está incompleta, la pregunta completa es;

La figura muestra un recipiente sellado en la parte superior por un pistón móvil. Dentro del recipiente hay un gas ideal a 1.00 atm. 20.0°C y 1.00 L.

"¿Cuál será la presión dentro del recipiente si el pistón se mueve hasta la marca de 1.60 L mientras se mantiene constante la temperatura del gas?"

Answer:

la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L manteniendo constante la temperatura del gas

Explicación:

Dado que;

P₁ = 1.00 atm

P₂ =?

V₁ = 1 L

V₂ = 1.60 L

la temperatura del gas se mantiene constante

sabemos que;

P₁V₁ = P₂V₂

por lo que hacemos la sustitución

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Por lo tanto, la presión dentro del recipiente será 0.625 atm si el pistón se mueve hasta la marca de 1.60 L mientras la temperatura del gas se mantiene constante

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

Set the initial location of kitten A on the left side of the field (designated as point A) at the origin, running east which is the positive direction. Kitten B starts at position , while kitten A’s beginning spot is .

Kitten A moves with a velocity of , and kitten B with . Their positions over time are described by

The collision occurs when the positions are the same, i.e. when . Solving this gives

Which results in approximately 6.8 seconds, considering significant figures.

Response:

x = 1.63 m

Details:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that the weight of the computer is largely applied to the belt instantaneously, we can implement the constant acceleration equation below

x =

where a = μk.g, thus

x = /2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m