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oksano4ka [1.4K]
2 years ago
9

Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo

año batea la cantidad de cuadrangulares del primer año más diez cuadrangulares, el tercer año bateó cinco cuadrangulares menos que el segundo año; si en total de cuadrangulares de los tres años es igual a 135 ¿cuántos cuadrangulares dio este joven en total por cada año?
Ayuda es para mañana :c​
Physics
2 answers:
Brilliant_brown [7]2 years ago
8 0

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

Tcecarenko [31]2 years ago
3 0

no se porque y tu si sabes porque yo no

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Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1
aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴  (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

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