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anastassius [24]
3 years ago
11

A zebra starts from rest and accelerates at 1.9 m/s how far has a zebra gone in five seconds

Physics
1 answer:
Maksim231197 [3]3 years ago
5 0
To that multiply 1.9 by 5 to get 9.5.  The zebra has gone 9.5 m.
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The beginning of the Phanerozoic is marked by what occurrence
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The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.
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10. A cyclist travels at 15m/s for 4 minutes. How far does she go?
stiks02 [169]

Answer:

speed = <u>distance</u>

time

15 = <u>X</u>

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4 years ago
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Zina [86]

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

distancia=\frac{1 segundo*90 km}{3600 segundos}

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

distancia= 25 metros

<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

6 0
4 years ago
Find the speed vfinal of the joined cars after the collision. mastering physics
Tanya [424]
<span>Px = 0 Py = 2mV second, Px = mVcosφ Py = –mVsinφ add the components Rx = mVcosφ Ry = 2mV – mVsinφ Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²) and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) simplifying Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²) Vf = (V/3)âš((cosφ)² + (2 – sinφ)²) Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ)) Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ)) using the identity sin²(Ď)+cos²(Ď) = 1 Vf = (V/3)âš1 + 4 – 2sinφ) Vf = (V/3)âš(5 – 2sinφ)</span>
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4 years ago
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