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Vilka [71]
4 years ago
6

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

that all the energy goes into the arrow, with what speed does the arrow leave the bow? (41.7 m/s)b. If the arrow is shot straight up, how high does it rise? (88.9 m)
Physics
1 answer:
Vlad [161]4 years ago
4 0

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

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Si está probando u motor para un nuevo modelo de coche ; este es capaz de pasar de 0 a 100 km/h en 7,5 segundos . Si el coche ti
RideAnS [48]

Answer:

La fuerza que realiza el motor es 2035 N.

Explanation:

Podemos encontrar la fuerza del motor usando la siguiente ecuación:

F = ma   (1)

En donde:

m: es la masa del coche = 550 kg

a: es la aceleración

Se puede calcular la aceleración mediante la siguiente ecuación cinemática:

v_{f} = v_{0} + at   (2)

En donde:

v_{f}: es la velocidad final del coche = 100 km/h

v_{0}: es la velocidad inicial del coche = 0

t: es el tiempo = 7,5 s

Resolviendo la ecuación (2) para "a" tenemos:

a = \frac{v_{f} - v_{0}}{t} = \frac{100\frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s} - 0}{7,5 s} = 3,70 m/s^{2}

Entonces, la fuerza es:

F = ma = 550 kg*3,70 m/s^{2} = 2035 N

Por lo tanto, la fuerza que realiza el motor es 2035 N.

Espero que te sea de utilidad!

7 0
3 years ago
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