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Irina-Kira [14]
3 years ago
15

An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad.

Physics
1 answer:
zaharov [31]3 years ago
7 0

gravitational force is the attraction force of earth on an object which is near the surface of earth

It will not depend on the velocity or acceleration of earth

So it will not change while an object is stationary or it is moving with some acceleration

So here we will say that force will remain the same

F = mg

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Which change would create light?
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Answer:

Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.

Explanation:

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Which statement about covalent bonds is true
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In covalent bonds the atoms share electrons.
7 0
2 years ago
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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work
amid [387]

Answer:

Its  answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

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6 0
2 years ago
A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo
Kitty [74]

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

v^2=u^2+2as

v^2=(7.3)^2+2\times 9.8\times 24  

v=\sqrt{523.69}

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

8 0
2 years ago
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