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Anettt [7]
3 years ago
9

An 82.0 kg spacewalking astronaut pushes off a 655 kg satellite, exerting a 95.0 N force for the 0.530 s it takes him to straigh

ten his arms. How far apart are the astronaut and the satellite after 1.0 min.?
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

41.41 m

Explanation:

When force F is applied on an object of mass m for time t and velocity v₁ is created

F X t = mv₁

F = 95 N , t = .53 s, m = 655 kg

95 x .53 = 655 x v₁

v₁ = .0768 m/s

Applying conservation of momentum on man and satellite

m₁ v₁ = m₂v₂

655 x .0768 = 82 xv₂

v₂ = .6134 m/s

their relative velocity

= .6134 + .0768

= .6902 ( they are in opposite direction )

After 60 second distance between them

= 60 x .6902 m

= 41.41 m

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Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

8 0
3 years ago
A stone is dropped from the top of a tower. What is its velocity after 3.0 seconds?
kap26 [50]

For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth. 

The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet.  That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.

3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s. 

That's the vertical component of its velocity.  The horizontal component is
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5 0
4 years ago
What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of
finlep [7]

L = r x p = rmv = mr²ω

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4 0
2 years ago
What label should mateo place in the marked cell? uses electrical energy provides electrical energy measures electric current co
ahrayia [7]

The product label which Mateo should place in the marked cell is that it: B. provides electrical energy.

<h3>What is a product label?</h3>

A product label can be defined as a paper-document with specific information about the chemical elements, nutrients, and other chemical compounds that are present in a product and its general use or function.

In this scenario, the product label which Mateo should place in the marked cell shown in the image attached below is that, this product was designed and developed to provide electrical energy, based on the electrical symbol for battery.

Read more on product label here: brainly.com/question/14446515

#SPJ4

4 0
2 years ago
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