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vfiekz [6]
4 years ago
5

The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay

. For this collision,
a. neither momentum nor kinetic energy is conserved.
b. only momentum is conserved.
c. both momentum and kinetic energy are conserved.
d. only kinetic energy is conserved.
Physics
1 answer:
Artemon [7]4 years ago
8 0

Answer:

b

Explanation:

The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.

This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.

 

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Un móvil con velocidad inicial de 19,8km/h adquiere una aceleración constante de 2,4m/s^2. Determine la velocidad y el espacio r
nata0808 [166]

Responder:

Velocidad = 41.5m / s

Espacio recorrida = 352.5 metros

Explicación:

Dado lo siguiente:

Velocidad inicial (u) = 19.8 km / h

Aceleración (a) = 2.4m / s ^ 2

Tiempo de viaje (t) = 15 s

A.) velocidad después de 15 s

Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s

Usando la ecuación: v = u + at, donde v es la velocidad

v = 5.5 + 2.4 (15)

v = 5.5 + 36

v = 41.5m / s

Espacio recorrida:

v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida

41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S

1722.25 = 30.25 + 4.8S

1722.25 - 30.25 = 4.8S

1692 = 4.8S S = 1692 / 4.8 S = 352.5m

8 0
3 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
Zanzabum

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

(b) 4.4\cdot 10^{-8} m/s^2

Since we know the cross-sectional area of the laser beam:

A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

6 0
3 years ago
A ship maneuvers to within 2.50 x 10^3 m of an islands 1.80 x 10^3 m high mountain peak and fires a projectile at an enemy ship
Neporo4naja [7]

Answer:

Explanation:

Distance between ship and enemy ship

= 500 + 610

= 3110 m

Range of projectile

R = u² sin2θ / g

= (250x 250 sin 150) / 9.8

= 3188m

The projectile falls within a distance of 3188 - 3110 = 78 m from enemy ship

Height of mountain = 1800 m

We shall find the height of projectile when its horizontal displacement is 2500m

x = 2500 , y = ?

u = 2500 ,

y = x / cos θ - .5 g x² /u²cos² θ

\frac{2500}{.2588} - \frac{.5 \times9.8\times2500\times2500}{250\times250\times.2588}

9660 - 7315 m

= 2345 m

It is within 545 m from mountain peak .

8 0
3 years ago
Electrons are continuously being knocked out of air molecules by cosmic ray particles from space. Once the electrons are free, t
nata0808 [166]

(a) 1.25\cdot 10^{-14} J

The change in potential energy of the electron is given by:

\Delta U=q E d

where

q=1.6\cdot 10^{-19}C is the magnitude of the electron's charge

E=150 N/C is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

\Delta U=(1.6\cdot 10^{-19}C)(150 N/C)(520 m)=1.25\cdot 10^{-14} J

(b) 78 kV

The potential difference the electron has moved through is given by

\Delta V=Ed

where

E=150 N/C is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

\Delta V=Ed=(150 N/C)(520 m)=78,000 V=78 kV

4 0
3 years ago
Struggling on this, really need help
Bad White [126]

Answer:

Guessing you just need help with the definition but if it's the question I can still help you.

7 0
3 years ago
Read 2 more answers
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