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solmaris [256]
3 years ago
10

A ball is tossed straight up from the surface of a small, sphericalasteroid with no atmosphere. The aball rises to a height equa

l tothe asteroid's raidus and then falls straight down toward thesurface of the asteroid.
1. What forces act on the ball while it is on the way up?

1) only a decreasing gravitational force that acts downward.
2) only an increasing gravitational forces that acts downward
3) only a constant gravitation force that acts downward
4) both a constant gravitational force that acts downward and adecreasing force that acts upward
5) no forces act on the ball

2. the acceleration of the ball at the top of its path is1) at its max vlue for the ball's flightb) Equal to the accleration at the surface of the asteroidc) equal to .5 the acceleration at the surface of the asteroidd) equal to 1/4 the acceleration at the surface of theasteroide) zero
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

1. 3) only a constant gravitation force that acts downward

2. b) Equal to the accleration at the surface of the asteroidc)

Explanation:

1.

  • In absence of atmosphere, the force that will act on the steroid is the force of gravity due to asteroid.
  • Gravitational force is a long range force acting always attractive in nature.
  • It is a contact-less field force which does not requires a medium.

Mathematically given as:

F=G.\frac{M_1.M_2}{R^2}

where:

G = gravitational constant

M_1\ \&\ M_2 are the mass of the two objects

R= radial distance between the two objects.

2.

The acceleration of the ball at the top height of the path is still under the influence of the gravity of the two masses so it will be equal to the acceleration due to gravity at the surface of the asteroid. Acceleration always remains constant.

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At which angle must a laser beam enter the water for no refraction to occur?
abruzzese [7]

Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).

Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story:  If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

5 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

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3 years ago
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ValentinkaMS [17]
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3 years ago
A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 3
galina1969 [7]

Explanation:

Internal energy = heat + work

U = Q + W

Since there's no change in volume (rigid walls), W = 0.

U = Q

U = n Cᵥ ΔT

U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

3 0
3 years ago
Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

5 0
3 years ago
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