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Nataly_w [17]
4 years ago
5

Hunter wants to measure the weight of a dumbbell. he should

Physics
1 answer:
pshichka [43]4 years ago
7 0
He should cleverly, using charm and subterfuge, entice the dumbbell slowly in the direction of where he had hidden the scale. As long as Hunter isn't obvious about it, the dumbbell will swallow his faux charm, and will remain naively oblivious to his true purpose.
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QUESTION 1
Sveta_85 [38]

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

7 0
3 years ago
Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law
Grace [21]

Answer:

Following are the answer to this question:

Explanation:

In option (a):

  • The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.  
  • Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

  • Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.  
  • Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.

8 0
3 years ago
2. How long will it take for a ball thrown vertically upward with an initial velocity of
KengaRu [80]

Answer:

13.33 seconds

Explanation:

At maximum height, the equation of motion becomes:

<em> v = u + at</em>

Since the object was thrown vertically, the initial velocity (u) is zero and the acceleration (a) becomes the acceleration due to gravity (10 m/s2). The equation becomes:

<em>v = at</em>

<em>v = 480 k</em>m/hr = 133.333 m/s

    10t = 133.333

       t = 133.333/10

           t = 13.33 seconds.

<em>The time for the ball thrown vertically with a velocity of 480 km/hr to reach the maximum height is </em><em>13.33 seconds</em><em>.</em>

6 0
3 years ago
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0
3 years ago
A stone whirled on a string experiences a centripetal acceleration of 10 m/s². If the string were shortened to half its length (
LUCKY_DIMON [66]

Answer:

a = 80 m/s²

Explanation:

10 = v²/r

a = (2v)²/(r/2)

a = 4v²•(2/r)

a = 8(v²/r)

a = 8(10)

a = 80 m/s²

4 0
3 years ago
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