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elixir [45]
3 years ago
10

Calculate the potential energy of a five KG object sitting at the top of a 2 m ramp

Physics
1 answer:
docker41 [41]3 years ago
5 0
5 x 10 x 200 = 10000
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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
3 years ago
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zheka24 [161]

Answer:

Friction force is independent of the direction of the contacting surfaces

Explanation:

It can go any way depending on how much force is being out on it.

5 0
3 years ago
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san4es73 [151]

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

7 0
3 years ago
Which statement correctly compares sound and light waves?
vitfil [10]
Light travels as transverse waves and can go through the vacuum of space, while sound has longitude waves and needs to travel through something solid. It can’t travel through a vacuum of space like light.
3 0
3 years ago
Read 2 more answers
The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium
Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz

Given:

Energy = 378 \frac{kJ}{mol}

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × 10^{3} = 6.63 × 10^{-34} × v

v = 111.31 × 10^{37} Hertz

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz


7 0
3 years ago
Read 2 more answers
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