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Juliette [100K]
2 years ago
8

A book with a mass of 1.2 kg sits on a bookshelf. If it has gravitational potential energy of 50 j, how high is the shelf ? The

acceleration of gravity is 9.8 m/s 2

Physics
1 answer:
Tatiana [17]2 years ago
5 0

Answer:

B

Explanation:

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an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart
viva [34]

Answer:

I think 75 m

Explanation:

tell if it was correct

5 0
3 years ago
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eimsori [14]

Explanation:

d =  \frac{1}{2} a {t}^{2}  + vt \\ d = 75 \\ t = 5 \\ v = 0 \\  \\ 75 =  \frac{1}{2}  \times 25a + 0 \\ a = 6 \frac{m}{ {s}^{2} }  \\

7 0
2 years ago
How do your shoe laces relate to a pulley
Margarita [4]
The grommets on your shoe (Holes through which you place the lace) act as the pulley. When you tighten at the top it restricts at the bottom and compresses around your foot. 

The same can be applied to a pulley; If you have the proper set up Pulling one string causes two actions; the first being the most obvious. And the second being its effect. The Grommets of your shoe are the pulley's; while your Shoelaces are just the rope with which you operate the pulley.
8 0
3 years ago
A block (6 kg) initially compresses spring #1 (k = 2000 N/m) by 60 cm from its equilibrium point. When the block is released, it
mart [117]

Answer:

block K = 29.39 J and spring #1   Ke = 360 J

Explanation:

In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work

        W_{fr}= Ef - E₀

Let's look for the energies

Initial

        E₀ = Ke = ½ k₁  x₁²

Final, this is just before starting to compress the spring

        Ef = Ke = ½ m v²

The work of the rubbing force is

       W_{fr}= -fr x

Let's write Newton's second law the y axis

       N-W = 0

      N = W

      fr = μ N

      fr = μ mg

Let's replace

      -μ mg x = ½ m v² - ½ k₁ x₁²

       v² = 2/m (½ k₁ x1₁² -μ mg x)

       v² = 2/6  (½ 2000 0.6²2 - 0.5  6  9.8 1) = 1/3 (360 - 29.4)

       v = 3.13 m / s

With this value we calculate the energy of the block

       K = ½ m v²

       K = ½  6  3.13²

       K = 29.39 J

Calculate eenrgy of the spring ke 1

      Ke = ½ k₁ x₁²

      Ke = ½ 2000 0.60²

      Ke = 360 J

4 0
3 years ago
A bus rolls to a stop along a horizontal road without the driver applying the brakes,
Nataly_w [17]

Answer:

should be d because friction allows things to go faster or slower

5 0
3 years ago
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