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pashok25 [27]
2 years ago
11

Can someone please solve this

Physics
1 answer:
Svetllana [295]2 years ago
7 0

Answer: i can see properly

Explanation:

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PLEASE HELPPP FASTTTTTTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!!In a paragraph of no less than five sentences, explain how these Russ
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<span>These Russian nesting dolls are a good analogy for the strongest magnet in the world, because the magnitis were put inside each other. The magnets would be fitted together. Because they are closer together, the coils of magnet would be stronger. They would be able to magnetize better.</span>
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3 years ago
What is escape velocity ​
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Escape velocity is the speed that an object needs to be traveling to break free of a planet or moon's gravity well and leave it without further propulsion. For example, a spacecraft leaving the surface of Earth needs to be going 7 miles per second, or nearly 25,000 miles per hour to leave without falling back to the surface or falling into orbit.
8 0
2 years ago
PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the
dedylja [7]

CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

5 0
2 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
lina2011 [118]

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

8 0
2 years ago
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