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nataly862011 [7]
4 years ago
12

PLEASE HELP!! WILL GIVE POINTS

Physics
1 answer:
AURORKA [14]4 years ago
3 0

Answer:

49 kg is the mass of the couch.

Explanation:

GPE = mgh

9800 = m * 10 * 20

9800 = 200m

m = (9800/20) = 49 m

Thenks and mark me brainliest :))

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Imagine you're standing on a skateboard and your friend pushes you to the left with a force of 100 N. Neglecting friction and ai
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

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A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23 m/s. If the ball was caught 42 m from the
ExtremeBDS [4]

Answer:

a) v_{0y} = 9.72 m/s

b) v_{0x} = 20.85 m/s

c) t = 2.01 s

d) h_{max} = 4.82 m

Explanation:

a) The initial vertical velocity is given by:

v_{0y} = v*sin(\theta)

Where:

θ: 25°

v: is the magnitude of the speed = 23 m/s

v_{0y} = 23 m/s*sin(25) = 9.72 m/s

b) The initial horizontal velocity can be calculated as follows:

v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s

c) The flight time can be calculated using the following equation:

v_{0x} = \frac{x}{t}

Where:

x: is the total distance = 42 m

t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s

d) The maximum height is given by:

v_{fy}^{2} = v_{0y}^{2} - 2gh_{max}

Where:

v_{fy}: is the final vertical velocity =0 (at the maximum heigth)

g: is the gravity = 9.81 m/s²

h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m

I hope it helps you!                                  

7 0
3 years ago
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