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Evgen [1.6K]
3 years ago
10

༼ つ ◕_◕ ༽つ༼ つ ◕_◕ ༽つ༼ つ ◕_◕ ༽つ☜(゚ヮ゚☜)(☞゚ヮ゚)☞☜(゚ヮ゚☜)(☞゚ヮ゚)☞( ´・・)ノ(._.`)

Physics
2 answers:
Kitty [74]3 years ago
8 0
༼ つ ◕_◕ ༽つ༼ つ ◕_◕ ༽つ༼ つ ◕_◕ ༽つ☜(゚ヮ゚☜)(☞゚ヮ゚)☞☜(゚ヮ゚☜)(☞゚ヮ゚)☞( ´・・)ノ(._.`)?
Vlada [557]3 years ago
4 0

Answer:

°∠°      °∡°      °×°

Explanation:

Yes. This is perfect. It is the best.

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Which physical property of matter depends on push
liberstina [14]

Answer:

I would say contact forces

Explanation:

This questions is kind of missing some info but I do know that contact forces need to be either pushed or pulled for a force to happen.

Hope this helped

4 0
3 years ago
A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pull
lora16 [44]

Answer:

E_l = 1.713 J

Explanation:

Given data:

mass of block is M_b = 0.84 kg

radius of block = 0.061 m

moment of inertia is 6.20 \times 10^{-3} kg m^2

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum  we have

M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 +  E_{loss}

0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times  0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l

solving for energy loss

E_l = 1.713 J

3 0
3 years ago
Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
Which of these BEST describes the concept of the rock cycle?
choli [55]
D  rocks are continually changing and type of rock mat be transformed into another type of appropriate process.

8 0
3 years ago
Which instrument, a cello or a violin, produces a higher-pitched sound? Explain your answer.
anastassius [24]
I'm pretty sure a violin produces a higher-pitched sound. I play the violin and it's pretty loud. (Hope this helps)
4 0
3 years ago
Read 2 more answers
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