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fomenos
3 years ago
6

4 kg of water was

Physics
1 answer:
morpeh [17]3 years ago
6 0

Given :

Mass =5kg

T

1

=20

∘

C,T

2

=100

∘

C

ΔT=100−20=80

∘

C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

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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}

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F_M=23.53N

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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

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F_j=73.53N

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8 0
7 months ago
A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

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radius(r) = 35 cm

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when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

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\omega_f = 585.37 \ rev/s

3 0
3 years ago
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Ugo [173]
Answer: A

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5 0
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Answer:

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7 0
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