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hram777 [196]
3 years ago
7

Slow twitch muscles are needed for?

Physics
2 answers:
kow [346]3 years ago
5 0

The answer to your question is C) Speed or Power Activities

ruslelena [56]3 years ago
4 0

Speed or power activities

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Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1
Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0
2 years ago
Use the concepts of kinetic energy and potential energy to describe the motion of a child on a swing. Why does the child need a
andrew-mc [135]
When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.

The child may need a push from time to time because friction with the air causes loss of energy.
4 0
2 years ago
Read 2 more answers
If the distance between two asteroids is doubled, the gravitational force they exert on each other will.
True [87]
The force will be 4 times smaller.
5 0
3 years ago
Sound waves are simultaneously interpreted through air (going through the external,
laila [671]
Not sure if this is right but I think it’s TRUE sorry if it’s wrong....
4 0
3 years ago
On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou
Feliz [49]

Answer:

Angular velocity, N_f = 242.36 rpm

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, m_{a} = 0.13 * \frac{M}{2} ( since it is 13% of the whole body and each arm is considered)

m_{a} = 0.13 * 37\\m_a = 4.81 kg

Mass of the trunk, m_{t} = M - 2m_{a}

m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2

(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2

The final moment of inertia of the person:

(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2

According to the principle of conservation of angular momentum:

(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm

3 0
2 years ago
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