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2 years ago

If you are doing a "rowing" motion, what muscle of the shoulder are you using?

1 answer:

6 0

The upper back muscles being worked while using a rowing machine .your upper trapezius and rhomboids located between your shoulder blades, and latissimus dorsi located beneath the armpits

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Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

myrzilka [38]

Given: Mass of earth Me = 5.98 x 10²⁴ Kg

Radius of earth r = 6.37 x 10⁶ m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r

but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²

GMe = 4π²r³/T²

T² = 4π²r³/GMe

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds or around 1.4 Hour

3 0

2 years ago

Q:-What is speed????

Margarita [4]

Explanation:

In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

SI unit: m/s, m s−1

s=d/t

5 0

2 years ago

Free points to those who rate this 5 stars and answer "Done". I don't think it lets more than 2 people answer so I guess the fir

chubhunter [2.5K]

Answer:

how do i rate it 5 stars?

Explanation:

8 0

2 years ago

Read 2 more answers

A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If

Gala2k [10]

Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

Let g = 9.81 m/s2

$\Delta E_p = mgH - mgh = mg(H - h) = 0.32*9.81*(2 - 1.2) = 2.51 J$

Since 1.8J of 2.51 J is due to work by air resistance, the rest of the energy (2.51 - 1.8 = 0.71 J) is would go to heating in the ground and ball when it bounces.

5 0

2 years ago

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init

AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

So the total mechanical energy of the system is

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

$E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J$

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

$E=U=\frac{1}{2}kx_{max}^2$

where $x_{max}$ is the maximum stretch. Solving for $x_{max}$ ,

$x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m$

So, 19.4 cm.

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

$U=0$