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3 years ago

What is gluteus maximum

Physics

2 answers:

kramer3 years ago

8 0

Any of three muscles in each buttock that move the thigh, the largest of which is the gluteus maximus

Orlov [11]3 years ago

5 0

Any of three muscles in each buttock that move the thigh, the largest of which is the gluteus maximus.

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If a piece of an object is dropped down vertically is the moment of inertia gonna be 0? And why?

Degger [83]

Answer:

The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.

No so sure

Explanation:

Hope it helps

5 0

3 years ago

A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st

wel

Answer:

the tension T2 when the rock is completely immersed is T2 = 29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V= T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 = 29.05 N

3 0

3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago

Which is a valid velocity reading for an object?

Brums [2.3K]

Answer:45 m/s north

Explanation:

8 0

3 years ago

A 2.4-m high 200-m2 house is maintained at 22Â°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch

kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is <u>$0.50</u> for 10 hours.

Option: a

<u>Explanation</u>:

"Energy discharged by air in every hour" can be determined by,

$\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

$\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}$

$\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Density of air } \rho=1.20 \times 200 \times 2.4$

$\text { Density of air } \rho=576 \mathrm{kg}$

∆T = 10 hours

$\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}$