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Inga [223]
3 years ago
15

What can be found in a compost bin?

Physics
2 answers:
Angelina_Jolie [31]3 years ago
6 0
D.All of the above is the correct Answer
BigorU [14]3 years ago
5 0
D. all of the above

Hope this helps!
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A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of
Alenkinab [10]

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/s^{2}= 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result

7 0
3 years ago
Engineers can determine properties of a structure that is modeled as a damped spring oscillator, such as a bridge, by applying a
Kazeer [188]

The value of spring constant and the oscillator's damping constant is

K= 6605.667008, b= 0.002884387

Explanation:

For Weakly damping spring oscillator

K/m = W_0^2     (at resonance)

K= mW_0^2

=0.206 * ( 2π * 28.5) ^2

=0.206 * (2π)^2 * (28.5)^2

K= 6605.667008

F = - bV

b= -F/V = -F/ -W_0 * m

=F/W_0 * m

= 0.438N / 2π * 28.5 * 0.848

b= 0.002884387

8 0
3 years ago
What is the weight of a ring tailed lemur that has a mass of 10 kg? Use w = mg (g = -9.8 m/s^2)
marusya05 [52]

m = 10 kg

g = -9.8 m/s2

w = m * g

w= -9.8 * 10

w= -98 N

4 0
3 years ago
If the vehiche has a speed of 24.0 m/s at point a what is the force of the track on the vehicle at this point?
ASHA 777 [7]

Question:

A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers.  If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the vehicle at this point?

<em>See attachment</em>

Answer:

33700 Newton

Explanation:

Given

m = 500kg\\v = 24.0m/s\\r=10m

First, we determine the forces acting on mass m.

They are: the force exerted by the track (Fn) and the weight of the vehicle (W)

So, the net force is:

F_{net} = F_n - W

W=mg

So:

F_{net} = F_n - mg

Make Fn the subject

F_n = F_{net} + mg

Fnet is calculated as:

F_{net} = F_c = \frac{mv^2}{r} --- i.e. the centripetal force

So:

F_n = F_{net} + mg

F_n = \frac{mv^2}{r} + mg

F_n = \frac{500 * 24^2}{10} + 500 * 9.8

F_n = 28800 + 4900

F_n = 33700N

7 0
3 years ago
gThe acceleration of gravity at the surface of Moon is 1.6 m/s2. A 5.0 kg stone thrown upward on Moon reaches a height of 20 m.
noname [10]

Answer:

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using newton's equation of motion,

v² = u²+2gs ..................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity at the surface of moon, s = height reached.

make u the subject of the equation,

u² = v²-2gs

u = √(v²-2gs)................ Equation 2

Note: As the stone is thrown up, v = 0 m/s, g is negative

Given: v = 0 m/s, s = 20 m, g = -1.6 m/s²

Substitute into equation 2

u = √(0-2×20×[-1.6])

u = √64

u = 8 m/s.

(b)

Using,

v = u+ gt

Where t = time of flight to reach the maximum height.

Make t the subject of the equation,

t = (v-u)/g................................... Equation 3

Given: v =  0 m/s, u = 8 m/s, g = - 1.6 m/s²

Substitute into equation 3

t = (0-8)/-1.6

t = -8/-1.6

t = 5 seconds.

6 0
3 years ago
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