You need about how many grams of protein per day?

A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl

soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

$\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s$

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂= I₂ ω₂

I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁ - m v r = I₂ ω₂

Iω₁ - m v r = ( I + m r²) ω₂

Now by putting the all values

Iω₁ - m v r = ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂

0.325 - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

$\omega_2=\dfrac{2\pi \times N_2}{60}$

N₂=20.05 rpm

3 0

3 years ago

A rocket is launched at an angle of 39◦ above the horizontal with an initial speed of 90 m/s. It moves for 7 s along its initial

vagabundo [1.1K]

Answer:

Y=1370.23m

Explanation:

The motion have two moments the first one the time the initial velocity is accelerating then when the engines proceeds to move as a projectile

$a=19 \frac{m}{s^{2} } \\voy=vo*sin(\alpha )\\voy=90*sin(39 )\\y_{o}=0m\\y_{f}=y_{o}+v_{oy}*t+\frac{1}{2}*a*t^{2}\\y_{f}=90*sin(39)*7s+\frac{1}{2}*19\frac{m}{s^{2} }*(7)^{2}\\y_{f}=861.97m$

Now the motion the rocket moves as a projectile so:

$v_{fy}=v_{iy}+a*t\\v_{fy}=90+9.8*7\\v_{fy}=158.6 sin(39)$

Now the final velocity is the initial in the second one

$v_{fy}^{2}=v_{fi}^{2}+2*a*yf \\\\a=g\\$

The maximum altitude Vf=0

$0=v_{fi}^{2}+2*a*yf \\\\yf=\frac{(158.6 sin(39))^{2} }{2*9.8\frac{m}{s^{2} } } \\yf=508.26m$

So total altitude is both altitude of the motion so:

$Y=508.2m+861.97m\\Y=1370.23m$

6 0

3 years ago

Older railroad tracks in the U.S. are made of 12 m-long pieces of steel. When the tracks are laid, gaps are left between the sec

Shkiper50 [21]

Answer:

0.005 m

Explanation:

length of steel (L°) = 12 m

initial temperature (T) = 16 degrees

expected temperature (T') = 50 degrees

We can find how large the gaps should be if the track is not to buckle when the temperature is as high as 50 degrees from the formula below

ΔL = ∝L°ΔT where

ΔL = expansion / gap
∝ = linear expansion coefficient of steel = $12x10^{-6} C^{-1}$
L° = initial length
ΔT = change in temperature

ΔL = $12x10^{-6}$ x 12 x (50-16) = 0.005 m

3 0

3 years ago

When analyzing the results of a scientific experiment, Kevin should base his conclusion on

elena-14-01-66 [18.8K]

The best answer would be b

5 0

2 years ago

Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 8.30

Lunna [17]

Answer:

a. = 93 N

b. = 94.9 N

Explanation:

Answer:

Explanation:

from the question we were given the following:

density of oil (ρ) = 8.3 x $10^{2}$ $\frac{kg}{m^{3} }$

radius of the output plunger (R) = 0.125 m

radius of the input piston (r) = 7.70 x $10^{-3}$ m = 0.0077 m

output force (F2) = 24500 N

input force (F1) = ?

acceleration due to gravity (g) = 9.8 $\frac{m}{s^{2} }$

difference in height of the plunger and piston (h) = 1.3 m

we can find the required input force when the piston and the plunger are at the same height using the equation below

$\frac{F2}{A2}$ = $\frac{F1}{A1 }$

F1 = ( $\frac{F2}{A2 }$ ) x A1

where A1 is the area of the input piston and A2 is the area of the output plunger

Area = π x $radius^{2}$

A2 = π x $0.125^{2}$ = 0.049

A1 = π x $0.0077^{2}$ = 0.000186

recall that from above F1 = ( $\frac{F2}{A2 }$ ) x A1

F1 = ( $\frac{24500}{0.049 }$ ) x 0.000186

F1 = 93 N

we can find the required input force when the height of the piston and the plunger are 1.3 m apart using the equation below

P2 = P1 + ρgh

where P = pressure = $\frac{F}{ A }$

therefore the equation above now becomes

$\frac{F2}{ A2 }$ = $\frac{F1}{ A1 }$ + ρgh

F2 = ( $\frac{F1}{ A1 }$ + ρgh ) x A2

F2 = ( $\frac{24500}{ 0.000186 }$ + ( 8.3 x $10^{2}$ [tex] x 9.8 x 1.3 ) ) x 0.049

F2 = 94.9 N

4 0

4 years ago