I can only offer you one which is mushroom. They're fungus' and they can't produce their own food. They get their food from decaying and dead animals and absorb it through their roots.
        
             
        
        
        
Answer:
The constant angular acceleration of the centrifuge = -252.84 rad/s²
Explanation:
We will be using the equations of motion for this calculation.
Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.
First of, we write the given parameters.
w₀ = initial angular velocity = 2πf₀ 
f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s
w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s
θ = 46 revs = 46 × 2π = 289.14 rad
w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)
α = ?
Just like v² = u² + 2ay
w² = w₀² + 2αθ
0 = 382.38² + [2α × (289.14)]
578.29α = -146,214.4644
α = (-146,214.4644/578.29)
α = - 252.84 rad/s²
Hope this Helps!!!
 
        
                    
             
        
        
        
1) the weight of an object at Earth's surface is given by 

, where m is the mass of the object and 

 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 

2) On Mars, the value of the gravitational acceleration is different:

. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: 

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 

<span>6) On Earth, the gravity acceleration is </span>

<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>

<span>
</span>
 
        
             
        
        
        
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore, 
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore, 
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
 
        
             
        
        
        
The answer to the question is:
75m/s
Just do 25*3