Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
Answer:
If transpiration didn't take place water would still be able to enter the roots of a plant
Explanation:
transpiration is the process of water leaving from living organisms to the atmosphere, therefore, if transpiration didn't occur the water would not transpire to the atmosphere and would remain in the root but water absorption would not change because it is a biological need for the living organism as such
Answer:
plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)
Explanation:
The electromagnetic spectrum is the system of frequencies that show electromagnetic radiation, respective wavelengths, and photon energies. Some examples of frequencies found on the electromagnetic spectrum are radio waves, microwaves, infrared, optical, ultraviolet, X-rays, and gamma-rays.
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g 
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g 
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
How energy is conserved
Em₀ = 
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)