OPTIONS :
A.) the force that the ball exerts on the wall
B.) the frictional force between the wall and the ball
C.) the acceleration of the ball as it approaches the wall
D.) the normal force that the wall exerts on the ball
Answer: D.) the normal force that the wall exerts on the ball
Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.
In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.
The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.
Answer:
10.52 m
Explanation:
The power radiated by a body is given by
P = σεAT⁴ where ε = emissivity = 0.97, T = temperature = 30 C + 273 = 303 K, A = surface area of human body = 1.8 m², σ = 5.67 × 10⁻⁴ W/m²K⁴
P = σεAT⁴ = 5.67 × 10⁻⁸ W/m²K⁴ × 0.97 × 1.8 m² × (303)⁴ = 834.45 W
This is the power radiated by the human body.
The intensity I = P/A where A = 4πr² where r = distance from human body.
I = P/4πr²
r = (√P/πI)/2
If the python is able to detect an intensity of 0.60 W/m², with a power of 834.45 W emitted by the human body, the maximum distance r, is thus
r = (√P/πI)/2 = (√834.45/0.60π)/2 = 21.04/2 = 10.52 m
So, the maximum distance at which a python could detect your presence is 10.52 m.
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
