Answer:
D)535.024N
Explanation:
Hello! The most important thing to solve this exercise is to make a good free body diagram of the sled on the hill, our coordinate system is aligned with the slope while the direction of the weight force will always go down, then divide the weight vector into components and we found Wy, I attached the drawing!
Wy=mgcos24
Wy=(59.7)(9.81)(cos24)=535.024N
Answer:
rate of heat exchange the animal and its environment. Feathers, vasoconstriction, Countercurrent heat exchanger and blubber are all mechanisms which restrict the transfer of heat from the animal body
Explanation:
Answer:
<em>The tension in the web is 0.017738 N</em>
Explanation:
<u>Net Force</u>
The net force exerted on an object is the sum of the vectors of each individual force applied to an object.
If the net force equals 0, then the object is at rest or moving at a constant speed.
The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.
A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:
T=W=m.g
The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:
The tension in the web is 0.017738 N
Answer:
.y₂ = 0.5704 m
, y2= 44.47 m,
Explanation:,
For this exercise we will use the kinematics relations
Ball
y₁ = y₀ + v₀₁ t
y₀ = 11.0 m
v₀₁ = 5.10 m / s
Pellet
y₂ = 0 + v₀₂ t - ½ g t²
V₀₂ = 39.0 m / s
At the meeting point the two bodies have the same height
y₁ = y₂
y₀ + v₀₁ t = v₀₂ t -1/2 g t²
11 + 5.1 t = 39 t - ½ 9.8 t²
4.9 t² - 33.9 t +11 = 0
.t2 - 6,918 + 2,245 = 0
t = [6,918 ±√ 6,918 2 - 4 2,245)] / 2
t = [6,918 ± 6.2356.] / 2
t₁ = 6.58 s
t₂ = 0.3412 s
Let's calculate the positions for each time
t₂ = 0.3412 s
y₂ = 39 t - ½ 9.8 t2
y₂ = 39 0.3112 - ½ 9.8 0.3412²
y₂ = 0.5704 m
.t1 = 6.58 s
.₂ = 39 6.58 - ½ 9.8 6.58 ^ 2
y2= 44.47 m
The masses of both objects and the distance between their centers.