Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
Answer:
5880lb-ft of work is done
Explanation:
The length of the heavy rope is given as 60ft and the weight per length is 0.7lb/ft.
Therefore, the total weight of the heavy rope is
60×0.7 =42lb.
The work done in pulling the heavy rope to the top of the building is w = Fd.
Where
F is force is measured in pounds;42lb
d is distance through which the heavy rope is to be pulled measured in feet; 140ft
w= 42lb×140ft= 5880lb-ft
Like charges repel, while opposite charges attract.
Answer:
A.3.64 m
Explanation:
Because
- v=(fλ)
- (1382)=(380)λ
- λ=3.637m~3.64m
<em>where</em><em> </em><em>,</em><em>v</em><em>=</em><em>velocity</em>
<em>f</em><em>=</em><em>frequency</em><em> </em>
<em>λ</em><em>=</em><em>wave</em><em> </em><em>length</em><em> </em>
