<span>The proper </span><span>battery cable connection when jumping two automotive batteries is : </span><span>(a) negative to negative / positive to positive.
</span><span>Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. </span>
<span>Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.
</span>
<span>Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.</span>
Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.
Answer:
600km/h as u are on a platform moving at the speed of 600 km/h where u are moving in relativity to the plane it's self.
