Answer:
B
Explanation:
That's the answer. Hope it helped!
Answer:
we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
Explanation:
given data
base = 3.60 m
speed u = 8 m/s
height = 1.70 m
to find out
check change in speed
solution
we know here formula for v that is
v² = u² - 2gh ............1 for upward speed
v² = u² + 2gh ............2 for projected speed
so here put all value and find v with h = 3.60 - 1.70 = 1.9 m
v² = 8² - 2(9.8) 1.9 = 26.76
v² = 8² + 2(9.8) 1.9 = 101.24
v = 5.173 m/s ..............3
v = 10.061 m/s ...................4
so change in speed form 3 and 4 equation
change in speed = v - u = 8 - 5.173 = 2.827 m/s .................5
change in speed = v - u = 10.061 - 8 = 2.061 m/s ..................6
so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
Answer:
Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.
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Answer:
The time for the entire race is 11.39 sec.
Explanation:
Given that,
Distance = 15 m
Remainder distance = 100 m
Suppose A sprinter begins a race with an acceleration of 3.4 m/s².
We need to calculate the time
Using equation of motion
Put the value in the equation
We need to calculate the final velocity of sprinter
Using equation of motion again
Put the value into the formula
We need to calculate the distance covers by sprinter
The sprinter need to covers only 85 m.
We need to calculate the time
Using formula of time
Put the value into the formula
We need to calculate the time for the entire race
Put the value into the formula
Hence, The time for the entire race is 11.39 sec.
