Answer:
1.68 s
Explanation:
From newton's equation of motion,
a = (v-u)/t.................................. Equation 1
Making t the subject of the equation
t =(v-u)g............................. Equation 2
Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.
Note: Taking upward to be negative and down ward to be positive,
Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²
t = (0-8.20)/-9.8
t = -8.20/-9.8
t = 0.84 s.
But,
T = 2t
Where T = time taken for the bowling pin to return to the juggler's hand.
T = 2(0.84)
T = 1.68 s.
T = 1.68 s
- Height (h) = 10 m
- Density (ρ) = 1000 Kg/m^3
- Acceleration due to gravity (g) = 10 m/s^2
- We know, pressure in a fluid = hρg
- Therefore, the pressure exerted by a column of fresh water
- = hρg
- = (10 × 1000 × 10) Pa
- = 100000 Pa
<u>Answer</u><u>:</u>
<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>
Hope you could understand.
If you have any query, feel free to ask.
Hey
Potential Difference given is : 2V
Resistance is : 2 ohms
By Ohm's Law, one can easily utilize the relation :
Where, { v , i , r } are the potential difference, current and Resistance Respectively.
Hence,
Hence, the Current is 1 Ampere
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a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
