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xeze [42]
3 years ago
6

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo

ng is required to reach this height?
For part A) I did: y=11m, g=9.81m/s^2, Vy0= 18 m/s, y0=0, (V^2)y=?

V^2 y= V^2 y0- 2g (y-y0)
--> V^2 y=18^2m/s-2(9.8)(11m-0m)
--> Vy=Square root ( 324m/s-215.6)= 10.4m/s

Part B) I started out with:
y=y0 + Vy0t - (1/2) gt^2
--> 11= 18t m/s-(1/2) 9.8t^2
--> -11 + 18t- 9.8t^2
Physics
1 answer:
aliina [53]3 years ago
5 0
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
 </span> 
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El kilogramo de fuerza, o kilopond, es una unidad de fuerza métrica gravitacional. Es igual a la magnitud de la fuerza ejercida sobre un kilogramo de masa en un campo gravitatorio de 9.80665 m / s².
7 0
3 years ago
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
arsen [322]

magnetic field due to a finite straight conductor is given by

B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)

here since it forms an equilateral triangle so we will have

\theta_1 = \theta_2 = 60 degre

also the perpendicular distance of the point from the wire is

r = \frac{a}{2\sqrt3}

now from the above equation magnetic field due to one wire is given by

B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)

B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)

B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

5 0
3 years ago
A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from
ladessa [460]

Answer:

P_1 = 1166.7 Watt

P_2 = 2000 Watt

Explanation:

Average power for the human sprinter is given as

Power = \frac{\Delta E}{\Delta t}

so we have

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}

P_1 = 1166.7 Watt

Average power for greyhound is given as

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}

P_2 = 2000 Watt

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3 years ago
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3 years ago
A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
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