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Kryger [21]
3 years ago
6

On the moon the acceleration due to gravity is about 1/6 that on earth. If A golfer on the moon impacted the same initial veloci

ty to the ball as she does on the earth, how much further with the ball go?
Physics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer: 6 times further

Explanation:

Initial velocity is the same that she uses on earth vertically and horizontally

Vertically we can say

V = -g t

If g is 1/6 earth gravity, then t is 6 times earth time.

Then, horizontally, with the same initial time, no incidence of gravity and knowing t

X= v t

Distance in the moon is 6 times distance in the earth.

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A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

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3 years ago
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In the drawing, water flows from a wide section of a pipe to a narrow section. In which part of the pipe is the volume flow rate
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Answer:

Volume flow rate is the same in both sections of the pipe

Explanation:

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3 years ago
A stone has a volume of 0.50 cm³ and a mass of
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The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m
vovangra [49]
The equation of the car is given by the equation,

                          x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

                  dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

                 dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2. 

If we substitute these values to the equation for displacement,

(0)   , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2)    , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters. 
3 0
3 years ago
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