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Citrus2011 [14]
3 years ago
14

As part of a NASA experiment, golfer Tiger Woods drives a golf ball on the moon, where g = 1.60 m/s2. He launches a golf ball wi

th a speed of 285 km/h, at an angle of 42° with the horizontal. What horizontal distance will his drive travel before landing back on the surface of the moon. Ignore the curvature of the moon.​
Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

Horizontal distance, R = vo2 sin(2x45)/g

v = 285 km/h = 79.17 m/s

R = 79.172 x 1/1.6

R = 3917 m

R = 3.9 km

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The Large Hadron collider (LHC) is a huge piece of equipment designed and built in order to make new scientific discoveries. The
Ket [755]

Answer:

D

Explanation:

"The Large Hadron collider (LHC) is a <em>huge piece of equipment designed and built in order to make new scientific discoveries.</em>"

3 0
3 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
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Kryger [21]

Answer:

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5 0
3 years ago
A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

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Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

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μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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aniked [119]

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7 0
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