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Nina [5.8K]
3 years ago
6

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws

a rock straight upwards with an initial speed of 15 m/s, while the other person throws their rock straight downwards with an initial speed of 15 m/s. If both of the rocks miss the building and continue down to the street,
a. How long will it take for the rock thrown upwards to reach the ground?
b. How long will it take for the rock thrown downwards to reach the ground?
c. How fast will the upward thrown rock be travelling just before it hits the ground? The downward thrown rock?
Physics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

C

Explanation:

Hope this helps!!!!

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Momentums equation is just p=mv
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Newton’s second law of motion says when a is applied to a it causes it to.
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4 years ago
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

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