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gizmo_the_mogwai [7]
3 years ago
9

a 62.0 kg curler runs into a stationary 78.1 kg curler and they hold on to each other. Together they move away at 1.29 m/s, west

. What was the original velocity of the 62.0 kg curler?
Physics
1 answer:
Lady_Fox [76]3 years ago
7 0
Momentum P is given by:
P = mv

Given: m₁ = 62 kg, m₂ = 78.1 kg, v₁ = ?, v₂ = 0 m/s, v₁₊₂ = 1.29 m/s

before:
P = m_1 v_1 * m_2 v_2 = m_1 v_1

after:
P = (m_1 + m_2)v_{1+2}

Since momentum is conserved, the momentum before the collision must be equal to the momentum after the collision.

P = m_1 v_1 = (m_1 + m_2)v_{1+2} \\ v_1 =  \frac{(m_1 + m_2)v_{1+2}}{m_1}

Solve for v₁.


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A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t
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Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

v_i = 0

also the acceleration of the stone is due to gravity which is given as

g = 32 ft/s^2

now we have

v_f = 180 ft/s

so from above equation

180^2 - 0 = 2(32)d

d = 506.25 ft

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3 years ago
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If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?
alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

4 0
3 years ago
Napoleon's army captured the island of San Salvador, then moved on to occupy Louisiana and close the port of New Orleans.
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A. true

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3 0
3 years ago
In the diagram, q1 = +6.39*10^_9 C and
ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

q_1=6.39\times 10^{-9} \ C

q_2=3.22\times 10^{-9} \ C

AP=(0.150+0.250)

      =0.40 \ m

BP=0.25 \ m

Now,

At point P, the electric potential will be:

⇒ V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}

By putting values, we get

⇒     =9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]

⇒     =259.695 \ Volt

4 0
3 years ago
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