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Sonbull [250]
3 years ago
7

What is tarzan's speed vf just before he reaches jane? express your answer in meters per second to two significant figures?

Physics
1 answer:
e-lub [12.9K]3 years ago
3 0
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)

Therefore
Ui = mgL(1-cos45)

Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)

Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2

According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s

<span>The speed f T after swing is 7.89 m/s</span>
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The center of the Milky Way galaxy lies in the direction of the _constellation, aboutlight-years away.
Elza [17]

Answer:

Sagittarius

Explanation:

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.

5 0
4 years ago
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Under electrostatic conditions, the electric field just outside the surface of any charged conductor
KatRina [158]

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

3 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
      = 39.6 N/C

Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
How would decreasing the amplitude of the sound waves from a television affect its intensity
Rainbow [258]
<span>The  sound will decrease in volume.</span>
5 0
3 years ago
deon is running at a velocity of 7.9 meters per second. deon has a total mass of 95 kilograms including his helmet, uniform, pad
brilliants [131]

Answer:

712.5 kg m/s

Explanation:

Work out the total momentum before the event (before the collision):

p = m × v

Massof Deon = 95kg

Velocity =7.5m/s

Mass of Chuck = 120kg

Velocity = 0m/s

Momentum of Deon before = 95 × 7.5 = 712.5 kg m/s

Momentum of Chuck before = 120 × 0 = 0 kg m/s

Total momentum before = 712.5+ 0 = 712.5 kg m/s

Working out the total momentum after the event (after the collision):

Because momentum is conserved, total momentum afterwards = 712.5 kg m/s

6 0
3 years ago
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