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Art [367]
3 years ago
7

An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is

thrown? Take g=10ms^2
A. 15m above the top of the building
B. 30 m below the top of the building
C. 15 m below the top of the building
D. 30 m above the building
Physics
1 answer:
Alik [6]3 years ago
8 0
We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>

<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

<span>d = Vo*t + 0.5g*t^2. </span>
<span>d = 10*1 + 5*1^2 = 15 m. <---- OPTION C</span>
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Answer:

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The magnitude of the charge is calculated as follows;

V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\where;\\\\k \ is \ coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2\\\\q = \frac{-1.5 \times 2\times 10^{-3}}{9\times 10^9 } \\\\q = -3.33 \times 10^{-13} \ C\\\\Magnitude \ of \ the\  charge, q = 3.33 \times 10^{-13} \ C

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3 years ago
Condyloid joints are, logically enough, created by joints at condyle bone markings.
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Answer: False

Explanation:

It's received in elliptical cavities.

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2 years ago
A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
mihalych1998 [28]
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2 years ago
A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s
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Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

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Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}

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3 years ago
A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire
soldi70 [24.7K]

Answer:

a)  v=4.0m/s

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Therefore

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v=4.0m/s

b)

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B=\frac{\sqrt{PR}}{vl}

B=\frac{\sqrt{4*0.35}}{4*0.10}

B=2.958T

5 0
2 years ago
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