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Oxana [17]
3 years ago
12

A 6.0 kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the inc

line, it's speed is vf. At what distance from the top of the incline is the speed of the block 0.5vf?
Physics
1 answer:
erica [24]3 years ago
8 0
Since 
<span>Vf^2 = 2*a*S </span>
<span>Given S=3.6m, thus </span>
<span>a = Vf^2/(2*3.6) </span>
<span>a = Vf^2/7.2 </span>

<span>Let d be the distance along the slope at which the velocity is 0.5Vf, then </span>
<span>(0.5Vf)^2 = 2*a*d </span>
<span>or </span>
<span>d = (0.5*Vf)^2/(2*a) </span>
<span>with a = Vf^2/7.2, we have </span>
<span>d = 0.9 m</span>
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What is 75 miles per hour (mi/h) in meters per second (m/s)? 1 mile = 1.6 km.
Sedbober [7]

Answer:33.33 m/s

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8 0
3 years ago
A copper transmission cable 140 {\rm km} long and 12.5 {\rm cm} in diameter carries a current of 115 {\rm A}.
SpyIntel [72]

Answer:

1. V=22.0407\ V

2. Q=9126837\ J

Explanation:

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current through the copper wire, i=115\ A

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<u>We know that the resistance of a wire is given as:</u>

R=\rho.\frac{l}{a}

where:

a= cross sectional area of the wire

R=1.68\times 10^{-8}\times\frac{140000}{\pi.\frac{0.125^2}{4} }

R=0.1917\ \Omega

1.

<u>Now from the  Ohm's law:</u>

V=R.i

V=0.1917\times 115

V=22.0407\ V

2.

<u>From the Joules law the thermal energy generated as an effect of electrical energy is mathematically given as:</u>

<u />Q=i^2.R.t

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Q=115^2\times 0.1917\times 3600 (∵ we have 3600 seconds in 1 hour)

Q=9126837\ J

5 0
3 years ago
Air is heated in a glass bottle. The heat energy added to the air is 2.0 × 104 joules. What is the change in internal energy of
Liono4ka [1.6K]
The change in internal energy of the gas is \Delta U = 2.0 \cdot 10^4 J.

In fact, the 1st law of thermodynamics states that the change in internal energy of a system is equal to the amount of heat given to the system (Q) plus the work done on the system (W):
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In this example, no work is done on the bottle so W=0, while the heat given to the system is Q=2.0 \cdot 10^4 J, so the change in internal energy of the gas is
\Delta U = Q = 2.0 \cdot 10^4 J
4 0
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