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sveta [45]
3 years ago
9

A construction crane lowers a beam into place at constant speed. Consider the work Wg done by gravity on the beam and the work W

T done by the tension in the cable on the beam. Which is true?
A) Wi is positive and Wr is positive.
B) We is positive and W'r is negative.
C) Wg is negative and W'r is negative.
D) Wi and W'r are both zero.
E) None of the above
Physics
1 answer:
aalyn [17]3 years ago
5 0

Answer:

Wg is positive and WT negative.

(Letters in options are all wrongly written).

Explanation:

Remember that the work of a force is the internal product between the force and the displacement W=F.d.

Since the displacement is downwards like the weight, the work done by gravity is positive, while the work done by the tension is negative since it points upwards.

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maksim [4K]
From that ragged, motley list of statements, only 'C' is true.
6 0
3 years ago
Select all of the answers that apply.
Kruka [31]
The answer is letter A. meteorite bombardment.
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Earth was bombarded continuously by the remnants of the dust and debris — like asteroids, meteors and comets — until it formed into a solid sphere, pulled into orbit around the sun and began to cool down during the Hellish period (4.5 to 3.8 billion years ago).

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6 0
3 years ago
Read 2 more answers
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
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Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
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Why does the night sky change at various times of the year ?  Here's how to
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The Earth spins once a day. You spin along with the Earth, and your clock is
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head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
Read 2 more answers
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
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