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A barge is 10 m wide and 60 m long and has vertical sides. The bottom of the boat is 1.2 m below the water surface. What is the

DENIUS [597]

Answer:

<h2>e. 7.1 MN approx.</h2>

Explanation:

Step one:

given data

density of water= 1000kg/m^3

the dimension of the barge

width= 10m

length= 60m

depth of the boat in the water= 1.2m

Hence the volume occupied by the boat is

volume=10*60*1.2

volume= 720m^2

Step two:

Required is the weight of the barge

we can first find the mass using the relation

density = mass/volume

mass= density*volume

mass= 1000*720

mass= 720000kg

Step three:

Weight =mg

g=9.81m/s^2

W=720000*9.81

W=7063200N

divided by 10^6

W=7.06MN

W=7.1MN approx.

5 0

3 years ago

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

4 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

Someone fires a slingshot at a target that is far enough away to take 1.4 seconds to reach. How far below does the target does t

Korolek [52]

There is not enough information to answer the question

6 0

3 years ago

If it took 125 seconds to complete 5 wave cycles, what is the period of the wave?

finlep [7]

The period of the wave is 25 s

<u>Explanation:</u>

To find the period of the wave, we have the formula

T= seconds/ no of cycle

T= 125/5

T= 25 s

The period of the wave is 25 s

4 0

4 years ago