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Crazy boy [7]
3 years ago
12

A Level is set up midway between two wood hubs that are about 300 ft. apart. The rod reading on hub A is 9.09 ft. and the readin

g on hub B is 5.65 ft. The level is moved so it is along the side of hub B and about 6 feet away from it. Readings are taken are taken from the new set up with the reading on hub A of 7.11 ft. and on hub B of 3.70 ft. (10 points) (see Section 2.11) a. What is the correct difference in elevation between the hubs A and B
Physics
1 answer:
Zarrin [17]3 years ago
4 0

Answer:

<em>The correct difference is 3.44 ft</em>

Explanation:

The correct difference is given as

D=R_A-R_B

D=9.09-5.65

D=3.44 ft

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What is the mass of a large ship that has a momentum of 1.60x10^9 kg*m/s and is moving at a velocity of 10m/s?
Elden [556K]

Answer:

160000000 kg.

Explanation:

p=mv

p=1.6x10^9

v=10m/s

rearrange and substitute:

(1.6x10^9)=m(10)

m=(1.6x10^9)/10

m= 1.6x10^8 kg.

7 0
3 years ago
9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
Korolek [52]

<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

=686/544.8

=1.259ms^2

4 0
3 years ago
A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?
ICE Princess25 [194]

Answer:

660V

Explanation:

V=IR

V=?,I=11A,R=60w

V=60×11

=660V

3 0
3 years ago
The unknown quantities to be determined are (a) the capacitive reactance, (b) the maximum and the rms voltages from the source,
Digiron [165]

The frequency f of the AC source is determined as 0.446 Xc.

<h3>Frequency of the AC source</h3>

The frequency of the AC source is calculated as follows;

Xc ≡ 1/2fC

where;

  • Xc is the capacitive reactance
  • f is frequency
  • C is capacitance

fC = 2Xc

f = 2Xc / C

f = (2Xc)/4.48

f = 0.446 Xc

Thus, the frequency f of the AC source is determined as 0.446 Xc.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

8 0
2 years ago
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