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Greeley [361]
3 years ago
5

The "lead" in pencils is a graphite composition with a Young’s modulus of about 1×1010N/m21×1010⁢N/m2. Calculate the change in l

ength of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long.
a) 0.03 mm
b) 0.1 mm
c) 0.3 mm
d) 1 mm
Physics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

b) 0.1 mm

Explanation:

Given that

E= 1 x 10¹⁰ N/m²

F= 4 N

d= 0.5 mm

L = 60 mm

We know that elongation due to force F given as

\Delta L=\dfrac{FL}{AE}

\Delta L=\dfrac{FL}{\dfrac{\pi d^2}{4}\times E}

\Delta L=\dfrac{4\times 60}{\dfrac{\pi \times 0.5^2}{4}\times 10^4}

ΔL = 0.12 mm

Therefore the answer is -

b) 0.1 mm

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How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?
quester [9]

Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

W=q\Delta V

where

W is the work done

q is the charge

\Delta V is the potential difference

In this problem, we have

q = 0.0050 C is the charge

\Delta V=9.0 V is the potential difference

Using the formula, we find the work done:

W=(0.0050 C)(9.0 V)=0.045 J

4 0
3 years ago
Read 2 more answers
An object is in uniform circular motion, tracing an angel at 30 degrees every 0.010 seconds. What's the period of this motion an
Neko [114]
Here's the rule you need to know
in order to answer this question:

                     1 full circle ==> 360 degrees .

Got that ?

Now you could set up a proportion:

     (30 degrees) / (0.01 second)  =  (360 degrees) / (time for full period)

Cross-multiply the proportion:

     (30°) · (period)  =  (360°) · (0.01 sec)

Divide each side by (30°) :    Period = (360° · 0.01 sec) / (30°)

                                                     =  (3.6° · sec) / (30°)

                                                     =  (3.6 / 30)  sec

                                                     =      0.12  sec .
___________________________________

Another way to look at it:

30°        takes    0.01 second
60°        takes    0.02 second
90°        takes    0.03 second
120°      takes    0.04 second
150°      takes    0.05 second
180°      takes    0.06 second
210°      takes    0.07 second
240°      takes    0.08 second
270°      takes    0.09 second
300°      takes    0.10 second
330°      takes    0.11 second
360°      takes   0.12 second

7 0
4 years ago
How do you calculate the mass of an object accelerating at22.35m/s2 with a force of 120N
Arte-miy333 [17]

Answer:

The mass of object is calculated as 5.36 kg

Explanation:

The known terms to find the mass are:

           acceleration of object (a) = 22.35 m/s^{2}

                        Force exerted (F) = 120N

                        mass of an object (m) = ?

From Newton's second law of motion;

                                   F = ma

                           or, 120 = m × 22.35

                          or, m= \frac{120}{22.35} kg

                           ∴ m = 5.36 kg

3 0
3 years ago
A force of 50 n acts on abody of mass 5 kg .calculate the acceleration produced​
Leokris [45]

Answer:

a = 10 m/s²

Explanation:

<u><em>Given:</em></u>

Force = F = 50 N

Mass = m = 5 kg

<u><em>Required:</em></u>

Acceleration = a = ?

<u><em>Formula:</em></u>

F = ma

<u><em>Solution:</em></u>

<em>Rearranging the formula for a</em>

=> a = F/m

=> a = 50/5

=> a = 10 m/s²

8 0
3 years ago
Two 1.1 kg masses are 1 m apart (center to center) on a frictionless table. Each has +10 JC of charge. What is the initial accel
sweet-ann [11.9K]

Answer:

acceleration = 0.8181 m/s²

Explanation:

given data

mass = 1.1 kg

apart d = 1 m

charge q = 10 μC

to find out

What is the initial acceleration

solution

we know that acceleration is

acceleration = \frac{force}{mass}   .................1

here force = k \frac{q1q2}{r^2}

here q1 q2 is charge and r is distance and Coulomb constant k = 9 × 10^{9} Nm²/C²

force = 9*10^{9} \frac{(10*10^{-6})^2}{1^2}

force = 0.9 N

so  from equation 1

acceleration = \frac{0.9}{1.1}

acceleration = 0.8181 m/s²

6 0
3 years ago
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