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ziro4ka [17]
3 years ago
13

We’ve already investigated this problem with one spring scale in Think About 34.1. Now, imagine you have two spring scales, A an

d B, connected at the end of the scale that doesn’t move. The end that moves of each spring scale (where you take readings from) is attached to a string that goes over a pulley and connects to a 1 kg mass for both spring scales A and B. 1 kg 1 kg Spring Scale A Spring Scale B (a) State what you think each spring scale will read in this situation. (b) Construct a logical argument that explains why the spring scales read
Physics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

Explained

Explanation:

1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2

2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.

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A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha
goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then  α =15.66°

8 0
3 years ago
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

<u>ac = 3.92 m/s²</u>

5 0
3 years ago
1 point
klio [65]

Answer:

0

10

20

30

40

50

60

70

80

90

100 g

0

100

200

300

400

500

8

9

108

Explanation:

ikkk

8 0
3 years ago
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
PolarNik [594]

Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

8 0
3 years ago
Is it better to wire a house using a series circuit or a parallel circuit?
Svetach [21]

Answer: its better to use parallel because, in parallel connection there will be more advantages than a series connection. and also the electronic devices are wired in series so thats why you should use parralel in house wiring

so its c.

because the parallel is wired through the whole house so if one of the circuits fail youŕe not screwed

Explanation:

5 0
3 years ago
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