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3 years ago

A stationary block of mass 35 Kg is suspended on a string.What is the tension in the string ?( Neglect the mass of the string)

Physics

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

Tstring = 343.35 [N].

Explanation:

Since the block is in equilibrium, we can say that the sum of the forces on the Y-axis or vertical is equal to zero.

∑Fy = 0

$T_{string}-W=0$

where:

Tstring = tension of the string [N]

W = weight of the block [N]

And the weight can be calculated multiplying the mass of the block by the gravity acceleration.

Now replacing:

$T_{string}=m*g\\T_{string}=35*9.81\\T_{string}=343.35[N]$

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An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard

UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,

velocity of the observer = 17 m/s

speed of the sound = 343 m/s

velocity of the source = 0 m/s

frequency emitted from the source = 2550 Hz

$f = f_0(\dfrac{v-v_0}{v-v_s})$

$f = 2550\times (\dfrac{343+17}{343-0})$

velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz

hence, the frequency heard by the observer is equal to 2677 Hz

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1. As the sound leaves the horn the sound waves are at first close to each other and as they move outwards they become further apart. The closer the sound waves are the louder the noise.

As the car gets the closer the sound waves get closer, so the horn becomes louder.

2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

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3 years ago

28. In the diagram above which symbol represents the light bulb ?

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28. It's B
29. It's A

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3 years ago

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The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

alukav5142 [94]

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

$1\ light\ year=3.72\times 10^{17}\ inches$

So, $4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches$

$4.218\ light\ year=1.56\times 10^{18}\ inches$

We need to convert 4.218 light-years barley corns.

Since, 1 barleycorn = 1/3 inch

$1\ inch=3\ barleycorn$

$1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn$

So, the nearest star to the Earth is at a distance of $4.68\times 10^{18}\ barleycorn$ . Hence, this is the required solution.

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3 years ago

An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field

lawyer [7]

Answer:

a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

Em₀ = U = e DV

final point. Where they hit the target

Em_f = K = ½ m v2

energy is conserved

Em₀ = Em_f

e ΔV = ½ m v²

ΔV = $\frac{1}{2}$ mv²/e (1)

If the speed of light is c and this is 100% then 1% is

v = 1% c = c / 100

v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

$K_e = K_p$

K_p = ½ m v_e²

K_p = $\frac{1}{2}$ 9.1 10⁻³¹ (3 10⁶)²

K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

ΔV = Kp / M

ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

e ΔV = ½ M v²

v² = 2 e ΔV / M

v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

v² = 49,035 10⁸

v = 7 10⁴ m / s

Relation

v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

v/c= 2.33 10⁻⁴

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3 years ago