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2 years ago

Show your work and resoning for the below requirement.

Physics

1 answer:

Leno4ka [110]2 years ago

4 0

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

- $T_{y}$ L + W₂ L + W₁ L / 2 = 0

T_{y} = W₂ + W₁ / 2

T_{y} = 120 + 150/2

T_{y} = 195 lb

we use trigonometry to find the cable tension

sin 30 = T_{y} / T

T = T_{y} / sin 30

T = 195 / sin 30

T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

T < 500 lb

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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0

2 years ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

3 0

2 years ago

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

8 0

2 years ago

A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu

kvasek [131]

Answer:

First Question

$E = 1.065*10^{-12} \ J$

Second Question

The wavelength is for an X-ray

Explanation:

From the question we are told that

The width of the wall is $w = 10\ fm = 10*10^{-15 }\ m$

The first excited state is $n_1 = 2$

The ground state is $n_0 = 1$

Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

$E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }$

Here h is the Planck's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

m is the mass of proton with value $m = 1.67 * 10^{-27} \ kg$

$E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }$

=> $E = 1.065*10^{-12} \ J$

Generally the energy of the photon emitted is also mathematically represented as

$E = \frac{h * c }{ \lambda }$

=> $\lambda = \frac{h * c }{E }$

=> $\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }$

=> $\lambda = 1.87*10^{-10} \ m$

Generally the range of wavelength of X-ray is $10^{-8} \to 1)^{-12}$

So this wavelength is for an X-ray.

8 0

2 years ago