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2 years ago

6

a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and

the displacement ?

1 answer:

Cloud [144]2 years ago

8 0

I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx

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Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h

ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

$Y+F=600$ ....(I)

$X=T$ .....(II)

On taking moment about base

$X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0$

Put the value into the formula

$X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0$

$3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30$

$1.598T=450-1.5(600-F)$ ....(III)

We need to calculate the force for ladder

$2F=600\trimes 2$

$F=600\ N$

We need to calculate the tension in the rope

From equation (3)

$1.598T=450-1.5(600-600)$

$1.598T=450$

$T=\dfrac{450}{1.598}$

$T=281.60\ N$

Hence, The tension in the rope is 281.60 N.

7 0

2 years ago

Read 2 more answers

The curved section of a horizontal highway is a circular unbanked arc of radius 740m. If the coefficient of static friction betw

fiasKO [112]

Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
$v \: = \: \sqrt{gr \tan( \alpha ) }$
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s

6 0

2 years ago

Read 2 more answers

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$

The energy of the system having mass 2m is,

$E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2$

For the two expressions mentioned above remember that the variables mean

m = mass

$\omega =$ Angular velocity

A = Amplitude

The energies of the two system are same then,

$E_m = E_{2m}$

$\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2$

$\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}$

Remember that

$k = m\omega^2 \rightarrow \omega^2 = k/m$

Replacing this value we have then

$\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}$

But the value of the mass was previously given, then

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}$

$\frac{A_1}{A_2} = 1$

Therefore the ratio of the oscillation amplitudes it is the same.

5 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 $m = 1.67 * 10^{-27} \ kg$

So

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

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

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

If a 10 meter ramp helps you move a 500 kg object up 1 meter. What was the mechanical advantage of the ramp?

sattari [20]

The mechanical advantage of an inclined plane is

(Length of the incline) / (its height)

= (10m) / (1m)

= 10 .

It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.

6 0

2 years ago