When treating the electron as a wave, the concept of electrons orbiting the nucleus allows for the distinct energy levels.The diameter of electron orbits matches the orbit radii which also discrete the energy levels.

The small number of deflected particles were repelled from powerful heavy positively charged central cores which is known as the atomic nucleus.

8 0

3 years ago

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago