If John mows 11.5 meters of lawn from east

Is there air resistance in space?........HELP!!!!!

Leokris [45]

There's no air in space, so there's no air resistance there.

5 0

3 years ago

Read 2 more answers

What part of the visible light spectrum produces the most light?

fomenos

Answer: all colors

.......

7 0

2 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

vodomira [7]

Answer:

A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs

7 0

3 years ago

Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength

soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

$E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J$

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

$V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s$

the shortest de Broglie wavelength for the electrons is given by;

$\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm$

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0

3 years ago