How long does it take the statue to reach the ground ?

A student performs Young’s double-slit experiment with slits that are separated by 0.160 mm. She observes the interference patte

Bogdan [553]

Answer:

Wavelenght is 4.53x10^-7m

Explanation:

Detailed explanation and calculation is shown in the image below

5 0

2 years ago

A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c

Dafna11 [192]

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball = 5kg

Speed of clay ball = 25m/s

mass of clay ball at rest = 10kg

speed of clay ball at rest = 0m/s

Unknown:

Velocity after collision = ?

Solution:

Since the balls stick together, this is an inelastic collision:

m1v1 + m2v2 = v(m1 + m2)

5(25) + 10(0) = v (5 + 10)

125 = 15v

v = 8.3m/s

7 0

2 years ago

Water changes from a liquid to a gas in the process of evaporation. Please select the best answer from the choices provided T F

const2013 [10]

The correct answer is T.

6 0

3 years ago

Read 2 more answers

A 5000.0-kg car goes from 0 m/s to 70 m/s in 3600 seconds. What is the accelera

antiseptic1488 [7]

given:

mass = 5000 kg

u (initial velocity) = 0 m/s

v (final velocity) = 70 m/s

time taken to change velocity = 3600 s

acceleration = v - u / t

a = 70 - 0 / 3600

a = 70 / 3600

a = 0.0194 m/s2 (approx)

given: mass = 5000 kg

acceleration = 0.0194 (found in 1st part)

force = mass * acceleration

f = ma

f = 5000*0.0194 = 97 N

therefore the acceleration will be 0.0194 m/s2

and the force involved in acceleration will be 97 N

4 0

2 years ago

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

7 0

2 years ago