INQUIRY IS A TYPE OF SCIENCE PROMPTED BY A QUESTION<br> TRUE OR FALSE

A bowling ball of mass 5.8 kg moves in a

evablogger [386]

Answer:

Velocity of the ping pong ball must be = V2= 6,035.34m/s

Explanation:

M1= momentum of the bowling ball

m1 = mass of the bowling ball= 5.8kg

v1= velocity of the bowling ball= 1.59m/s

M2= momentum of the ping pong ball

m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg

v2= velocity of the ping pong ball

Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s

Momentum of the ping pong ball = M2= M1= m2v2

= 0.001528 *v2= 9.222

v2= 9.222/0.001528= 6,035.34 m/s

7 0

4 years ago

A ball is dropped off a building and falls past a window that is 2.2m long. If it takes .28s for the ball to cross the window wh

Sidana [21]

0.616 is the distance from the top
of the building to the top of the window.

8 0

2 years ago

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

$\lambda$ = Wavelength

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

Hence, the answer is b) twice the energy of each photon of the red light.

7 0

3 years ago

Read 2 more answers

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

AleksAgata [21]

Answer:

A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

A_res = 2A

A_resultant = 2 .01

A_resulting = 0.2 m

8 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$