Ruby is in miami an texts her cousin, Xavier, in Seattel. her clock says 1:00 am

30. The slope of a velocity-time graph will give

CaHeK987 [17]

Answer:

The acceleration

Explanation:

The slope of velocity time graph is the acceleration

5 0

3 years ago

The human eye is most sensitive to green light of wavelength 505 nm . Experiments have found that when people are kept in a dark

liq [111]

1. $5.94\cdot 10^{14} Hz$

The frequency of a photon is given by:

$f=\frac{c}{\lambda}$

where

c is the speed of light

$\lambda$ is the wavelength

The wavelength of the photon in this problem is

$\lambda=505 nm=5.05\cdot 10^{-7}m$

So, the frequency of the photon is

$f=\frac{3\cdot 10^8 m/s}{5.05\cdot 10^{-7} m}=5.94\cdot 10^{14} Hz$

2. $3.94\cdot 10^{-19}J, 2.46 eV$

The energy of a photon is given by

$E=hf$

where

h is the Planck constant

f is the frequency of the photon

The frequency of the photon in this problem is

$f=5.94\cdot 10^{14} Hz$

so its energy in Joules is

$E=(6.63\cdot 10^{-34}Js)(5.94\cdot 10^{14}Hz)=3.94\cdot 10^{-19}J$

And since

$1 eV = 1.6\cdot 10^{-19}J$

The energy in eV is

$E=\frac{3.94\cdot 10^{-19} J}{1.6\cdot 10^{-19}J/eV}=2.46 eV$

7 0

3 years ago

A 15.00 kg particle starts from the origin at time zero. Its velocity as a function of time is given by = 8t2î + 5tĵ where is in

Elden [556K]

Answer:

Explanation:

I will assume the equation reads:

v = 8t²î + 5tĵ

The velocity v is the time derivative of the position x.

$x = \int\limits^t_0 {v} \, dt = \int\limits^t_0 {8t^{2}\hat i + 5t\hat j} \, dt = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j |^t_0 = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j - \frac{8}{3} \hat i - \frac{5}{2} \hat j\\ x = \frac{8}{3} (t^{3} - 1 )\hat i + \frac{5}{2} (t^{2} - 1 )\hat j$

4 0

4 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter