I'm not sure tbh, but if it helps, heres a game that helped me understand bases acids and neutrals for chem in 8th grade! http://static.lawrencehallofscience.org/scienceview/scienceview.berkeley.edu/html/showcase/flash/jui... I know it looks cheesy, but it helps! Chem is something youll have until college, and it helps to learn about it. You can make drinks using acids, bases, and neutrals!

8 0

3 years ago

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

3 0

3 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago