<span>A. </span>Let’s
say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
<span>v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
<span>The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>
Setting the origin to be the end corner of the
counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is
negative since mug is going down)
<span>1.50m = v0x t
----> v0x= 1.50/t</span>
<span>-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s</span>
Calculating for v0x:
v0x = 3.10 m/s
<span>B. </span>v0x
is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
<span>vy = v0y + at where a=-g
(negative since going down)</span>
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø =
56.87˚<span> below the horizontal</span>
Radio waves are the waves with the lowest energy in the electromagnetic spectrum. X-rays and gamma rays are the highest. Sound is not part of the electromagnetic spectrum.
Answer:
I think it is the last one.
Explanation:
I am not sure because i am stuck on this one, too.
Answer:
beta particles
Explanation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Given mass = 14.0 g
Molar mass = 137 g/mol
![\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%20of%20cesium%7D%3D%5Cfrac%7B14.0g%7D%7B137g%2Fmol%7D%3D0.102moles)
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number
of particles.
1 mole of cesium contains atoms =
0.102 moles of cesium contains atoms =
The relation of atoms with time for radioactivbe decay is:
![N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}](https://tex.z-dn.net/?f=N_t%3DN_0%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5E%7B%5Cfrac%7Bt%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7D)
Where
=atoms left undecayed
= initial atoms
t = time taken for decay = 3 minutes
= half life = 30.0 years =
minutes
The fraction that decays : ![1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}](https://tex.z-dn.net/?f=1-%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7B3%7D%7B1.577%5Ctimes%2010%5E7%7D%7D%3D1.32%5Ctimes%2010%5E%7B-7%7D)
Amount of particles that decay is = ![0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}](https://tex.z-dn.net/?f=0.614%5Ctimes%2010%5E%7B23%7D%5Ctimes%201.32%5Ctimes%2010%5E%7B-7%7D%3D0.81%5Ctimes%2010%5E%7B16%7D)
Thus
beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.