Answer:
a) v₀ = 32.64 m / s
, b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = 
t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
I think it is False
hope this helps :3
Answer:
Explanation:
If you add or subtract a proton from the nucleus, you create a new element. If you add or subtract a neutron from the nucleus, you create a new isotope of the same element you started with. In a neutral atom, the number of positively charged protons in the nucleus is equal to the number of orbiting electrons.
Answer:
The type of decay illustrated by the equation is a Beta decay.
Option B is correct.
Explanation:
Complete Question
What type of decay is illustrated by the equation below?
²¹⁴₈₃Bi → ²¹⁴₈₄Po + ⁰₋₁e
- alpha decay
- beta decay
- positron emission
- electron capture
Solution
The type of radioactive decay is usually discernable from studying the products and reactants (the parent nucleus/atom, the daughter nucleus and the emitted or absorbed particles) of the radioactive decay.
The changes in atomic and mass numbers is accounted for in the particular particle that that is emitted and subsequently termed the type of radioactive decay that is going on.
For example, this equation represented shows that the Bi atom splits into a Po and a Beta particle. Hence, it is easy and straight forward to see that this radioactive decay is to relaese a Beta particle and the decay is a Beta decay.
(Note that, just as presented in the equation, a Beta particle has a mass number of 0 and an atomic number -1, kind of similar to an electron).
- An alpha decay releases an alpha particle.
- A beta decay releases a Beta particle.
- A positron emission releases a positron.
- An electron capture involves an electron attacking the parent atom/nucleus or the reactant.
Hope this Helps!!!
