Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
= M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water
Answer:
<h2>
E = 2.8028*10⁻¹⁹ Joules</h2>
Explanation:
The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo
h = planck's constant
fo = threshold frequency
Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹
h = 6.626× 10⁻³⁴ m² kg / s
Substituting this value into the formula to get the energy E
E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴
E = 28.028*10¹⁴⁻³⁴
E = 28.028*10⁻²⁰
E = 2.8028*10⁻¹⁹ Joules
If it is a matter of which way you are going you could lean forward. It would help to put all the weight opposite of where you are falling.
B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
Answer:
a) Vi = 137.2 m/s
b) h = 960.4 m
Explanation:
a)
In order to find the initial speed we will use first equation of motion:
Vf = Vi + gt
where,
Vf = Final velocity = 0 m/s (since ball stops at highest point)
Vi = Initial Velocity = ?
g = - 9.8 m/s² (negative sign for upward moyion)
t = time interval = 14 s
Therefore,
0 m/s = Vi + (-9.8 m/s²)(14 s)
<u>Vi = 137.2 m/s</u>
<u></u>
b)
Now, we use second equation of motion to find height (h):
h = Vi t + (1/2)gt²
h = (137.2 m/s)(14 s) + (1/2)(-9.8 m/s²)(14 s)²
h = 1920.8 m - 960.4 m
<u>h = 960.4 m</u>
