Answer:
72 kg
Explanation:
The computation of the mass of the person is shown below:
As we know that
Water displaced weight = Raft weight + person weight
P_w.V = M_r + M_p
where
P_w denotes the water density i.e. 1000 kg/m^3
V denotes the volume of the water in m^3
Now
1000 × 0.224 = 152 + M_p
M_p = 224 - 152
= 72 kg
For this case, the first thing you should do is write the kinematic motion equation of the block.
We have then:
vf = vo + a * t
Where,
vf: Final speed.
vo: Initial speed.
a: acceleration.
t: time.
Substituting the values:
(16) = (0) + a * (16)
Clearing the acceleration:
a = 16/16 = 1m / s ^ 2
Note: the other data for this case are not used in this problem.
answer:
The acceleration of the box is 1m / s ^ 2
Rock climbing. Free diving. Sky diving. Dog sledding.
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
