Odpowiedź :
Odpowiedź:
1.Dane:
h=20m
g=10m/s^2
Oblicz:
V=?
Rozwiązanie:
\begin{gathered} < var > Ek=Ep\\\\ \frac{1}{2}mV^2=mgh\\\\ mV^2=2mgh\\ V^2=2gh\\ V=\sqrt{2gh}\\ V=\sqrt{2*10*20}\\ V=\sqrt{400}\\\\ V=20\frac{m}{s} < /var > \end{gathered}
<var>Ek=Ep
2
1
mV
2
=mgh
mV
2
=2mgh
V
2
=2gh
V=
2gh
V=
2∗10∗20
V=
400
V=20
s
m
</var>
2.Dane:
h=90m-20m=10m
Oblicz:
V=?
Rozwiązanie:
\begin{gathered} < var > Ek=Ep\\\\ \frac{1}{2}mV^2=mgh\\\\ mV^2=2mgh\\ V^2=2gh\\ V=\sqrt{2gh}\\ V=\sqrt{2*10*10}\\ V=\sqrt{200}\\\\ V\approx14,15\frac{m}{s} < /var > \end{gathered}
<var>Ek=Ep
2
1
mV
2
=mgh
mV
2
=2mgh
V
2
=2gh
V=
2gh
V=
2∗10∗10
V=
200
V≈14,15
s
m
</var>
3.Dane:
V1=12m/s
V2=20m/s
Oblicz:
h=?
Rozwiązanie:
\begin{gathered} < var > Ek1=Ep+Ek2\\\\ \frac{1}{2}mV1^2=mgh+\frac{1}{2}mV2^2\\\\ V1^2=2gh+V2^2\\ 2gh=V1^2-V2^2\\ 2gh=20^2-12^2\\ 2gh=400-144\\ 2gh=256\\\\ h=\frac{256}{2g}\\\\ h=\frac{256}{2*10}\\\\h=\frac{256}{20}\\\\h=12,8m < /var > \end{gathered}
<var>Ek1=Ep+Ek2
2
1
mV1
2
=mgh+
2
1
mV2
2
V1
2
=2gh+V2
2
2gh=V1
2
−V2
2
2gh=20
2
−12
2
2gh=400−144
2gh=256
h=
2g
256
h=
2∗10
256
h=
20
256
h=12,8m</var>
4.Dane:
V=4m/s
Oblicz:
h=?
Rozwiązanie:
\begin{gathered} < var > Ek=Ep\\\\ \frac{1}{2}mV^2=mgh\\\\ \frac{1}{2}V^2=gh\\\\ V^2=2gh\\\\ h=\frac{V^2}{2g}\\\\ h=\frac{4^2}{2*10}\\\\ h=\frac{16}{20}\\\\ h=0,8m < /var > \end{gathered}
<var>Ek=Ep
2
1
mV
2
=mgh
2
1
V
2
=gh
V
2
=2gh
h=
2g
V
2
h=
2∗10
4
2
h=
20
16
h=0,8m</var>
5.Rozwiążę te zadanie dla 1m ponieważ nie ma podanej wysokości:
Dane:
h=1m
g=10m/s^2
< var > \eta=75\% < /var ><var>η=75%</var>
Oblicz:
V=?
Rozwiązanie:
\begin{gathered} < var > \eta Ep=Ek\\\\ 75\%Ep=Ek\\\\ 0,75*mgh=\frac{mV^2}{2}\\\\ 0,75*gh=\frac{V^2}{2}\\\\ 0,75*2gh=V^2\\\\ V^2=0,75*2*10*1\\\\ V^2=0,75*20\\\ V^2=15\\ V=\sqrt{15}\approx 3,87\frac{m}{s} < /var > \end{gathered}
<var>ηEp=Ek
75%Ep=Ek
0,75∗mgh=
2
mV
2
0,75∗gh=
2
V
2
0,75∗2gh=V
2
V
2
=0,75∗2∗10∗1
V
2
=0,75∗20
V
2
=15
V=
15
≈3,87
s
m
</var>
