Carnivores like lions, wolves, cheetahs and leopards are also found in temperate grasslands. Other animals of this region include deer, prairie dogs, mice, jack rabbits, skunks, coyotes, snakes, foxes, owls, badgers, blackbirds, grasshoppers, meadowlarks, sparrows, quails, hawks and hyenas.
Answer:
Cool question! First step is to find the time taken to fall
57
m
, then to find the horizontal velocity needed to cover
24
m
in that time. In this case the answer is
7.0
m
s
−
1
.
Explanation:
This is a less typical projectile motion question, but it's still projectile motion. This means the horizontal and vertical directions can be considered separately. We assume that the initial vertical velocity,
u
y
=
0
m
s
−
1
, and we are trying to find the required initial horizontal velocity,
u
x
.
To find the time taken to fall
57
m
:
s
=
u
t
+
1
2
a
t
2
Since
u
=
0
, we can rearrange this to:
t
=
√
2
s
a
=
√
2
⋅
57
9.8
=
3.41
s
The horizontal velocity will be constant (ignoring air resistance), so to cover
24
m
in
3.41
s
will be given by:
v
=
s
t
→
u
x
=
24
3.41
=
7.0
m
s
−
1
Answer link
Explanation:
define the term specific heat capacity of water
Answer:
a) t =12[s]; b) x = 348[m]
Explanation:
We can solve this problem using the following kinematics equations:
a)
where:
vf = final velocity = 12 [m/s]
vo= initial velocity = 6 [m/s]
a = acceleration = 0.5[m/s^2]
t = time [s]
Now clearing the time t, we have:
![t=\frac{v_{f} -v_{o} }{a} \\t = \frac{12-6}{0.5} \\t=12[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_%7Bf%7D%20-v_%7Bo%7D%20%7D%7Ba%7D%20%5C%5Ct%20%3D%20%5Cfrac%7B12-6%7D%7B0.5%7D%20%5C%5Ct%3D12%5Bs%5D)
b)
We can calculate the displacement for the first 12 [s] then using the equation for the constant velocity we can calculate the other displacement for the 20[s].
![v_{f}^{2}= v_{o}^{2}+2*a*x_{1} \\therefore\\x_{1} =\frac{v_{f}^{2}-v_{o}^{2}}{2*a} \\x_{1} =\frac{12^{2}-6^{2}}{2*.5}\\x_{1} =108[m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20%20v_%7Bo%7D%5E%7B2%7D%2B2%2Aa%2Ax_%7B1%7D%20%5C%5Ctherefore%5C%5Cx_%7B1%7D%20%3D%5Cfrac%7Bv_%7Bf%7D%5E%7B2%7D-v_%7Bo%7D%5E%7B2%7D%7D%7B2%2Aa%7D%20%5C%5Cx_%7B1%7D%20%3D%5Cfrac%7B12%5E%7B2%7D-6%5E%7B2%7D%7D%7B2%2A.5%7D%5C%5Cx_%7B1%7D%20%3D108%5Bm%5D)
The we can calculate the second displacement for the constant velocity:
![x_{2} =x_{o}+v*t_{2} \\ x_{2} =0+12*(20)\\x_{2} =240[m]](https://tex.z-dn.net/?f=x_%7B2%7D%20%3Dx_%7Bo%7D%2Bv%2At_%7B2%7D%20%5C%5C%20x_%7B2%7D%20%20%3D0%2B12%2A%2820%29%5C%5Cx_%7B2%7D%20%3D240%5Bm%5D)
x = x1 + x2
x = 108 + 240
x = 348[m]
