Answer:
Explanation:
Work is defined as the scalar product of force and distance
W=F•d
Given that
F = 8.5i + -8.5j. +×-=-
F=8.5i-8.5j
d = 2.5i + cj
If the work in the practice is zero, then W=0
therefore,
W=F•ds
0=F•ds
0=(8.5i -8.5j)•(2.5i + cj)
Note that
i.i=j.j=k.k=1
i.j=j.i=k.i=i.k=j.k=k.j=0
So applying this
0=(8.5i -8.5j)•(2.5i + cj)
0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)
0=21.25-8.5c
Therefore,
8.5c=21.25
c=21.25/8.5
c=2.5
Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
<u></u>
(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
<u></u>
(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>
<span>Direction and magnitude it is also </span>determined<span> by the gravitational acceleration any impacts or interruptions are ignored. </span>
Answer:
How much kinetic energy does a 4 Kg cat have while running at 9 m/s?
its 5 J of kinetic energy.
Explanation:
