Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Answer:
x_{cm} = 4.644 10⁶ m
Explanation:
The center of mass is given by the equation
= 1 /
∑
Where M_{total} is the total masses of the system,
is the distance between the particles and
is the masses of each body
Let's apply this equation to our problem
M = Me + m
M = 5.98 10²⁴ + 7.36 10²²
M = 605.36 10²² kg
Let's locate a reference system located in the center of the Earth
Let's calculate
x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]
x_{cm} = 4.644 10⁶ m
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
(a)
M = 1.898 x 10^27 kg
(b)
v = 13.74 km/s
(c) E = 0.28 N/kg
Explanation:
Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s
Radius, r = 6.71 x 10^8 m
G = 6.67 x 10^-11 Nm^2/kg^2
(a) 


M = 1.898 x 10^27 kg
(b) Let v be the orbital velocity


v = 13739.5 m/s
v = 13.74 km/s
(b) The gravitational field E is given by


E = 0.28 N/kg
Answer:
According to the Conservation of Momentum,
Momentum of the gun = momentum of the bullet
M(gun)×V(gun)=m(bullet)×v(bullet)
4kg × V = 0.3kg × 600m/s²
V = (0.3 × 600)/4 = 45 m/s
The recoil velocity on the gun is <em><u>45 m/s</u></em>
<h3><u>45 m/s</u> is the right answer.</h3>