Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
b) The escalar product a·b is:
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
![(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5Ccdot%20b%20%3D%20%5B%288.6%20%2B%209.3%29i%20%2B%20%285.1%20%2B%209.5%29j%5D%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20%2817.9i%20%2B%2014.6j%29%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20305.17)
d) The component of a along the direction of b is:
I hope it helps you!
Answer:
Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects, and is inversely proportional to the square of the distance between them. The factors G, M, and r are the same for all masses at the surface of the Earth.
Answer:
C
Explanation:
the image formed is real but inverted and magnified
you can remember it by R I M
hope this helps
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
Answer:
V₀ = 5.47 m/s
Explanation:
The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 3.04 m
θ = Launch Angle = 41.7°
V₀ = Minimum Launch Speed = ?
g = 9.81 m/s²
Therefore,
3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)
V₀² = 3.04 m/(0.10126 s²/m)
V₀ = √30.02 m²/s²
<u>V₀ = 5.47 m/s</u>
