Answer:
D = 104.4 m
Explanation:
We are given two displacement vectors. One in north direction other in east direction. We know that north and east directions are perpendicular to each other. Hence, the displacements vectors are also perpendicular to each other. Therefore, there resultant can be found by using Pythagora's Theorem like rectangular components method.
D = √(Dₓ² + Dy²)
where,
D = Magnitude of vector sum of both displacements = ?
Dₓ = Magnitude of Displacement Vector in east direction = 30 m
Dy = Magnitude of Displacement Vector in North Direction = 100 m
Therefore,
D = √[(30 m)² + (100 m)²]
<u>D = 104.4 m</u>
Answer:
2.63m
Explanation:
In order to use this expression, we must calculate the values of Velocity (V0) and Velocity(Ve). The initial velocity has only a horizontal component V(0) = V(0x), the final velocity also has a horizontal component since the ball it's at a point of trajectory, V(e) = V(ef). No forces act in the horizontal direction so the momentum of the ball in this direction is conserved, hence, V(0) = V(e)
Therefore, h(e) = h(0) = 2.63m
Amplitude: the height of the wave<span>, measured in meters
</span><span>Wavelength: the distance between adjacent crests, measured in meters
</span>
Answer:
<em>The rock reaches a maximum height of 442.225 m and was thrown at a speed of 93.1 m/s</em>
Explanation:
<u>Vertical Launch</u>
The rock is launched up with a certain speed vo, it reaches its highest point at a height h when the speed is zero, and then it stars a free-fall motion until it hits the ground at time t.
The maximum height is given by
The rock takes the same time to reach this point as to return to the ground, thus we can calculate the maximum height
The rock was thrown up at a speed
Answer:
A = 0.325 Bq
Explanation:
given,
half life of ¹⁴C = 5730 years
fixed fraction 1.30 × 10⁻¹² of ¹²C
half life = 5730 years
T_{1/2} = 5730 x 365 x 24 x 60 x 60
= 1.807 x 10¹¹ s
radioactive decay constant
λ =
λ =
λ = 3.835 x 10⁻¹² /s
number of atom
mass m = 1.30 g
n = 6.524 x 10²²
Number of ¹⁴C atoms in 1.3 g of sample N ' =
= 1.30 × 10⁻¹² x 6.524 x 10²²
= 8.482 x 10¹⁰
Required activity, A = λ N
A = 3.835 x 10⁻¹² x 8.482 x 10¹⁰
A = 0.325 Bq