They are positive and remain inside the nucleus.
Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
<u>D = - 0.5i - 0.25j</u>
Therefore,
<u>Dx = -0.5</u>
<u>Dy = -0.25</u>
First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used is
For constant acceleration:
a = v,final - v,initial /t
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s After 2.8 seconds, the speed of the blue car is 10.64 m/s.
Answer:
b liquids
Explanation:
this is because liquids take the shape of their container while solids stay solid and do not change shape and solids, liquids and gases all have definite volume.
Answer:
85.8 m/s
Explanation:
We know that the length of the circular path, L the plane travels is
L = rθ where r = radius of path and θ = angle covered
Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt
where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path
Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,
θ = θ₀ + ωt = 0 + ωt = ωt
So, v = rdθ/dt + θdr/dt
v = rω + ωtdr/dt
v = (r + tdr/dt)ω
v = (r + tdr/dt)v'/r
v = v' + tv'/r(dr/dt)
v = v'[1 + t(dr/dt)/r]
Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)
So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]
v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]
v = 110 m/s[1 + 11.0 s(-0.02/s)]
v = 110 m/s[1 - 0.22]
v = 110 m/s(0.78)
v = 85.8 m/s
