Answer:
Explanation:
Given:
Initial θ = 0 rad (from rest)
Final θ = 14.3 rad
Time, t = 5 s
B.
Angular velocity, w = dθ / dt
= (14.3 - 0)/5
= 2.86 rad/s
A.
Acceleration, ao = dw/dt
Initial angular velocity, wi = 0 rad/s (from rest)
Final angular velocity, wf = 2.86 rad/s
a = (2.86 - 0)/5
= 0.572 rad/s^2
Answer:
minimum angle is 128.69°
Explanation:
given data
player velocity with respect ground v1 = 3.5 m/s
ball velocity with respect himself v2 = 5.6 m/s
to find out
smallest angle
solution
we know ball velocity with respect field will be
ball velocity = v1 +v2
ball velocity = 3.5 + 5.6 = 9.1m/s
we consider angle that player hit ball is θ
then by as per figure triangle
cosθ =
cosθ =
θ = 51.31
so minimum angle is 180 - 51.31 = 128.69°
The reason as to why the substage condenser does not need to be included in computing the magnification and the only component needed is the ocular lens and the objective lenses is because the condenser is only responsible for gathering light and it does not contribute with the magnification of the object under the microscope.
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
One side of the wave changes speed before the other side, causing the wave to move