Solution
x(t) = 8 cos t, x(5π/6)= 8 cos(<span>5π/6)
</span>cos(5π/6)=cos(3π/6 + 2π/6 )=cos(π/3 +π/2)= - sin π/3 (cos (x+<span>π/2)= -sinx)
</span>x(t) = -8sin <span>π/3 = - 4 .sqrt3
</span>v(t) = -8sint = -8sin (π/3 +<span>π/2)= -8 cosπ/3 </span>(sin (x+π/2)= cosx)
v(t) =<span> -8 cosπ/3 = -8/2= - 4
</span>a(5π/6) = - 8cost = -(- sin π/3)= 4 .<span>sqrt3
</span>a(5π/6) = 4 .<span>sqrt3</span>
Answer:
Explanation:
The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.
As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.
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^
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
