Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
Yes, it is very helpful.
Explanation:
It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.
I hope this helps!
Answer:
6.1 × 10^9 Nm-1
Explanation:
The electric field is given by
E= Kq/d^2
Where;
K= Coulombs constant = 9.0 × 10^9 C
q = magnitude of charge = 1.62×10−6 C
d = distance of separation = 1.53 mm = 1.55 × 10^-3 m
E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2
E= 14.58 × 10^3/2.4 × 10^-6
E= 6.1 × 10^9 Nm-1
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
