Answer:
a) 0.462 m/s^2
b) 31.5 rad/s
c) 381 rad
d) 135m
Explanation:
the linear acceleration is given by:

the angular speed is given by:

to calculate how many radians have the wheel turned we need the apply the following formula:

the distance is given by:


Answer:103 pounds
Explanation:
Given
width of window 
height of window 
standard atmospheric pressure 
Also 
Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .



The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.
At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.
They would both be measuring what they see correctly.
The synapse is actually the link between 2 neurons. Now when
an action potential contacts the synaptic knob of a neuron, the voltage-gate
calcium channels are unlocked, resulting in an influx of positively charged
calcium ions into the cell. This makes the vesicles containing
neurotransmitters, for example acetylcholine, to travel towards the
pre-synaptic membrane. When the vesicle arrives at the membrane, the contents
are released into the synaptic cleft by exocytosis. Neurotransmitters disperse
across the space, down to its concentration gradient, up until it reaches the
post-synaptic membrane, where it connects to the correct neuroreceptors. Connecting
to the neuroreceptors results in depolarisation in the post-syanaptic neuron as
voltage-gated sodium channels are also opened, and the positively charged
sodium ions travel into the cell. When adequate neurotransmitters bind to
neuroreceptors, the post-synaptic membrane overcame the threshold level of
depolarisation and an action potential is made and the impulse is transmitted.
Explanation:
(a) Frequency of radar energy, 
The relation between frequency and wavelength is given by :




or

(b) If wavelength, 




or
f = 1.2 GHz
Hence, this is the required solution.