Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
Distance = 2AU / tan1.0
If you mean 1.0 is in degrees, then Distance = 114.58 AU
Answer:
Angle with the +x axis is θ = 79.599degree
Then the velocity of owner = 1.235m/s
Explanation:
Given that the mass of dog is m1 =26.2 kg
velocity of dog is u1 = 3.02 m/s (north)
mass of cat is m2 = 5.3 kg
velocity is u2 = 2.74 m/s (east )
Mass of owner is M = 65.1 kg
Consider the east direction along +x axis andnorth along +y
momentum of dog is Py = m1 x u1
= 79.124 kg.m/s (j)
momentum of cat is Px = m2 x u2
= 14.522 kg.m/s (i)
Then the net magnitude of momentum is P = (Px2 + Py2)1/2
= 80.445
Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree
Then the velocity of owner is v = P / M = 1.235 m/s
Answer:
The direction of the B-field is in the +y-direction.
Explanation:
The corresponding formula is
This means, we should use right-hand rule.
Our index finger is pointed towards +x-direction (direction of velocity),
our middle finger should point towards the direction of the B-field,
and our thumb should point towards the +z-direction (direction of the force).
Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.
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.
