Answer:
groups
Explanation:
vertical sections are called groups
Answer:
Friction acts in the opposite direction to the motion of the truck and box.
Explanation:
Let's first review the problem.
A moving truck applies the brakes, and a box on it does not slip.
Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.
The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.
Answer:
a)
b)
c)
d) Δs=4428ft
Explanation:
From the exercise we know the equation of position
a) To calculate the velocity we need to derivate the equation of position
So, v(0) is:
b) To find the two times where the particle stops A and B we need to solve the quadratic equation:
Solving for t
and
c)
d) Δs=4428ft-0ft=4428ft
<span>Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:
</span><span>(3 kg)(9.8 m/s^2) = -k(-0.38 m)
</span>k =<span> 77.4
</span><span>Then use k to find the new displacement, again using Hooke's law:
(7 kg)(9.8 m/s^2) = -(77.4)x
x = -0.89 m</span>
Answer:
I = 2.33 µA
Explanation:
given,
Effective resistance (R)= 300 Ω
Voltage of signal (V)= 700 µV
current flow in the antenna = ?
Using ohm's law
V = I R
where I is the current flow
I = 2.33 x 10⁻⁶ A
I = 2.33 µA
Hence, the current flow in the antenna downloaded is equal to I = 2.33 µA