Answer:
1.06 m
Explanation:
The kinetic friction of the grass on the sled is the product of friction coefficient and the normal force, which equals to gravitational force. Let g = 9.81 m/s2:

So the deceleration caused by kinetic friction on the 60 kg sled is

If the sled on a speed of 2.5m/s and then subjected to a deceleration of 2.943 m/s2, then we can use the following equation of motion to find out the distance traveled by the sled before rest:

where v = 0 m/s is the final velocity of the sled when it stops,
= 2.5m/s is the initial velocity of the can when it hits, a = -2.943 m/s2 is the deceleration, and
is the distance traveled, which we care looking for:


Answer:
West
Explanation:
We can solve the problem by using the right-hand rule. We can apply the rule as follows:
- Index finger: direction of the current
- middle finger: direction of the magnetic field
- thumb: direction of the force exerted on the wire
By applying the rule to this situation, we have:
- index finger: upward (current)
- middle finger: north (magnetic field)
- thumb: west (force)
So, the direction of the force exerted on the wire is to the west.
A dog with four legs is an intact, marginally redundant, canine quadruped.
Answer:

Since we have identical diodes we can use the equation:

And replacing we have:
Since we know that 1 mA is drawn away from the output then the real value for I would be

And for this case the value for
would be:

And the output votage on this case would be:

And the net change in the output voltage would be:

Explanation:
For this case we have the figure attached illustrating the problem
We know that the equation for the current in a diode id given by:
![I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}](https://tex.z-dn.net/?f=%20I_D%20%3D%20I_s%20%5Be%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D%20-1%5D%20%5Capprox%20I_S%20e%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D)
For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode
and each voltage is the same v for each diode, so then:

Since we have identical diodes we can use the equation:

And replacing we have:

Since we know that 1 mA is drawn away from the output then the real value for I would be

And for this case the value for
would be:

And the output votage on this case would be:

And the net change in the output voltage would be:
