The net force on the cart is
Force = (mass) x (acceleration) .
We know the mass = 1.5 kilogram, but we need to
calculate the acceleration.
Acceleration = (change in speed) / (time for the change)
Change in speed = (ending speed) minus (beginning speed)
= ( 0 ) - ( 2 m/s ) = -2 m/s
Acceleration = (change in speed) / (time for the change)
= ( -2 m/s ) / (3 sec)
= - 2/3 m/s² .
Finally ... Net force = (mass) x (acceleration)
= (1.5 kilogram) x ( - 2/3 m/s²)
= ( - 3/2 x 2/3 ) (kg-m/s²)
= - 1 newton .
The negative sign on the force means the force is applied
opposite to the direction the cart is moving. That gives
negative acceleration (slowing down), and the cart
eventually stops.
The current is defined as the amount of charge Q that passes through a certain point of a wire in a time

:

But the charge Q flowing in the wire is just the charge of a single electron, e, times the number of electrons N:

so the current can be rewritten as

Using

,

(charge of one electron), and

, we find the current:

Therefore correct answer is A).
Answer:
(1) The resistivity of the rod at 20 °C is 8.652×10^-6 ohm-meter.
(2) The temperature coefficient of resistivity at 20 °C is 0.00125/°C
Explanation:
(1) Resistance at 20 °C = V/I = 14/18.7 = 0.749 ohm
Length = 1.7 m
Diameter (d) = 0.5 cm = 0.5/100 = 0.005 m
Area = πd^2/4 = 3.142×0.005^2/4 = 1.96375×10^-5 m^2
Resistivity at 20 °C = resistance × area/length = 0.749×1.96375×10^-5/1.7 = 8.652×10^-6 ohm-meter.
(2) Resistance at 92 °C = V/I = 14/17.2 = 0.814 ohm
Temperature coefficient at 20 °C = (0.814/0.749 - 1) ÷ (92 - 20) = (1.09 -1) ÷ 72 = 0.09 ÷ 72 = 0.00125/°C
Answer:
.
Explanation:
In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.
In this example, the spring toy is the projectile. Refer to the diagram attached.
- The spring toy started with a velocity of
, meaning that its initial KE is non-zero. - On the way up, the KE of this spring toy gets converted to GPE.
- At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
- As the toy returns to the ground, the GPE of this toy gets converted back to KE.
- The GPE of this toy is zero when the toy is on the ground.
In other words, the kinetic energy (KE) of this toy would be
when it is at the top of the trajectory and GPE is maximized.
Since the KE of this toy at the top of the trajectory would be
, the velocity of this toy at that moment would also be
.