If we assume that the acceleration is constant, we can use on the kinematic equations:
Vf = Vi + a*t = 15 + 3*4 = 27 m/s
<h2>Answer:</h2>
The refractive index is 1.66
<h2>Explanation:</h2>
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum
.
Refractive index of medium = speed of light in vacuum / speed of light in medium
So
RI = 1/0.6 = 5/3 or 1.66
Answer:
False. the system does not complete the circle movement
Explanation:
For this exercise, we must find the rope tension at the highest point of the path,
-T - W = m a
The acceleration is centripetal
a = v² / R
T = ma - mg
T = m (v² / R - g)
The minimum tension that the rope can have is zero (T = 0)
v² / R - g = 0
v = √ g R
Let's find out what this minimum speed is
v = √ 9.8 1
v = 3.13 m / s
We see that the speed of the body is less than this, so the system does not complete the movement.
Answer:
The voltage on the secondary is 12 V while the current is 0.5 A.
Explanation:
A transformer works by changing the level of the voltage and current on a circuit using a magnetic field and two coils. The ratio by wich they are changed is dependant on the ratio of turns between the primary and secondary of the transformer. In this case we have a ratio for the voltage of:
ratio = (turns on the secondary)/(turns on the primary)
ratio = 100/1000 = 0.1
So in this case the voltage delivered to the primary will be multiplied by 0.1. We can now calculate the voltage on the secondary:
Voltage secondary = Voltage primary* ratio = 120*0.1 = 12 V
The transformer maintains roughly the same power output on both sides, since the power output on a electric circuit is given by the product of the voltage by the current on that circuit, to maintain the same power when the voltage has been droped the current must be raised by the same ratio. So we have:
Current secondary = Current primary*(1/ratio) =0.05*(1/0.1) = 0.5 A
Answer:
false
Explanation:
the temperature of an object is a measure of the average KINETIC ENERGY of the molecules in the object (not potential energy)
