The answer is 68 F. i hope this helps
Answer:

Given:
Mass (m) = 6.8 kg
Speed (v) = 5.0 m/s
To Find:
Kinetic energy (KE)
Explanation:
Formula:

Substituting values of m & v in the equation:




Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem

Put the value into the formula


We need to calculate the magnitude of the charge q₃
Using formula of net force

Put the value into the formula






Hence, The value of charge q₃ is 40.46 μC.
Hi there! :)
Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:

In parallel:

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:

2.
We can use Ohm's law to solve for the current in the circuit.

3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

Answer:

Explanation:
First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.
Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:
