Answer:
194,400 joules of kinetic energy.
Explanation:
Remember that to calculate the Kinetic energy you need to use the next formula:

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.
Give u = start velocity
v = end velocity
v = u + at
50 = 400 + a*30
30a = -350
a = -116.67 m/
**Why the accecleration is negative number**
Because displacement, velocity, and acceleration are VECTOR QUANTITIES.
Vector Quantity must have direction.
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
Study and practice of environmental sustainable computing or It. such as monitors, printers, storage devices, and networking and communication systems-efficiently and effectively with minimal impact on the environment.
I think it is either A or B. I’m mostly leaning towards B.