Answer:
Acceleration=3.95
Explanation:Use the formula a=m/f
a=128.6/32.5
a=3.95
According to Newton's first law of motion, it is the natural tendency of all moving objects to continue in motion in the same direction that they are moving ... unless some form of unbalanced force acts upon the object to deviate its motion from its straight-line path.
Hope this helped, have a great day!
<span>31.3 m/s
Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know.
Distance the water balloon travels at velocity v for time t
d = vt
Total time required for the entire trip is double since the balloon goes up, then goes down
t = 2v/a
Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2
t = 2v/9.8
100 = vt
Substitute 2v/9.8 for t in the 2nd formula
100 = v(2v/9.8)
Solve for v.
100 = v(2v/9.8)
100 = 2v^2/9.8
980. = 2v^2
490 = v^2
22.13594 = v
So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that
2*22.13594 / 9.8 = 4.51754
So it will take 4.51754 second for the balloon to hit the ground after being launched.
4.51754 * 22.13594 = 100
And during that time it will travel 100 meters horizontally.
But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so
sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
