The change in speed of this object is 3m/s
According to Newton's second law;
F = ma
F = mv/t
Given the following parameters
Force F = 8.0N
mass m = 16kg
time t = 4.0s
Required
speed v
Substitute the given parameters into the formula
v = Ft/m
v = 8 * 6/16
v = 48/16
v = 3m/s
Hence the change in speed of this object is 3m/s
Learn more here: brainly.com/question/19072061
It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)
It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.
Answer:
Distance from start point is 72.5km
Explanation:
The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

Replacing the given values in the equations, the distance is calculated.
Here is your answer:
First find the notations:
2×10^-3
=0002
And...
2.5×10^4=25000
Then divide:
0002÷25000=8E-9
Your answer:
=8 x 10-8
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = 
v²
v = √( 2K /
)
lets relate the cross-sectional area A of the beam to its diameter D;
A =
πD²
now, we substitute for v and A
n = I /
πeD² ×√( 2K /
)
n = 4I/π eD² × √(
/ 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³