Answer:10842.33m/s
Explanation:
F=qvBsine
V=f/(qBsine)
V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)
V=10842.33m/s
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Weight = (mass) x (gravity)
Weight = (7.0 kg) x (gravity)
On Earth, where (gravity) is roughly 10 N/kg . . .
Weight = (7.0 kg) x (roughly 10 N/kg)
Weight = roughly 70 Newtons
That's <em>B </em>on Earth.
It would be some other number on other bodies.
The kinematic equations are used to <span>quantify motion in the case of uniform acceleration.
The other name is :
SUVAT equations, where the letters signify:
displacement (s),
initial velocity (u),
final velocity (v),
acceleration (a), and
time (t).
There are three equations are attached in the picture: </span>