Answer:velocity = 7.26 * 10^6 m/sec
Explanation:The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
r2 is the radius of the sphere = 2 cm = 0.02 m
Substitute with the givens in the equation to get the value of the velocity
Hope this helps :)
Answer:
1.10 m/s
Explanation:
Linear speed is given by
Kinetic energy is given by
Potential energy
PE= mgh
From the law of conservation of energy, KE=PE hence
where m is mass, I is moment of inertia,
is angular velocity, g is acceleration due to gravity and h is height
Substituting m2-m1 for m and 0.5l for h,
for
we obtain
and making v the subject
For the rod, moment of inertia
and for sphere
hence substituting 0.5L for R then
For the sphere on the left hand side, moment of inertia I
while for the sphere on right hand side,
The total moment of inertia is therefore given by adding
Substituting
for I in the equation
Then we obtain
This is the expression of linear speed. Substituting values given we get
Answer:
Explanation:
Acceleration is the time rate of change of velocity.
Acceleration and velocity are vectors
If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.
There are 3 hours or 10800 seconds between 10 AM and 1 PM
a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²
There are 14400 seconds between 10 AM and 2 PM
The velocity changes are still the same
a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
