Answer:
The shell lands 963 m away from the cliff’s base.
Explanation:
Projectile Motion
Consider an object that is propelled from a height yo above ground level with an initial velocity vo and angle with respect to the horizontal. The height of this object at any time t can be calculated by
Meanwhile, the horizontal distance it covers at any instant t is given by
The shell is projected at 100 m/s at a 30° angle, starting from a height of 50 m. We must determine when the shell impacts the ground after being launched. Substituting the known values into the formula:
By simplifying and rearranging the equation
This leads to a quadratic equation in terms of t, which yields two real solutions:
Only the positive solution is considered since time cannot be negative. Now that we have the duration of the flight, we can compute how far from the cliff's base the shell hits the ground
The shell lands 963 m away from the cliff’s base.
Answer:
Explanation:
Given that the charge starts from rest and at point B has gained kinetic energy K = 4.8 J
we can apply the principle of conservation of mechanical energy, indicating that the increase in kinetic energy equals the reduction in electrostatic potential energy.
Thus, we can represent this as
Response:
a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°
Clarification:
To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).
a)
The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.
To compute the time, we utilize the following formula with the known values:
We have and
, allowing us to simplify the equation to:
Now, we can calculate for t:
We know and
The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:
Which results in:
t=0.495s
This time helps us evaluate the horizontal distance the mug traverses. Since:
Solving for x, we have:
Substituting the known values yields:
x=(1.5m/s)(0.495s)
This calculates to:
x=0.7425m
b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:
The initial vertical velocity being zero simplifies our calculations:
Thus, we can determine the final velocity:
Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:
This leads to:
Now, we ascertain the velocity components:
and
Next, we find the speed by calculating the vector's magnitude:
Yielding:
|V|=5.08m/s
Lastly, to ascertain the impact direction, we apply the equation:
Response:
Clarification:
Refer to the diagram indicating the charges on the specified sphere (see attachment).
The electric field at the stated positions is
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We understand that electric potential correlates with the electric field through
V = Ed
A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Thus,
Vc = (kQ / c²) × c
Vc = kQ / c
As a result, the total charge Q consists of +q, -q, and +q
Hence, Q = q - q + q = q
V = kq / c
B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have
V = kQ/r
V = kQ / b, noting that r = b
So, Q = q
V = kq / b
C. At r = a
Following from equation 1:
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Thus,
V = Ed = 0 V
Consequently, the electric potential at the solid sphere's surface is 0.
D. At r = 0
The electric potential can be determined by
V = kq / r
As r approaches 0,
V = kq / 0
V approaches infinity.
