A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to
1 answer:
Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
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Answer:
Explanation:
Since the wire is infinitely long, we will use Gauss' Law:
We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.
In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to
The left-hand side of the Gauss' Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.
where R is the radius of the imaginary cylinder.
Finally, Gauss' Law gives
The vector expression is
As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss' Law. In the end, what matters is the charge density of the wire and the distance from the wire.
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch