The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
1 answer:
1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
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Answer:
Explanation:
The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s
therefore, the total duration of the upwards motion above the peak of the pole is provided as
now we have
this indicates the speed at the flagpole's top
at this point we have
the height of the flagpole is stated as
The rocket's acceleration is described here as
now recognizing that
we integrate both sides
given that the rocket is accelerating for a duration of t = 10 s
thus, we have
consequently, after t = 10 s, the rocket will achieve a speed of 130 m/s in an upward direction
Response:
Explanation:
Let T denote the tension.
By employing Newton's second law to analyze the bucket's downward motion, we have:
mg - T = ma
A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:
TR = Iα
Rearranging gives: TR = Ia/R
This leads to T = Ia/R²
Substituting this expression for T back into the previous equation yields:
mg - T = ma
mg - Ia/R² = ma
Consequently, we find that mg = Ia/R² + ma
Therefore, a (I/R² + m) = mg
This results in: a = mg / (I/R² + m)
Next, we aim to express T as:
mg - T = ma
which simplifies to mg - ma = T
Rearranging gives mg - m²g / (I/R² + m) = T
Thus, we arrive at: mg - mg / (1 + I / m R²) = T
For part (b), T = Ia/R²
and for part (c), the moment of inertia of a hollow cylinder calculates to:
I = 1/2 M (R² - (R² / 4))
This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²
Thus, I / R² = 3/8 M
When we substitute, we find a = mg / (3/8 M + m)
and subsequently T = Ia/R²
= 3/8 MR² × mg / (3/8 M + m) × 1/R²
Results in:
Response:
The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.
We can utilize Gauss's law to solve this problem
Ф = ∫ e. dA = / ε₀
The flow of the field is directly correlated to the charge within it. Consequently, placing a Gaussian surface beyond the non-conductive spherical shell means the flow will be zero since the sphere’s charge equals the charge induced in the shell, resulting in a net charge of zero. This evaluation shows that the shell effectively obstructs the electric field.
According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.