If the velocity of the particle is uniform, then the path of the particle must be a

The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall

ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is also acting downward as it is creating friction between ball and air molecules, so creating hindrance in upward motion

External force to throw ball upward

So

Net Force = Upward force - Air friction - Weight

Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

So

Net Force = Weight - Air friction

Since ball is going downward, so weight is greater than net force which is in this case is air friction which is pulling ball upward.

4 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

_____ have a nearly circular orbit.

bezimeni [28]

B. Asteroids have a nearly circular orbit.

4 0

3 years ago

Find the acceleration that can result from a net force of 20 N exerted on a 2.8-kg

vladimir2022 [97]

Force=mass*acceleration
F=ma

20=2.8a

a=20/2.8

a=7.14 m/s^2

6 0

3 years ago

Axis Powers used the Anti-Comintern Pact to hide their aggressive intentions with a false pledge to stop the spread of communism

lesantik [10]

What is the question here? if it is true or false?

5 0

3 years ago

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If the velocity of the particle is uniform, then the path of the particle must be a​

If the velocity of the particle is uniform, then the path of the particle must be a