All categories

3 years ago

When an object is moving with uniform circular motion, the centropetal acceleration of the object?

2 answers:

5 0

Yes this is possible if the speed is changing. An object moves in a circular path at a constant speed. I hope this helps my friend.

skelet666 [1.2K]3 years ago

3 0

Answer:

The correct answer is "is directed toward the center of motion".

Explanation:

When an object moves in a uniform circular motion, the centrifugal acceleration of the object is directed toward the center of the motion. This acceleration is the only acceleration of the object experiences when it has constant velocity on a circular path. This causes the body to be attracted to the center of the trajectory by a centripetal force that prevents the body from entering a rectilinear trajectory.

Have a nice day!

You might be interested in

1. Is this circuit SERIES or PARALLEL?

Setler79 [48]

Answer:

Series

Explanation:

Because I listen to my science teacher

3 0

3 years ago

One beaker contains 100 mL of pure water and second beaker contains 100 mL of seawater. The two beakers are left side by side on

Harman [31]

Answer: The beaker containing pure water has decreased more.

Explanation:

In both cases, the decrease of water level is due to evaporation. We know that evaporation is a surface phenomenon. In the case of salt water, the salt molecules somewhat hinders the evaporation process of the water molecules and hence the salt water evaporates at a slower rate than pure water.

Hence, pure water level falls more.

7 0

3 years ago

Read 2 more answers

IF YOU ANSWER ALL THE QUESTIONS THEN I WILL GIVE YOU BRAINEST!!! 20 POINTS!!!!

Firdavs [7]

Answer:

1.) A, 2.) D, 3.) C

Explanation:

Hope this Helps! :D

5 0

3 years ago

The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the

lara [203]

As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
So, a = 7/6
a = 1.17
As the graph is decreasing, value of acceleration would be negative.
So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!

7 0

3 years ago

In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way i

lara [203]

Answer:

Approximately $1.44\times 10^3 \; \rm N \cdot m^{-1}$ assuming that the spring has zero mass.

Explanation:

Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.

On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force $\mathbf{F}$ when its displacement is $\mathbf{x}$ , then its force constant would be:

$\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}$ .

The goal here is to find the expressions for $F$ and for $x$ . By Hooke's Law, the spring constant would be ratio of these two expressions.

Let $T$ represent the time period of this oscillation. With the chair alone, the period of oscillation is $T = 1.00\; \rm s$ .

For a simple harmonic oscillation, the angular frequency $\omega$ can be found from the period:

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

Let $A$ stands for the amplitude of this oscillation. In a simple harmonic oscillation, both $\mathbf{F}$ and $\mathbf{x}$ are proportional to $A$ . Keep in mind that the spring constant $k$ is simply the opposite of the ratio between $\mathbf{F}$ and $\mathbf{x}$ . Therefore, the exact value of $A$ shouldn't really affect the value of the spring constant.

In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time $t$ would be:

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

The restoring velocity at time $t$ would be:

$\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)$ .

The restoring acceleration at time $t$ would be:

$\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)$ .

Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time $t$ would be:

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

Apply Hooke's Law to find the spring constant, $k$ :

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

Again, $\omega$ stands for the angular frequency of this oscillation, where

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

Before proceeding, note how $A$ was eliminated from the ratio (as expected.) Additionally, $t$ is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to $k$ :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}$ .

Side note on the unit of $k$ :

$\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}$ .

6 0

3 years ago