Which best describes a circuit in series?

a 3.2 kg durian fruit is pushed across the table.If the acceleration of the durian is 3.1 m/s/s to the right,what is the net for

Gala2k [10]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.2 × 3.1

We have the final answer as

Hope this helps you

4 0

3 years ago

Read 2 more answers

What is the approximate velocity of the object at 5 seconds ? .

Gnoma [55]

Answer:

do you have an image?

Explanation:

3 0

3 years ago

what will be the speed of a solid sphere of mass 2.0 kilograms and radius 15.0 centimeters when it reaches the bottom of a incli

Yuki888 [10]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

What do you need in order to change the velocity of an object?

Elden [556K]

Answer:

Forces affect how objects move. They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity.

Explanation:

5 0

2 years ago

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$