Distance, such that displacement over time is taken into account doesn't change regardless of the length in x meters or mass in kg
So the force will remain constant given 16 units

6 0

3 years ago

DaniilM [7]

$\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$

$\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}$

$\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}$

⚡ Let $\rm{\vec{a}}$ and $\rm{\vec{b}}$ are the two vectors .

✍️ We have know that,

$\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}$

Where,

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}$

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}$

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}$

$\rm{\red{\therefore}}$ [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

$\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}$

Where,

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}$

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}$

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}$

$\rm{\red{\therefore}}$ [2] The cross product of two vectors is “ <u>ab</u> ” .

3 0

3 years ago

a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

#SPJ1

3 0

2 years ago