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Ulleksa [173]
3 years ago
9

Look at the image to answer the question correctly.

Physics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

1-  b: 2- a : 3- c : 4- d

Explanation:

it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.

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Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?
cluponka [151]

Answer:

Torque, \tau=0i+0j-70k

Explanation:

It is given that,

Force acting on the particle, F=-10j\ N

Position of the particle, r=(7i+5j)\ m

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

\tau=r\times F

\tau=(7i+5j)\times (-10j)

The cross product of vectors is given by :

\tau=\begin{pmatrix}0&0&-70\end{pmatrix}

or

\tau=0i+0j-70k

So, the torque on the particle about the origin 0i+0j-70k. Hence, this is the required solution.

6 0
3 years ago
 A cat napping on a window sill demonstrates which scientific concept?
ser-zykov [4K]
The answer is inertia becuase it means nothing happening or is doing nothing
8 0
3 years ago
Read 2 more answers
What is the correct explanation of Newton's Third Law of Motion?
viva [34]

Answer:

a

Explanation:

it explains the most, and it is the correct theorem

8 0
3 years ago
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A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the
elena55 [62]

Answer:

(a) t = 5.66 s

(b) t = 8 s

Explanation:

(a)

Here we will use 2nd equation of motion for angular motion:

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (23.25 rad)(2)/(1.45 rad/s²)

t = √(32.06 s²)

<u>t = 5.66 s</u>

<u></u>

(b)For next 3.7 rev

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (46.5 rad)(2)/(1.45 rad/s²)

t = √(64.13 s²)

<u>t = 8 s</u>

7 0
3 years ago
A single conservative force F = (7.0x - 11) N, where x is in meters, acts on a particle moving along an x axis. The potential en
Umnica [9.8K]

Answer:

(a) 34.6429J

(b) -1.57 m

(c) 4.71 m

Explanation:

The derivative of the potential energy with respect to the position is equal to the negative of the force, so:

-\frac{dU}{dx}=F\\ dU=-Fdx

Then, if we integer both sides, we get:

∫dU = -∫(7x - 11)dx

U=\frac{-7}{2}x^{2} + 11x + c

we know that U is equal to 26 J when x is zero, so:

U=\frac{-7}{2}x^{2} + 11x + c

26=\frac{-7}{2}(0)^{2} + 11(0) + c

26=c

Finally, the equation for the potential energy is:

U(x)=\frac{-7}{2}x^{2} + 11x + 26

Therefore, the maximum positive potential energy is the energy when x is equal to 11/7. That is because the equation of U is the equation of a parable, and the vertex in a parable is given by:

x=\frac{-b}{2a} = \frac{-11}{2(-7/2)} =\frac{11}{7}

Where b is the number beside x and a is the number beside x^{2}, Then, the value of maximum U is:

U(11/7)=\frac{-7}{2}(11/7)^{2} + 11(11/7) + 26

U(11/7)=34.6429J

On the other hand, the negative and positive values of x where the potential energy is equal to zero is calculated as:

U(x)=\frac{-7}{2}x^{2} + 11x + 26

0=\frac{-7}{2}x^{2} + 11x + 26

if we solve this using the quadratic equation, we get:

x =\frac{-11+\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=-1.5747

x =\frac{-11-\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=4.7175

Finally, the negative and positive values of x where the potential energy is equal to zero are -1.5747 and 4.7175 respectively.

3 0
3 years ago
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