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schepotkina [342]

4 years ago

9

What is the velocity of the mass at a time t? you can work this out geometrically with the help of the hints, or by differentiat

ing the expression for r' (t) given in the introduction. (figure 2)

1 answer:

Alex777 [14]4 years ago

6 0

The angle between r (t = o) and r (t), is diresctly related to the direction pf the verlocity vector which is when t = 0, r(t = 0) = R(i^)θ = ωt
So, the speed of the mass at time (t0 is v(t) = ωR (v(t) in terms of ω and R)

the answer is v(t) = − Rωsin(ωt) + Rωcos(ωt)j^

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Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angle

wariber [46]

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

$q$ = magnitude of charge on each particle = 5 μC = 5 x 10⁻⁶ C

$r$ = distance of each particle from center of circle = 2 m

$E$ = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

$E = \frac{kq}{r^{2} }$

inserting the values

$E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}$

From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.

So net electric field at center is only due to charge q₂ direction towards positive x-direction

So

$E_{res}$ = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

7 0

3 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago

The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P

bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

$a=\dfrac{P}{m}$

$a=\dfrac{181\ N}{68\ kg}$

$a=2.67\ m/s^2$

(c) P = 352 N

$a=\dfrac{P}{m}$

$a=\dfrac{352\ N}{68\ kg}$

$a=5.17\ m/s^2$

Hence, this is the required solution.

5 0

3 years ago

A force of 31 N south is used to accelerate an object uniformly from rest to 8 m/s in 22 s. What is the mass of the object?

harina [27]

this question is centered around newtons second law f=ma

First you need to find the acceleration

$acceleration = \frac{8 - 0}{22}$

that gives you 0.36 recurring(on both numbers) m/s²

mass= force ÷ acceleration

$mass = \frac{31}{0.36recurring}$

mass= 85.25 kg

if you want to check

f=ma

85.25 × 0.36 recurring (remember its on both numbers not just 6) = 31 N

4 0

3 years ago

Read 2 more answers

When a wave hits an object,energy from the wave is both absorbed and reflected off the object

poizon [28]

I believe the statement is true. Have a good day.

5 0

3 years ago