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3 years ago

True or False: Exercise is an underrated stress reliever and can be used to 25

Physics

2 answers:

Andreyy893 years ago

5 0

Answer:

I think it is true I'm not saying it is but if you get another person who says its true say true

Explanation:

uranmaximum [27]3 years ago

3 0

I believe the answer is true

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Ifa truck starts from rest and it has acceleration of 4 m/s for 5 second

yan [13]

Answer:

Distance 50m

final velocity 20ms^-1

$s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times t + \frac{1}{2} \times 4 \times {5}^{2} = 50m$

$v = u + at \\ v = 0 + 4 \times 5 \\ v = 20ms^{ - 1}$

6 0

3 years ago

Two polarizers are oriented at 68 ∘ to one another. unpolarized light falls on them. part a what fraction of the light intensity

Radda [10]

Your answer is.07 hope this helped

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3 years ago

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If the thrower takes 0.90 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

Marysya12 [62]

Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled

Let
α = the angular acceleration
ω = the final angular velocity

The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²

The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s

If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s

Answer: 13.963 rad/s

8 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

Usimov [2.4K]

Answer:

Explanation:

7 0

3 years ago

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