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zalisa [80]
2 years ago
14

the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s

− 2 sin t 1 3 cos t, where t is measured in seconds.
Physics
1 answer:
12345 [234]2 years ago
4 0

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

If you need to learn more about displacement click here:

brainly.com/question/28370322

#SPJ4

The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

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kaheart [24]

Answer:

Sometimes may cause involuntary responses like twitching

Explanation:

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3 years ago
A man stands at the edge of a cliff and throws a rock downward with A speed of 12.0 m s . Sometimes later it strikes the ground
kvasek [131]

B) 48.0 m/s

We can actually start to solve the problem from B for simplicity.

The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation

v^2 -u^2 = 2as

where

v is the final velocity

u = 12.0 is the initial velocity (positive since we take downward as positive direction)

a=g=9.8 m/s^2 is the acceleration of gravity

s = 110 m is the vertical displacement

Solving for v, we find the final velocity (and so, the speed of the rock at impact):

v = \sqrt{u^2+2as}=\sqrt{12^2+2(9.8)(110)}=48.0 m/s

A) 3.67 s

Now we can find the time of flight of the rock by using the following suvat equation

v=u+at

where

v = 48.0 is the final velocity at the moment of impact

u = 12.0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time it takes for the rock to reach the ground

And solving for t, we find

t=\frac{v-u}{a}=\frac{48.0-12.0}{9.8}=3.67 s

6 0
3 years ago
Particles q1, q2, and q3 are in a straight line.
natima [27]

The net force on q2 will be 1.35 N

A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.

Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.

We have to find the net force on q2

At first we will find Force due to q1

F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²

F = 450 × 10⁻³

F₁ = 0.45 N (+)

Now we will find Force due to q2

F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²

F = 1800 × 10⁻³

F₂ = 1.8 N (-)

So net force (F) will be

F = F₂ - F₁

F = 1.8 - 0.45

F = 1.35 N

Hence the net force on q2 will be 1.35 N

Learn more about force here:

brainly.com/question/25573309

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8 0
2 years ago
I tried to solve for it 3 times but I still got it wrong. HELP?!?!
QveST [7]
The answer is = 793.00m
3 0
4 years ago
Compare the characteristics of 4d orbitals and 3d orbitals and compared the following sentences. Check all that apply
charle [14.2K]

Answer :

<em>(b) 4d orbitals would be larger in size than 3d orbitals</em>

<em>(e) 4d orbitals would have more nodes than 3d orbitals</em>

Explanation :

As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.

Hence, 4d orbitals would be larger in size than 3d orbitals.

Number of nodes is any orbital is n - 1 where, n is principal quantum number.

So, number of orbital in 4d is 3.

And number of orbital in 3d is 2.

So, options (b) and (e) are correct.

8 0
3 years ago
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