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2 years ago

An example of constant velocity

1 answer:

4 0

Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.

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A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

3 0

3 years ago

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

Juli2301 [7.4K]

They made me do it I don’t even know what to say I’m so sorry

7 0

3 years ago

A pendulum is suspended from the cusp of a cycloid cut in rigid support. The path described by the pendulum bob is cycloidal and

oksano4ka [1.4K]

Answer:

Verified that he oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude.

Explanation:

Starting from the first principle for the derivation and to prove that the oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude. The mathematical manipulations was applied, trigonometric identities was also applied.The steps and explanation are shown in the attachment.

5 0

3 years ago

A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is m

iVinArrow [24]

Answer:

The magnitude of the magnetic field will act in a direction towards me.

Explanation:

When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.

7 0

3 years ago

As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks nor

Anastaziya [24]

Answer:

Option A

Solution:

As per the question:

The distance covered by the woman in the North direction, d = 3000 m

Time taken to travel in North direction, t = 25.0 min = 1500 s

Velocity of woman in the south direction, v = 2.00 m/s

Time taken in the south direction, t' = 60.0 min = 3600 s

Now,

The distance covered in the south direction, d' = vt' = $2.00\times 3600 = 7200\ m$

Now, the total displacement is given by:

D = d' - d = 7200 - 3000 = 4200 m in South

(a) Average velocity of the woman in the whole journey is given by:

$v_{avg} = \frac{Total\ displacement}{Total\ time} = \frac{4200}{t + t'}$

$v_{avg} = \frac{4200}{1500 + 3600} = 0.8235\ m/s$ ≈ 0.824 m/s South

6 0

3 years ago