What is the magnitude of the maximum acceleration experienced by the rider?

Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

7 0

2 years ago

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

tatuchka [14]

Explanation:

It is given that,

Magnetic field, B = 0.5 T

Speed of the proton, v = 60 km/s = 60000 m/s

The helical path followed by the proton shown has a pitch of 5.0 mm, p = 0.005 m

We need to find the angle between the magnetic field and the velocity of the proton. The pitch of the helix is the product of parallel component of velocity and time period. Mathematically, it is given by :

$p=v_{||}\times T$

$p=v\ cos\theta\times \dfrac{2\pi m}{Bq}$

$cos\theta=\dfrac{pBq}{2\pi mv}$

$cos\theta=\dfrac{0.005\times 0.5\times 1.6\times 10^{-19}}{2\pi \times 1.67\times 10^{-27}\times 60000}$

$\theta=50.58^{\circ}$

So, the angle between the magnetic field and the velocity of the proton is 50.58 degrees. Hence, this is the required solution.

3 0

2 years ago

Question 1

Vesna [10]

Answer:

1) g = 4π² / m, 3) xaxis the length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

w = √ g / l

angular velocity, frequency and period are related

w = 2π f = 2π / T

we substitute

T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

T² = 4π² l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

y = (4π² / g) l

so the slope is

m = 4π² / g

clearing

g = 4π² / m

where the slope can be found with the values of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

m = 4pi2 / g

where it ends up reaching the floor

g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

5 0

2 years ago

At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w

aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s
Momentum of car A + B = 15,000 x 2m/s

So combining the two, will give us the equation:
15,000/5,000 = 3
3 x 2 =6m/s

6 0

2 years ago

Which objects have both potential and kinetic energy? Check all that apply.

wariber [46]

A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.

3 0

2 years ago

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