A roller coaster uses the track in this picture. Where will the roller coaster car have the most potential energy?

Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is

irga5000 [103]

Answer:

x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

F = ma

F = q E

a = qE / m

a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

Vf² = Vo² -2 a x

0 = Vo² - 2 a x

x = Vo² / 2a

x = 3000²/ 2100

x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

F -W = 0

F = w = mg

E q = mg

E = mg / q

E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

E = 9.8 10⁵ C

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

7 0

4 years ago

If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?

joja [24]

Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: $g=-9.81~m/s^2$ , and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
$v_f^2 = v_i^2 + 2gS$
where $v_i=4~m/s$ is the initial vertical velocity of the athlete, $v_f=0$ is the vertical velocity of the athlete at the maximum height (and $v_f=0~m/s$ at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
$S= \frac{v_f^2-v_i^2}{2g}=0.82~m$

3 0

3 years ago

A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to

Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

$\tau = I\alpha$

Where $\tau$ is the torque

$I$ is the moment of inertia

$\alpha$ is the angular acceleration

But, the angular acceleration is given by

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

Now, to determine the angular speed, we first determine the Torque $\tau$ and the moment of inertia $I$ .

Here, The torque is given by,

$\tau = rF$

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴ $I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0

3 years ago

A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro

LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock = 0.22kg

Initial velocity = 20m/s

Distance moved = 10m

Unknown:

Initial kinetic energy of the rock = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2}$ x 0.22 x 20² = 44J

6 0

3 years ago

The acceleration of a particle is directly proportional to the square of the time t. When t =0, the particle is at x =24 m. Know

AlexFokin [52]

Answer:

x(t) = ⅟₁₀₈t⁴ + 10t + 24

v(t) = ⅟₂₇t³ + 10

Explanation:

a(t) = C₁t²

velocity is the integral of acceleration

v(t) = ⅓C₁t³ + C₂

position is the integral of velocity

x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃

x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃

C₃ = 24

x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24

72 = 108C₁ + 6C₂

C₂ = 12 - 18C₁

v(6) = 18 = ⅓C₁6³ + C₂

18 = 72C₁ + C₂

18 = 72C₁ + (12 - 18C₁)

6 = 54C₁

C₁ = 1/9

C₂ = 12 - 18(1/9)

C₂ = 10

4 0

3 years ago