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3 years ago

In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?

Physics

2 answers:

Jobisdone [24]3 years ago

7 0

Explanation:

The equation of motion of an object is given by :

$v_x^2=v_{ox}^2+2ax(x-x_o)$

Where

$v_x$ is velocity of a particle at position x

$v_{ox}$ is the velocity at position x = 0

x is the position of an object

$x_o$ is position at t = 0

So, the correct option is (b) "velocity at position x, velocity at position x=0, position x, and the original position". Hence, this is the required solution.

tatuchka [14]3 years ago

6 0

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

$v_{x}^{2}$ = $v_{ox}^{2}$ +2 a x (x - x₀)

$v_{x}$ = velocity at position "x"

$v_{ox}$ = velocity at position "x = 0 "

x = final position

$x_{o}$ = initial position of the object at the start of the motion

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A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance

just olya [345]

Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

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2 years ago

What is the weight of an object (mass = 60 kilograms) on Mars, where the acceleration due to gravity is 3.75 meters/second2?. Se

Shalnov [3]

Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N

<span>Option D.</span>

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4 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A force of 0.2 Newtons is required to slide a book across the table. The book accelerates at 0.11 m/s squared. What is the mass

adelina 88 [10]

My calculations state, not rounding, the mass is 1.8

3 0

3 years ago

If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a

PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s $w_o$ = 12 rad/s

Angular Velocity at time 0.15s $w_f$ = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( $w_f$ - $w_o$ ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0

3 years ago