All categories

2 years ago

How much time does it take for a spinning baseball with an angular speed of 38 rad/s to rotate through 15 degrees

Physics

1 answer:

Nadusha1986 [10]2 years ago

3 0

Answer:

the time of motion of the ball is 6.89 ms.

Explanation:

Given;

angular speed, ω = 38 rad/s

angular distance, θ = 15 degrees

Angular distance in radian;

$\theta = 15^0 \times\frac{2\pi \ rad}{360^0} = 0.2618 \ rad$

Time of motion is calculated as;

$time = \frac{angular \ distance}{angular \ speed} \\\\t= \frac{\theta}{\omega} = \frac{0.2618 \ rad}{38 \ rad/s} \\\\t = 6.89 \ \times 10^{-3} \ s\\\\t = 6.89 \ ms$

Therefore, the time of motion of the ball is 6.89 ms.

You might be interested in

Kayla draws the image shown as part of her physical science homework.

AnnZ [28]

What the question for this assessment

8 0

2 years ago

Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degree

Nataliya [291]

Answer:

High pressure inside the giant planet

Explanation:

As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.

Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.

If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.

Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.

7 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

2 years ago

You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th

Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

$a=\frac{dv}{dt}\\$

if a=constant and we integrate the last equation

$v(t)=v_{o}+a*t$

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

8 0

2 years ago

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

3 years ago