Ask question

Ask question

All categories

2 years ago

7

A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov

ing along the same line at a speed of 0.10 m/s. After the collision, both balls continue moving in the same, original direction, and the speed of the 0.100 kg ball is 0.26 m/s. What is the new velocity of the first ball?

1 answer:

zzz [600]2 years ago

4 0

Answer:

0.22m/s

Explanation:

The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.

momentum of an object = mass* velocity

Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;

Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;

Solve for x.

You might be interested in

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

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

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

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

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a

AURORKA [14]

Answer:

<h2>0.069 N, in the X direction</h2>

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B= 0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

5 0

2 years ago

Elements in group 2 are all called alkaline earth metals. What is most similar about the alkaline earth metals? how many protons

Rom4ik [11]

Elements in group 2 are called alkaline earth metals, the most similarity about the alkaline metals is which chemical properties they have.

<h2>Further Explanation </h2><h3>Periodic table </h3>

Periodic table is a table that contains elements arranged in columns called groups and rows called periods.
Elements are arranged based on physical and chemical properties such that elements in the same group will have similar physical and chemical properties.

<h3>Chemical families </h3>

Based on the chemical properties elements belong to a family of elements sharing similar chemical or physical characteristics.
Examples of common chemical families include; alkali metals, alkaline-earth metals, halogens and noble gases among others.

<h3>Alkaline-earth metals </h3>

These are elements that are found in group 2 of the periodic table. Alkaline-earth metals include, Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.

Properties of Alkaline-earth metals

Alkaline earth metals have a valence of two since they form ions by losing two electrons from their outermost energy levels.
Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive compared to alkali metals.
Alkaline-earth metals are highly metallic and are good conductors of electricity.
They react with water and steam to form metal hydroxide and metal oxides respectively
They react with air to form metal oxides
Reactivity of alkaline metals depends on the ease of losing electrons, thus the reactivity increases down the group as the number of energy levels increases.
Additionally, alkaline-earth metals have low electronegativities and low electron affinities

Keywords: Chemical families, alkaline-earth metals, reactivity

<h3>Learn more about: </h3>

Chemical families: brainly.com/question/1358941
Alkaline-earth metals: brainly.com/question/8498732
Properties of alkaline-earth metals: brainly.com/question/11116789
Reactivity of metals: brainly.com/question/7101478

Level: High school

Subject: Chemistry

Topic: Periodic table and chemical families

Sub-topic: Alkaline-earth metals

4 0

2 years ago

Read 2 more answers

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0

2 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

bazaltina [42]

Answer:

The current in the rods is 171.26 A.

Explanation:

Given that,

Length of rod = 0.85 m

Mass of rod = 0.073 kg

Distance $d = 8.2\times10^{-3}\ m$

The rods carry the same current in the same direction.

We need to calculate the current

I is the current through each of the wires then the force per unit length on each of them is

Using formula of force

$\dfrac{F}{L}=\dfrac{\mu_{0}I^2}{2\pi d}$

$\dfrac{mg}{L}=\dfrac{\mu_{0}I^2}{2\pi d}$

Where, m = mass of rod

l = length of rod

Put the value into the formula

$I^2=\dfrac{mgd}{\mu L}$

$I^2=\dfrac{0.073\times9.8\times8.2\times10^{-3}}{2\times10^{-7}}$

$I=\sqrt{29331.4}$

$I=171.26\ A$

Hence, The current in the rods is 171.26 A.

5 0

2 years ago