How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller

Answers

Answer 1

Answer:

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

Now, let's consider moment of inertia = I and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia = (five times larger)

angular speed = ω/5 (five times smaller)

The kinetic energy of a spinning body is given as:

$K.E.=(1)/(2) I. w^2$

On substituting the values, we will get:

$K.E.= (1)/(2) (5I) ((w)/(5) )^2 \n\nK.E. =(1)/(10) I. w^2$

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

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What is suedo force.

Answers

What is pseudo force?

A pseudo force, also called a fictitious force or an inertial force, is an apparent force that acts on all bodies whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?

Answers

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^(-6) T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it takethe ball to reach the highest point?

Answers

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the same diameter. The rods are now joined end-to-end to form a rod 90.0 cm long. If the temperature is now raised from 15°C to 90°C, what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is 2.4 × 10-5 K-1 and that of steel is 1.2 × 10-5 K-1.

Answers

Answer:

0.9 cm

Explanation:

The computation in the increase in the length of the joined rod is shown below:

As we know that

Increase in length = increase in the length of aluminum rod + increase in The length of steel rod

$= 10cm * 2.4e - 5* (90-15) + 80cm* 1.2e - 5* (90-15)$

= 0.9 cm

We simply added the length of aluminium rod and length of steel rod so that the length of the joined rod could come and the same is to be considered

Final answer:

The increase in length of the joined rod when the temperature is raised from 15°C to 90°C is 0.090 cm.

Explanation:

To determine the increase in length of the joined rod when the temperature is raised from 15°C to 90°C, we need to use the formula for linear expansion: AL = aLAT, where AL is the change in length, AT is the change in temperature, and a is the coefficient of linear expansion. First, we need to calculate the change in temperature for each rod: ΔT = 90°C - 15°C = 75°C. For the aluminum rod, using a linear expansion coefficient of 2.4 × 10-5 K-1 and a length of 10.0 cm, we can calculate the change in length using the formula: ALaluminum = (2.4 × 10-5 K-1)(10.0 cm)(75°C) = 0.018 cm. Similarly, for the steel rod, using a linear expansion coefficient of 1.2 × 10-5 K-1 and a length of 80.0 cm, we can calculate the change in length: ALsteel = (1.2 × 10-5 K-1)(80.0 cm)(75°C) = 0.072 cm. Since the rods are joined end-to-end, the total change in length of the joined rod is the sum of the individual changes: ΔL = ALaluminum + ALsteel = 0.018 cm + 0.072 cm = 0.090 cm. Therefore, the increase in the length of the joined rod is 0.090 cm.

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"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electron in a uniform field. Calculate the magnitude of the electric force felt by the proton. Express your answer in newtons.(F = ? )Calculate the proton's acceleration.
( a= ? m/s2 )

Calculate the proton's speed after 1.40 {\rm \mu s} in the field, assuming it starts from rest.
( V= ? m/s )"

Answers

To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

$F=qE$

Here

q= Charge

E = Electric Force

$F=(1.6*10^(-19)C)(2750N/C)$

$F = 4.4*10^(-16)N$

PART B) Rearrange the expression F=ma for the acceleration

$a = (F)/(m)$

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

$a = (4.4*10^(-16)N)/(1.67*10^(-27)kg)$

$a = 2.635*10^(11)m/s^2$

PART C) Acceleration can be described as the speed change in an instant of time,

$a = (v_f-v_i)/(t)$

There is not $v_i$ then

$a = (v_f)/(t)$

Rearranging to find the velocity,

$v_f = at$

$v_f = (2.635*10^(11))(1.4*10^(-6))$

$v_f = 3.689*10^(5)m/s$

Final answer:

The magnitude of the electric force felt by the proton is 4.4 x 10^-16 N. The proton's acceleration is 2.64 x 10^11 m/s^2. The proton's speed after 1.40 μs in the field is 3.70 x 10^5 m/s.

Explanation:

The charge of a proton is 1.6 x 10-19 coulombs and the electric field strength is 2750 N/C. Therefore, the magnitude of the electric force felt by the proton is (1.6 x 10-19 C)(2750 N/C) = 4.4 x 10-16 N. The mass of a proton is approximately 1.67 x 10-27 kilograms. Therefore, the proton's acceleration is (4.4 x 10-16 N)/(1.67 x 10-27 kg) = 2.64 x 1011 m/s2. Since the proton starts from rest, its initial velocity (u) is 0. Therefore, the proton's speed after 1.40 μs is v = (2.64 x 1011 m/s2)(1.40 x 10-6 s) = 3.70 x 105 m/s.

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