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3 years ago

7

A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When

each has fallen 1 m, they both have the same?

1 answer:

S_A_V [24]3 years ago

5 0

Answer:

time of fall and the final velocity

Explanation:

the mass of solid ball is more than the mass of hollow ball.

According to the third equation of motion

v² = u² + 2gh

As the final velocity v does not depend on the mass of the object, so the final velocity of both the ball is same.

According to the first equation of motion

v = u + gt

As v is same for both the balls, the time is also same for both the balls.

So, they both have same time of fall and final velocity.

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A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r

kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = $K.E=\frac{1}{2}I\omega^2$

where,

I is the moment of inertia = mr²

on substituting the values, we get

$K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2$

or

K.E = 0.0075 J

3 0

4 years ago

If a circle was flattened by pushing down on it, it would most likely form into the shape of ----------- which has ----------- f

murzikaleks [220]

The question is looking for "ellipse" and "two" to fill in the blanks.

8 0

4 years ago

Read 2 more answers

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

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

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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3 years ago

A person suffering from anaemia gets tired

avanturin [10]

Answer: yes.

Explanation:

4 0

3 years ago

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m

Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}$

F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a = 8.20 x 10⁻⁸

$a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}$

a = 9 x 10²² m/s²

c) speed of the electron

$a =\dfrac{v^2}{r}$

$9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}$

v² = 4.77 x 10¹²

v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}$

T = 1.53 x 10⁻¹⁶ s

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3 years ago

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