Which type of galaxy can be commonly shaped by collisions or the gravitational pull of other galaxies?

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2 m/s to 4 m/s?

mash [69]

Answer:

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

Explanation:

The momentum of an object is given by the product between its mass (m) and its velocity (v):

$p=mv$

Let's apply this formula to calculate the initial momentum and final momentum of the ball:

- initial momentum:

$p_i = m v_i = (6 kg)(2 m/s)=12 kg m/s$

- Final momentum:

$p_f = m v_f = (6 kg)(4 m/s)=24 kg m/s$

So, the correct answer is

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

3 0

3 years ago

Read 2 more answers

Write a difference between force and work?

Nesterboy [21]

Answer:

force is a derived quantity

work is a fundamental quantity

2.Force is measured in Newton why work is measured in Newton per seconds (N/s)

Explanation:

no

3 0

3 years ago

Read 2 more answers

A motorist drives north for 35.0 minutes at 85.0 kph. He then stops for 15.0 minutes. The motorist then drives 130.0 km in 2.0 h

o-na [289]

Answer:

A. 216.36 $\frac{km}{h}$

B. 96.56 $\frac{km}{h}$

Explanation:

Let $s_{1}$ be the distance in first part.

$s_{1}$ = velocity × time

$s_{1}$ = 85 × $\frac{35}{60}$

$s_{1}$ = 49.58 km

Let $s_{2}$ be the distance in first part.

$s_{2}$ = 130 km

Average velocity = $\frac{Total displacement}{Total time}$

When second leg of the trip is

A. Toward north

Average velocity = $\frac{[tex] s_{1} + s_{2}$ }{Total time} [/tex]

Average velocity = $\frac{130+49.58}{0.25+0.58}$

Average velocity =216.36 $\frac{km}{h}$

B. Toward south

Average velocity = $\frac{[tex] s_{1} - s_{2}$ }{Total time} [/tex]

Average velocity = $\frac{130-49.58}{0.25+0.58}$

Average velocity =96.56 $\frac{km}{h}$

3 0

3 years ago

Unless an object at rest is acted upon by a force, it stays at rest due to its

Lelu [443]

There are a lot of things about this whole statement that bother me,
but I'm pretty sure you're not interested in those. You just want a word
to glue onto the end and complete the statement. OK. Use "inertia".

4 0

3 years ago