Name three methods that can be used to change the overall density of an object.

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.

PtichkaEL [24]

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

angular acceleration = angular speed /time
angular acceleration = 12.9/2.98 = 4.329rad/s²

(b) Find the angle in radians through which it rotates in this time interval.

angular speed = 2x3.14xf
12.9rad = 2 x3.14

rad = 6.28/12.9
rad = 0.487

Now we convert rad to angle

1 rad = 57.296°
0.487 = unknown angle
unknown angle =57.296 x 0.487 = 27.9°

The angle in radians = 27.9°

8 0

3 years ago

A. Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with

zzz [600]

(a) 10 GHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

(a). 1 MHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

a. Frequency is the measure of number of times a same thing will be repeated in a given time interval for a given time. And wavelength is the measure of distance between two successive crests or troughs. So wavelength and frequency are inversely proportional to each other. And velocity of light is the proportionality constant.

So frequency of microwave radiation = Speed of light/Wavelength of radiation

Frequency = $\frac{3*10^{8} }{3*10^{-2} }$

Frequency = $10^{8+2} = 10^{10}=10 GHz$

So 10 GHz is the frequency of microwave radiation.

b). As microwave is a part of light waves, so it will be experiencing the speed of light.

As the speed is 3* $10^{8}$ m/s and the distance between the two mountains is given as 50 km, then time can be calculated as

Time = Distance/Velocity

Time = $\frac{50*10^{3} m}{3*10^{8} }=16.67*10^{3-8}=16.67*10^{-5}$

So time = 0.167 ms.

Thus, 0.167 ms is required by the microwave to travel between two mountains.

6 0

3 years ago

A mountain climber weighs 42.0 N. If he climbs a hill 100m high. Calculate the work done in joules

quester [9]

Answer:

The answer is 4200 J.

Explanation:

The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :

W = F×D

F = 42N

D = 100m

W = 42 × 100

= 4200 J

5 0

2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

Maximum height reached by the helicopter:

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

height fallen freely by Austin:

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

Velocity just before opening the parachute:

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

Time taken by the helicopter to fall:

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

Now the height fallen in the remaining time using parachute:

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

Now the height of Austin above the ground when the helicopter crashed on the ground:

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

3 years ago

Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)

OlgaM077 [116]

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

 (weight) x (distance from the pivot) on one side
is equal to
 (weight) x (distance from the pivot) on the other side.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

 (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).

So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

#1).
(400 N) in the middle-seat, facing (200 N) in the end-seat.

 (400) x (1) = (200) x (2)

#2).
(200 N) in the middle-seat, facing (100 N) in the end-seat.

 (200) x (1) = (100) x (2)

#3).

On one side: (300 N) in the end-seat (300) x (2) = 600

On the other side:
 (400 N) in the middle-seat (400) x (1) = 400
 and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . 600

These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.

5 0

3 years ago

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