Uranium 235 can be split if it is hit by:

The shanghai maglev train is capable of reaching speeds of up to 350km h-1 what is the speed in m/h

LenaWriter [7]

Answer:

The shanghai maglev train is capable of reaching speeds of up to 217.48 in mph

217.48 mph

Explanation:

8 0

3 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that

andreev551 [17]

Answer:

a) 1450watts

b) 564watts

c) 1.11

Explanation:

Power consumed = IV

I is the current rating

V is the operating voltage

If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;

a) For blow-dryer

Operating voltage = 120V

Its current rating = 12A

Power consumed = IV

= 120×12

= 1440watts

b) For vacuum cleaner:

Operating voltage is the same as that of blow dryer = 120V

Its current rating = 4.7A

Power consumed = IV

= 120×4.7

= 564watts

c) Energy used = Power consumed × time taken

Energy used = Power × time

Energy used by blow dryer = 1440×20×60

= 1,728,000Joules

Energy used up by vacuum cleaner = 564×46×60

= 564×2760

= 1,556,640Joules

Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11

4 0

3 years ago

Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di

Goshia [24]

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v = 250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

6 0

2 years ago

Your tired 10 lb. baby cousin is cradled in your arms as you're walking back and forth across a 15m room to get her to fall asle

Mashcka [7]

Work done = Force x Distance

Force = 10 lb = 44.5 N

Work Done = 44.5 N x 15 m

= 667.5 N-m

6 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago