All categories

3 years ago

Lol i'm going to fail please help

Physics

1 answer:

Novosadov [1.4K]3 years ago

6 0

Answer:

Explanation:

The frequency is 16.0 Hz. That means that 16 of these waves can pass a single point in 1 second. We are given frequency and wavelength. The equation that relates them is

$f=\frac{v}{\lambda}$ where f is frequency, v is velocity, and λ is wavelength. Putting all this together:

$16.0=\frac{v}{97.5}$ and solving for velocity,

v = 16.0(97.5) so

v = 1560 m/s. This wave can travel 1560 meters in a single second!!! Now that we know this velocity, we can use it in a proportion to find our unknown, which is how long, t, it will take to hear this sound 11000m away. (11 km is 11000m):

$\frac{1560m}{1s}=\frac{11000}{t}$ and cross multiply to get

1560t = 11000 so

t = 7.1 seconds

You might be interested in

A pendulum makes 60 vibrations in 15 secs what’s its frequency

o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

cycle or vibrations=60

tame taken = 15

now,

frequency = no. of cycle / time taken

= 60/15

=4 hertz

hence, the its frequency = 4hertz

8 0

3 years ago

Graphs do not need to have a title.true or false

katovenus [111]

The answer for this question should be "false".

3 0

3 years ago

Read 2 more answers

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

Need help asap pls

ozzi

Detailed Explanation:

1) Rusting of Iron

4Fe + 3O2 + 2H2O -> 2Fe2O32H2O

Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4

Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4

2) Fermentation of sucrose…

C12H22O11 + H2O -> 4C2H5OH + 4CO2

Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12

Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12

Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.

All the Best!

8 0

2 years ago

Read 2 more answers

A rubber ducky is placed 20 cm from a thin convex lens with a focal length of 15 cm. Which statement correctly describes the nat

blondinia [14]

I believe the answer is B, a real and inverted image is formed on the side of the lens opposite the rubber ducky. The focal length is 15 cm and therefore the center of curvature (2F) will be 30 cm. When the object is placed between F and 2F (in this case 20 cm) in front of a convex lens, an inverted, real image is formed on the other side of the lens.

6 0

3 years ago

Read 2 more answers