All categories

4 years ago

Hi can you please help me, im really stuck

1 answer:

3 0

I have the answer for A. Since there is blockage in the ear canal, some sound waves may not be able to get through or travel as quickly so you would have trouble hearing

You might be interested in

The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to

DedPeter [7]

Answer:

$9.4\cdot 10^{10} m$

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

$\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}$

where

$r_o$ is the distance of the new object from the sun (orbital radius)

$T_o=180 d$ is the orbital period of the object

$r_e = 1.50\cdot 10^{11} m$ is the orbital radius of the Earth

$T_e=365 d$ is the orbital period the Earth

Solving the equation for $r_o$ , we find

$r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m$

3 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Help :((((((((((((((((((((

Ann [662]

3-6 seconds time interval is the object slowing down.

The correct option is C.

<h3>What is a time interval?</h3>

The time interval is the span of time among two specified times. To put it another way, it is the amount of time that has passed between the event's start and finish.

<h3>What are different time intervals?</h3>

The time interval is the length of time that the aim uses to gather data and determine values. The critical overview can be one or more seconds, minutes, hours, days, weeks, or months. The period must be greater than zero and positive. When providing minutes, the amount of minutes must divide evenly by 60.

To know more about Time interval visit:

brainly.com/question/28238258

#SPJ13

The complete question is -

During which time interval is the object slowing down ?

a- 8-10 seconds

b- 6-8 seconds

c- 3-6 seconds

d- 0-3 seconds

4 0

1 year ago

Blake is setting up his tent at a renaissance fair. If the tent is 8 feet tall and the tether can be staked no more than 2 feet

Mariana [72]

The stake, height and tether length of the tent form a right angle triangle where the tether length is the hypotenuse.
Applying Pythagoras theorem:
length² = height² + (stake distance)²
length = √(8² + 2²)
length = 8.5 feet

3 0

3 years ago

Match the object with its characteristic.

Triss [41]

What object do you need to match

8 0

3 years ago