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3 years ago

5

Bees can see into the ultraviolet region of the electromagnetic spectrum. Compared with humans, bees can sense electromagnetic w

aves that have

2 answers:

PIT_PIT [208]3 years ago

6 0

The electromagnetic spectrum spans a variety of radiation types including x-rays all the way through to radio waves. Humans are able to only see in the visible part of the electromagnetic spectrum (390-700nm) but bees can see ultraviolet radiation (10-100nm). Lower wavelength electromagnetic waves have more energy. Hence, compared with humans, bees can sense electromagnetic waves that have more energy.

Anna35 [415]3 years ago

6 0

The answer is D, “a higher frequency and a shorter wavelength”.

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

3 years ago

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea

Semmy [17]

Answer:

${\rm K} = 2.4\times 10^{-19}~J$

Explanation:

The electric field inside a parallel plate capacitor is

$E = \frac{Q}{2\epsilon_0 A}$

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

$F = qE = \frac{qQ}{2\epsilon_0 A}$

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

$C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60$

Plugging this identity into the force equation above gives

$F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q$

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is $1.6 \times 10^{-19}$

Therefore, the kinetic energy is $2.4\times 10^{-19}$

8 0

4 years ago