Which type of electromagnetic radiation cannot be focused? A. Gamma rays
B. X-rays
C. Ultraviolet
  D. Infrared

Answers

Answer 1

Answer: X-Rays and Gamma Rays

Related Questions

Sound waves can't travel through A. an elastic material.B. a liquid.(C. a solid.)D. a vacuum.I answered C and it's not correct so i dont know what else it could be please help
How could rescue workers use squeezing or compressing to get energy to their flashlights during rescue missions?
Cheryl has a mug that she says is made up of matter. Heather says that the hot chocolate inside the cup is made up of matter, too. Keaton says that the steam coming from the cup is also made up of matter. Mikayla says that none of it is matter because matter is too small to see. Which best describes who is correct?
How dense the medium is in the compression part of the wave, and how
A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applied to the string, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s, what is the moment of inertia of the pulley

A 3.8 kg object is lifted 12 meters. approximately how much work is performed during the lifting?

Answers

Work done is equivalent to potential energy.

W = mgh. m is mass in kg, g is acceleration due to gravity = 9.81 m/s², h is height = 12 m

= 3.8 * 9.81 * 12

= 447.336 Joules

W = mgh.

m is mass in kg,

g is acceleration due to gravity = 9.81 m/s²,

h is height = 12 m

= 3.8 * 9.81 * 12

= 447.336 Joules

An oscillating bock-spring system has a mechanical energy of 1.0 J, an amplitude of 0.10 m, and a maximum speed of 1.2 m/s. Find (a) the force constant of the spring, (b) the mass, and (c) the frequency of oscillation.

Answers

Answer:

a) F = 20 N

b) m = 1.39 kg

c) f = 1.909 Hz

Explanation:

Given

E = 1 J

A = 0.1 m

vmax = 1.2 m/s

a) F = ?

b) m = ?

c) f = ?

Solution

a) We apply the equation

E = 0.5*k*A²

then

k = 2*E/A²

k = 2*1 J/(0.1 m)²

k = 200 N/m

then we use the equation

F = kA

F = (200 N/m)(0.1 m)

F = 20 N

b) We use the formula

E = K + U

if U = 0 J

then

E = K = 0.5*m*v²

⇒ m = 2*K/v²

m = 2*1 J/(1.2 m/s)²

m = 1.39 kg

c) we apply the equation

f = (1/2π)√(k/m)

then

f = (1/2π)√(200 N/m/1.39 kg)

f = 1.909 Hz

Answer:

a) $k = 200.016\,(N)/(m)$ , b) $m = 1.389\,kg$ , c) $f = 0.524\,hz$

Explanation:

a) The maximum speed of the oscillating block-spring system is:

$v_(max) = \omega \cdot A$

The angular frequency is:

$\omega = (v_(max))/(A)$

$\omega = (1.2\,(m)/(s) )/(0.1\,m)$

$\omega = 12\,(rad)/(s)$

The mass of the system is:

$E = (1)/(2)\cdot m\cdot v_(max)^(2)$

$m = (2\cdot E)/(v_(max)^(2))$

$m = (2\cdot (1\,J))/((1.2\,(m)/(s) )^(2))$

$m = 1.389\,kg$

The spring constant is:

$\omega = \sqrt{(k)/(m) }$

$k = \omega^(2)\cdot m$

$k = (12\,(rad)/(s) )^(2)\cdot (1.389\,kg)$

$k = 200.016\,(N)/(m)$

b) The mass is:

$m = 1.389\,kg$

c) The frequency of oscillation is:

$\omega = 2\pi\cdot f$

$f = (2\pi)/(\omega)$

$f = (2\pi)/(12\,(rad)/(s) )$

$f = 0.524\,hz$

The water at the top of a waterfall has _____ energy.potential
kinetic
heat
metabolic

Answers

I believe the answer isPOTENTIAL ENERGY. The water at the top of the waterfall has Potential Energy.Potential Energy is defined as the energy that is stored when an object is atrest. At the top of the waterfall, water will have maximum potential energy. Whenthe water falls down, it is will already have kinetic energy which is energythat is produced through continuous motion. When it is finally in the bottom,water will have maximum kinetic energy.

Answer:

The water at the top of a waterfall has __potential___ energy.

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery. if the plates are pulled apart to a separation of 1.63 cm , suppose the battery remains connected while the plates are pulled apart. part a what is the capacitance?

Answers

Answer:

at the beginning: $2.3\cdot 10^(-10) F$

when the plates are pulled apart: $1.1\cdot 10^(-10) F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 (A)/(d)$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^(-12) F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^(-3) m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^(-12)F/m)(0.210 m^2)/(8.15\cdot 10^(-3) m)=2.3\cdot 10^(-10) F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^(-12)F/m)(0.210 m^2)/(0.0163 m)=1.1\cdot 10^(-10) F$

What does the object on the screen below model?A. Molecules
B. The orbit of the moon around the Earth
C. The orbits of the planets around the Sun
D. Electronic circuits

Answers

Answer:

A, molecules

Explanation:

Molecules were viewable around the time period of the computers shown on the screen

Answer:

correct option is (A)........

Use the H-R diagram to answer the following questions.What is the approximate temperature of the supergiant Betelgeuse?

Which star on this diagram is the most similar to the Sun in luminosity and temperature?