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3 years ago

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A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 j

oule·seconds. What is the frequency of the photon?
5.13 × 1019 hertz

5.60 × 1017 hertz

2.45 × 1017 hertz

2.45 × 10-12 hertz

7.80 × 10-19 hertz

2 answers:

suter [353]3 years ago

4 0

The correct answer is: "A. 5.13 x 10^19 hertz"

Wittaler [7]3 years ago

4 0

The answer is A, 5.13 x 10^19 hertz

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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on

g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J

6 0

3 years ago

Science<br> Write two examples of heating appliances

bearhunter [10]

Answer:

a} <u>Infrared radiant heaters.</u>

<u>b} Fan heaters.</u>

Explanation:

I hope this is okay, should I give more.

4 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

A flat uniform circular disk (r= 2.00m,

dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0

3 years ago

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter

valentinak56 [21]

Wee can use here kinematics

as we know that

$y = v*t + \frac{1}{2} at^2$

for shorter tree we know that

$y = 0 + \frac{1}{2}*9.8 * 2^2$

$y = 19.6 meter$

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

$y = v*t + \frac{1}{2} at^2$

$39.2 = 0 + \frac{1}{2}*9.8 * t^2$

$t = 2.83 s$

so the time taken is 2.83 s

4 0

2 years ago