All categories

3 years ago

How small are the wavelengths of gamma ray radiation?

1 answer:

3 0

Gamma rays are the highest energy EM radiation and typically have energies greater than 100 keV, frequencies greater than 1019 Hz, and wavelengths less than 10 picometers.

You might be interested in

a stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes th

Ksju [112]

Answer:

$v=\sqrt{2gh}$

Explanation:

We could use conversation of energy. Total distance the stone will cover will be

$h=\frac{1}{2} g t^{2}$

the Final velocity will be

$v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}$

5 0

2 years ago

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

astraxan [27]

Answer:

Acceleration = 2.35 m/ $s^{2}$

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = $\frac{mg}{2}$ = 0.5 mg

Friction along incline = umg cos30° = mg $0.3\times \frac{\sqrt{3}}{2}$

Friction along incline = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = $\frac{net force}{mass}$ = 0.24 g = 2.35 m/ $s^{2}$

The height of incline = 8 m

Length of the inclined edge = 16 m

$v^{2}=u^{2}+2as$

$v^{2}= 2\times 0.24 \times 9.8\times 16$

v= 8.67 m/s

5 0

3 years ago

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

2 years ago

A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy

MrRa [10]

The kinetic energy is $\frac 1 2 m v^2$ and the height of the building doesn't matter at all.

$E = \frac 1 2 m v^2 = \frac 1 2 (2)(5)^2 = 25$ joules

8 0

2 years ago

Using poiseulli assumption derive an expression for the rate of flow of a liquid through a horizontal tube in terms of a, l, p a

Marizza181 [45]

Answer:

im sorry im just trying this app please dont report

Explanation:

please im begging you maam/sir

7 0

2 years ago