Find the x-component and y-component of 65.0 and 15.3 PLEASE HELPPPP

Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this

Crank

Energy of gamma rays is given by equation

$E = h\nu$

here we know that

h = Planck's constant

$\nu = frequency$

now energy is given as

$E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}$

$E = 7.52 \times 10^{-13} J$

now by above equation

$E = h\nu$

$7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu$

$\nu = 1.14 \times 10^{21} Hz$

now for wavelength we can say

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}$

$\lambda = 2.63 \times 10^{-13} m$

3 0

3 years ago

According to Kepler, what do all bodies in orbit around another have in common?

dolphi86 [110]

D.
All follow an elliptical path that has two foci, rather than a circular path.

Kepler found paths are elliptical, not circular

3 0

3 years ago

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

pentagon [3]

Answer:

2.877 m/s

Explanation:

According to the laws of conservation of linear momentum,

the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)

Let both players are tackled and moving in V velocity

M and m - masses of the players
U and u - velocities of them respectively (both velocities are towards east direction )

momentum before impact = momentum after impact

→MU + →mu = →(M+ m )v

91.5 * 2.73 + 63.5 * 3.09 = (91.5 + 63.5) * V

→V = 2.877 m/s (To East)

3 0

3 years ago

Read 2 more answers

How do you calculate the potential energy by it's height and weight? and is there any other way to solve it?

White raven [17]

Answer:

PE = mgh

Explanation:

m is mass, g is gravitational constant (9.81) and h is height

5 0

3 years ago

An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately 12.77 seconds.

c) The speed of the package as it lands is approximately 145.51 m/s.

d) The path of the package based on a stationary frame of reference is parabolic

e) The path of the package as seen from the plane is directly vertical downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

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

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0

2 years ago