The mass of an object with 500j of kenetic energy moving with a velocity of 5 m/s is

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

5 0

3 years ago

A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the

Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0

3 years ago

CAN ANY OF YALL ANSWER DIS PLS!!!!!!!! I AM GIVING 20 POINTS BTW!!!!!!!

sveticcg [70]

PHYSICAL CHANGES :
Melting an ice cube.
Boiling water.
Mixing sand and water.
Breaking a glass.
CHEMICAL CHANGES :
Digesting food.
Cooking an egg.
Heating sugar to form caramel.

4 0

3 years ago

Which of the following best describes a plane?

zloy xaker [14]

On that list of choices, 'C' is the only "example" of a plane.
None of the choices "describes" a plane.

8 0

3 years ago

Question C) needs to be answered, please help (physics)

Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.

3 0

3 years ago