WRONG ANSWERS WILL BE REPORTED

As part of a NASA experiment, golfer Tiger Woods drives a golf ball on the moon, where g = 1.60 m/s2. He launches a golf ball wi

Elanso [62]

Answer:

Horizontal distance, R = vo2 sin(2x45)/g

v = 285 km/h = 79.17 m/s

R = 79.172 x 1/1.6

R = 3917 m

R = 3.9 km

8 0

3 years ago

What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and

Firdavs [7]

The sun emits electromagnetic waves with a power of

4.0 ∗ 10 (26) W.

6 0

2 years ago

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted p

jarptica [38.1K]

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

7 0

3 years ago

When is Ampère’s Law valid? Evaluate all of the following statements, and explain your reasoning. a. When there is a high degree

Reil [10]

Answer:

Check Explanation

Explanation:

Ampere's Law gives the expression for the magnetic field produced by current in a current carrying material.

Mathematically, it is explained that the product of permeability and current is equal to the sum of the dot product of the magnetic field in the direction of the length elements and the length elements.

μ₀I = ∫ B.dl (note that this integral is around the entire loop of current carrying material)

This law is valid for infinite current carrying wires because the magnetic field around such infinite current carrying wire is symmetrical.

The magnetic field around a finite length of wire is non-symmetrical and the current isn't continuous.

For Ampere's Law to work, the magnetic field around the wire must be symmetric, for straightforward computation and the current in the wire needs to be continuous, not cut short at the edges which makes the field around the wire non-symmetrical.

So, this condition is interlinked.

Hence, it is evident that only option A is absolutely correct.

Options C and D on their own aren't totally correct, but in conjunction with option E, they are correct.

Hope this Helps!!!

8 0

3 years ago

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

$a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}$

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

$v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}$

Finally, since $\sin60\°=\frac{\sqrt{3} }{2}$ and $\sin45\°=\frac{\sqrt{2} }{2}$ , we can conclude that:

$t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}$

In words, the cannonball b spends more time in the air than cannonball a.

5 0

2 years ago