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A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w

astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\$

So to find t that is the same time for all the motion

$t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s$

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

$Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}$

5 0

2 years ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion?

gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- A planet's distance from the sun will not be the same in six months. --> using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

- A planet's speed as it moves around the sun will not be the same in six months. --> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. --> using the 3rd law. In fact, the ratio $\frac{a^3}{T^2}$ (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Is not supported by any Kepler's law.

8 0

2 years ago

Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted

34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium