Answer:
0.00000019631 meters
Explanation:
Rearrange Fg=(Gm1m2)/r^2 for r
You get r=sqrt((Gm1m2)/Fg)
G=6.674*10^-11
r=sqrt((6.674*10^-11)(0.13)(0.2)/45)
r=0.00000019631 meters, or 0.00019631 mm
The so-called "velocity-time" graph is actually a "speed-time" graph. At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.
'Velocity' is a speed AND a direction. Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion. It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
so my take is that you'll never see a true "velocity-time" graph.
At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
each instant. The graph of uniform circular motion, for example, would have a straight
horizontal line for speed, and a 'sawtooth' wave for direction.
Answer:
the charge carriers have an energy 2.8 10⁻¹⁹ J
Explanation:
The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.
The energy of a photon is given by the Planck expression
E = h f
the speed of light, wavelength and frequency are related
c = λ f
we substitute
E =
a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m
we calculate the energy
E = 6.626 10⁻³⁴ 3 10⁸/700 10⁻⁹
E = 2.8397 10⁻¹⁹J
therefore the charge carriers have an energy 2.8 10⁻¹⁹ J,
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)

