Plz help

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

Ket [755]

Hope this can help u

7 0

3 years ago

Menciona un ejemplo de la vida cotidiana donde se apliquen las cuatro fuerzas fundamentales de la naturaleza.

Gekata [30.6K]

Yesterday was the day that we got canceled on it so we had a lot

4 0

3 years ago

A student wish to measure the gravitational acceleration g. She does it by releasing a small lead ball from rest and measures th

Goryan [66]

Answer:

(9.64 +- 0.86) m/s^2

Explanation:

The generic motion equation for constant acceleration is

$x = X0 + v0 * t + \frac{1}{2}*a * t^2$

Where

X0: initial position

v0: initial speed

a: acceleration

t: time

If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

$x = \frac{1}{2}*a * t^2$

And the acceleration can be obtained as:

$a = 2*\frac{x}{t^2}$

Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

$g = 2*\frac{100}{144^2} = 9.64e-3 \frac{mm}{ms^2} = 9.64 \frac{m}{s^2}$

For the uncertainty we have to calculate the relative uncertainties first

For the distance (100 * 0.3)/100 = 0.3%

For the time (100 * 3)/144 = 2.08%

For multiplications or divisions the relative uncertainties are added

0.3% + 2.08% + 2.08% = 4.46%

We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.

5 0

3 years ago

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between the

Sever21 [200]

Answer:r=5.824 mm

Explanation:

Given

Charge $q_1=70 nC$

$q_2=70 nC$

Force between them $F=1.30 N$

Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$

5 0

3 years ago

Find the work w1 done on the block by the force of magnitude f1 = 60.0 n as the block moves from xi = -3.00 cm to xf = 1.00 cm

Norma-Jean [14]

By definition, the work done by a force is given by:
$W1 = F1 * d
$
Where,
F: magnitude of force
d: distance traveled.
Substituting values we have:
$W1 = F1 * (xf - xi)

W1 = 60 * (0.01 - (-0.03))
$
$W1 = 60 * (0.01 + 0.03)

W1 = 60 * (0.04)

W1 = 2.40 J
$
Answer:
the work w1 done on the block by the force of magnitude f1 = 60.0 n is:
W1 = 2.40 J

6 0

3 years ago