What is the lowest and highest mass the triple beam balance can measure

storchak [24]

The closer to the equator, the hotter the climate will be.

6 0

4 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago

100 POINTS!!! :D

charle [14.2K]

Explanation:

... in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

4 0

3 years ago

Read 2 more answers

A musician on a unicycle is riding at a steady 6.2 m/s while playing their well tuned oboe at an A above middle C (440 Hz). They

Verizon [17]

Answer:

The frequency of the beat is $f_t = 8Hz$

Explanation:

From the question we are told that

The speed of the ride $v = 6,2 m/s$

The frequency of the oboe is $f = 440Hz$

The temperature outside is $T = 27^oC$

Generally the speed of sound generated in air is mathematically evaluated as

$v_s = 331 + 0.61 T$

Substituting value

$v_s = 331 + 0.61 *27$

$v_s = 347.47 \ m/s$

The frequency of sound(generated by the musician on he park) getting to the musicians on the unicycle is mathematically evaluated as

$f_a = \frac{v_s }{v_s - v} f$

substituting values

$f_a = \frac{347.47 }{347.47 - 6.2} * 440$

$f_a = 448Hz$

Since the musician on the park is not moving the frequency of sound (from the musicians riding the unicycle )getting to him is = 440Hz

The beat frequency these musician here is mathematically evaluated as

$f_t = f_a - f$

So $f_t = 448 - 440$

$f_t = 8Hz$

3 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

4 years ago