An object is placed 1.0 cm from a convex mirror with a focal length of 4.0 cm.

The Sun emits a solar flare directly aimed at Earth. Which instrument on an orbiting satellite would detect the flare first?

Tatiana [17]

Answer:

X-ray Telescope

Explanation:

X-ray Telescope which can be either hard or soft is an instrument on an orbiting satellite that would detect the flare first.

This is because X-ray Telescope is technically designed to detect and determine X-rays or radio and optical emissions from flares, from places or points beyond Earth's atmosphere.

4 0

3 years ago

Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

When does an object falling vertically through the air reach terminal velocity?

zlopas [31]

Answer:

a. when the acceleration of the objects become negative

7 0

3 years ago

6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the

kkurt [141]

Answer:

tough

ques

Explanation:

5 0

2 years ago

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

Alinara [238K]

Answer:

The angular separation between the red and violet components of the spectrum that emerges from the glass is 1.72°

Explanation:

Given:

Refractive index for red light $n _{1} = 1.620$

Refractive index for violet $n' _{1} = 1.660$

Refractive index for air $n_{2} = 1$

Incident angle $\theta _{1} =$ 28.30°

According to the snell's law,

$n_{1} \sin \theta _{1} = n_{2} \sin \theta _{2}$

For red light,

$1.620 \times \sin 28.30 = 1 \sin \theta _{2}$

Where $\theta _{2} =$ transmitted angle for red light

$\theta _{2} =$ 50.18°

For violet light,

$1.660 \times \sin 28.30 = 1 \sin \theta' _{2}$

Where $\theta' _{2} =$ transmitted angle for violet light

$\theta' _{2} =$ 51.9°

Angular separation between red and violet light is given by,

$\theta ' _{2} - \theta _{2} = 51.9 -50.18 =$ 1.72°

Therefore, the angular separation between the red and violet components of the spectrum that emerges from the glass is 1.72°

6 0

3 years ago