All categories

2 years ago

Which planets are visible without a telescope

1 answer:

3 0

The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.

You might be interested in

1) what is a calorie?<br>

Bogdan [553]

I read it’s a unit of energy

3 0

2 years ago

1. A baseball is thrown horizontally at 45 m/s. The ball slows down at a rate of 5 m/s?

meriva

Answer:

Explanation:

a )

initial velocity u = 45 m/s

acceleration a = - 5 m/s²

final velocity v = 0

v = u - at

0 = 45 - 5 t

t = 9 s

b )

s = ut - 1/2 at²

= 45 x 9 - .5 x5x 9²

405 - 202.5

202.5 m

2 )

a )

s = ut + 1/2 a t²

u = 0

s = 1/2 at²

= .5 x 9.54 x 6.5²

= 201.5 m

b )

v = u + at

= 0 + 9.54 x 6.5

= 62.01 m / s

a )

acceleration = (v - u) / t

= (34 - 42) / 2.4

= - 3.33 m /s²

b )

v² = u² - 2 a s

34² = 42² - 2 x 3.33² s

s = 27.41 m

c )

Average velocity

Total displacement / time

= 27.41 / 2.4

= 11.42 m /s

4 )

a )

v = u + at

v = 0 + 3 x 4

= 12 m /s

b )

s = ut + 1/2 a t²

= o + .5 x 3 x 4²

= 24 m

8 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST FILE IS ATTACHED

Ainat [17]

Answer:

chemical, electrical

mechanical, electrical

heat, light

7 0

3 years ago

Nvhfbvhefvhabjbnvjhhjaqiuv.....

Agata [3.3K]

Answer:

I don’t understand it

Explanation:

If anyone can the. Let me know

5 0

3 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago