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2 years ago

A star’s parallax angle is 1.0. How far away is the star in light years?

Physics

2 answers:

larisa [96]2 years ago

7 0

Distance = 2AU / tan1.0

If you mean 1.0 is in degrees, then Distance = 114.58 AU

valkas [14]2 years ago

6 0

The distance is 114.58

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Phương trình trạng thái tổng quát của khí lí tưởng diễn tả là

tatyana61 [14]

Expla

pV/T=const goodluke

6 0

2 years ago

The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter

Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

6 mg

The value of h when the normal force is zero

3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0

3 months ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

2 years ago

PLS HELP QUICK!!!!!!!!

vazorg [7]

Position 1 I believe because that is when the potential energy is released on the downfall

5 0

2 years ago

Read 2 more answers

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

7 0

2 years ago