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3 years ago

What is the distance to a star whose parallex is 0.1 sec?

Physics

1 answer:

Arte-miy333 [17]3 years ago

6 0

Answer:

$30.86\times 10^{13} km$

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

$d=\frac{1}{P}$

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

$d=\frac{1}{0.1}\\d=10 parsec$

And also know that,

$1 parsec=3.086\times 10^{13} km$

Therefore the distance of the star is $30.86\times 10^{13} km$ away.

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if an object is moving with a velocity of 25m/s, an acceleration of 5m/s and has a travel time of 5s, what is the final velocity

miv72 [106K]

V = at + V0

where v0 is the initial speed, a is the acceleration and t is the time.

So:

v = 5m/s^2*5s + 25m/s = 50 m/s

4 0

3 years ago

A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

6 0

3 years ago

A plane flies from alphaville to betaville and then back to alphaville. when there is no wind, the round trip takes 4 hours and

kykrilka [37]

Refer to the diagram shown below.

d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.

With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)

With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5
$\frac{d}{v+100} + \frac{d}{v-100} =5 \\ \frac{2vd}{v^{2}-100^{2}} =5 \\ 2vd = 5(v^{2}-10^{4})$ (2)

From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi

Answer: 1199 miles (nearest integer)