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Diano4ka-milaya [45]

3 years ago

13

The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at a

bout 14 km/s when it went by Pluto. If it maintained this speed, how long would it take New Horizons to reach the nearest star, Proxima Centauri, which is about 4.3 light-years away? (Note: It isn’t headed in that direction, but you can pretend that it is.)

1 answer:

LenKa [72]3 years ago

7 0

Answer:

2.91 x 10¹² sec

Explanation:

d = distance of nearest star, Proxima Centauri = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m

v = speed of new horizon probe = 14 km/hr = 14000 m/s

t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?

Using the equation

d = v t

Inserting the values given

4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

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Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

$p_A = p_B$

Pressure is given by the ratio between force F and area A, so we can write

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

The force exerted on each piston is just equal to the weight of the corresponding mass: $F=W=mg$ , where m is the mass and g is the gravitational acceleration. So the equation becomes

$\frac{m_A g}{A_A}=\frac{m_B g}{A_B}$

Now we can rewrite the mass as the product of volume, V, times density, d:

$\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}$

We also know that $A_B = 2.0 m^2\\A_A = 1.0 m^2$

So we can further re-arrange the equation (and simplify g as well):

$\frac{V_A d_A}{1}=\frac{V_B d_B}{2}$

$\frac{d_A}{d_B}=\frac{V_B}{2V_A}$

We are also told that block B has bigger volume than block A: $V_B > V_A$ . However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0

3 years ago

A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k

VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

5 0

3 years ago

Satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to orbit at an

Volgvan

Answer:

0.63

Explanation:

We are given that

Radius of earth, $R_E=6370 km$

Radius of orbit A, $R_A=7590+6370=13960 km$

Radius of orbit B, $R_B=6370+15800=22170 km$

We have to find the ratio of the potential energy of satellite B to that of satellite A in orbit.

Potential energy of orbit A= $U_A=-\frac{GM}{R_A}=-\frac{GM}{13960}$

Potential energy of orbit B= $U_B=-\frac{GM}{R_B}=-\frac{GM}{22170}$

$\frac{U_B}{U_A}=\frac{13960}{22170}=0.63$

Hence,the ratio of the potential energy of satellite B to that of satellite A in orbit=0.63

8 0

4 years ago

Which example shows separation of different wavelengths of light due to diffraction?

Veseljchak [2.6K]

Answer:

spectrum

Explanation:

The spectrum formed different colors due to refraction of light

8 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

4 years ago