To Express The Distance Between The Milky Way Galaxy And Other Galaxies The Most Appropriate Unit Of Measure Is The..?

Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is th

postnew [5]

Answer:a

Explanation: they are all positive

4 0

2 years ago

Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef

Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μ $F_n$ where μ is the coefficient of friction, and $F_n$ is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

$F_n=mg$ so

$F_n=(105)(9.8)$ so the weight of the sled is

$F_n=$ 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0

2 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a circular orbit and a low orbit near the surface as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

2 years ago

An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

Center of Curvature:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

The Radius of Curvature:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

Focal Length:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

$f = \frac{R}{2}$

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

$f = \frac{24\ cm}{2}$

f = 12 cm

8 0

2 years ago

2. In each of A, B, C, & D below, a sphere is moving from left to right next 20 points

laila [671]

Answer:

zero

rank the magnitude of the average velocity over the first 2 second

8 0

2 years ago