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aksik [14]
2 years ago
8

Many immigrants lived in​

Physics
1 answer:
Fudgin [204]2 years ago
7 0
California I think not sure
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What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?
stiks02 [169]

p=mv

p=0.005kg×3.0m/s

p= 0.015kgm/s

4 0
2 years ago
In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0
3 years ago
A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu
Zielflug [23.3K]
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
7 0
3 years ago
Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.
dezoksy [38]
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr) 
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
5 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
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