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2 years ago

Calculate the average orbital speed of Ceres in

1 answer:

8 0

] Ceres is composed of rock and ice and is estimated to comprise approximately one third of the mass of the entire asteroid belt. Ceres is the only object in the asteroid belt known to be rounded by its own gravity (though detailed analysis was required to exclude 4 Vesta). From Earth, the apparent magnitude of Ceres ranges from 6.7 to 9.3, peaking once every 15 to 16 months,[21]hence even at its brightest it is too dim to be seen with the naked eye except under extremely dark skies.

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A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t

Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

$10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]$

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

$v_{f}^{2} =v_{o}^{2}-2*a*x$

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

$0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]$

4 0

2 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

2 years ago

Why is the crank on a meat grinder larger than the crank on a pencil

MArishka [77]

It takes more work to use a meat grinder

4 0

2 years ago

Read 2 more answers

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$