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2 years ago

What and where is the asteroid belt? Please ANSWER THIS

Physics

2 answers:

Mrrafil [7]2 years ago

7 0

Answer:

The asteroid belt (sometimes referred to as the main asteroid belt) orbits between Mars and Jupiter. It consists of asteroids and minor planets forming a disk around the sun. It also serves as a sort of dividing line between the inner rocky planets and outer gas giants.

Explanation:

Hope this helps! :)

Rudiy272 years ago

3 0

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

4 0

3 years ago

A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If

Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0

2 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

Learn more about centripetal and angular acceleration here:

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8 0

3 years ago

10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m

Mariulka [41]

Work formula:

$W = Fd\cos(\theta)$

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

$W = (50N)(1.0)(0) = 0$

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

$W = (50 N)(1.0m)(1)$

$W = 50$

total work = 50 joules

6 0

2 years ago

A basketball player is 4.22 m from

max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

3 0

2 years ago

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