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2 years ago

14

The distance between twin satellites that were originally 150 meters apart are now 300 meters apart. Which best describes the gr

avitational force between the twin satellites?

1 answer:

Marina CMI [18]2 years ago

3 0

Answer:

B or It decreased by 1/4.

Explanation:

Answer for edgenuity

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The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co

motikmotik

Explanation:

The position vector r:

$\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}$

The velocity vector v:

$\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}$

The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

$\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}$

$\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}$

5 0

1 year ago

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

patriot [66]

If I have done my math correctly, your answer is 67.5

4 0

2 years ago

Read 2 more answers

Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass

pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

4 0

2 years ago

In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north

Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

$\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}$ (1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

$F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}$

5 0

2 years ago