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Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se

ANTONII [103]

Answer:

Density

Explanation:

Each of the three liquids have different densities. Have you ever tried to mix oil and water? They don't mix because they have different densities.

3 0

3 years ago

Read 2 more answers

Two manned satellites approaching one another at a relative speed of 0.150 m/s intend to dock. The first has a mass of 2.50 ✕ 10

Reika [66]

Answer: $u_{1}=-0.075m/s$ and $u_{2}=0.500m/s$

Explanation:

An elastic collision is one in which both the total kinetic energy of the system and the linear momentum are conserved. That is, during the collision there is no sound, heat or permanent deformations in the bodies as a result of the impact.

Now, in the case of the satellites described here, we have:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (1) Conservation of momentum

$\frac{1}{2}m_{1}v_{1}^{2} +\frac{1}{2}m_{2}v_{2}^{2} =\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2}$ (2) Conservation of kinetic energy

Where:

$m_{1}=2.5(10)^{3}kg$ is the mass of the first satellite

$m_{2}=7.5(10)^{3}kg$ is the mass of the second satellite

$v_{1}=0.150m/s$ is the initial velocity of the first satellite

$v_{2}=0m/s$ is the initial velocity of the second satellite (we are told it is at rest)

$u_{1}$ is the final relative velocity of the first satellite

$u_{2}$ is the final relative velocity of the second satellite

Now, as we know the second satellite is at rest before the collision, equations (1) and (2) change to:

$m_{1}v_{1}=m_{1}u_{1}+m_{2}u_{2}$ (3)

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2}$ (4)

Solving this system of equations we have the equations for $u_{1}$ and $u_{2}$ :

$u_{1}=\frac{v_{1}(m_{1}-m_{2})}{m_{1}+m_{2}}$ (5)

$u_{2}=\frac{2m_{1}v_{1}}{m_{1}+m_{2}}$ (6)

Substituting the known values on both equations:

$u_{1}=\frac{0.150m/s(2.5(10)^{3}kg-7.5(10)^{3}kg)}{2.5(10)^{3}kg+7.5(10)^{3}kg}$ (7)

$u_{1}=-0.075m/s$ (8) This is the final relative velocity of the first satellite

$u_{2}=\frac{2(2.5(10)^{3}kg)(0.150m/s)}{2.5(10)^{3}kg+7.5(10)^{3}kg}$ (9)

$u_{2}=0.500m/s$ (10) his is the final relative velocity of the second satellite

7 0

4 years ago

A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity rela

ASHA 777 [7]

Answer:

Velocity of Gulf Stream= $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

Explanation:

Given

speed of ship relative to Gulf stream= $25^\circ$ west of north at a speed of 4 m/s

speed of ship relative to earth= $5^\circ$ west of north at a speed of 4 .8 m/s

Let $V_S$ be the velocity of the ship relative to the earth and $V_{sg}$ be the velocity of the ship with respect to the Gulf stream

Let the west direction be negative x axis and north direction be positive x axis

Now

$V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s\\$

magnitude of the velocity is given by

$V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s\\$

Let the velocity of gulf stream makes an angle $\theta$ with the positive y axis we have

$\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ$

Velocity of Gulf Stream $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

8 0

3 years ago

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

Over [174]

There are few equations that link acceleration and velocity.

One of them is derived from the definition of acceleration - is the rate of change of velocity in respect with time

a = v-u/t
at=v-u

a = acceleration
v=velocity at the end
u=velocity at the start
t=time

Now put everything you know and solve for time.

6t = 42 -0
6t = 42
t = 42 / 6
t = 7 second

It takes 7 seconds for the roccet to reach a velocity of 42 m/s

Hope it helps :).

4 0

3 years ago

a bucket of mass 1.90 kg is whirled in a vertical circle of radius 1.45 m . at the lowest point of its motion the tension in the

NeTakaya

The speed of the bucket is v = 4.61 m/sec

For the free body diagram, and apply Newton's second law:

T - mg = m(v2/r)

Plug in known values,

25 - 1.00*g = 1.00(v2/1.40)

Solve for v,

v = 4.61 m/sec

What is motion?

Motion is the change of position or direction of a body over time. Movement along a line or curve is called a translation. Movement that changes the direction of the body is called rotation. In both cases, all points on the body have the same speed (directional speed) and the same acceleration (time rate of change of speed). The most common motion combines both translation and rotation.

There are 6 types of motion

Oscillatory Motion.

Rotational Motion.

Translational Motion.

Periodic Motion.

Circular Motion.

Linear Motion.

Uniform Motion.

Non-Uniform Motion.

To learn more about motion, refer;

brainly.com/question/3421105

#SPJ4

3 0

1 year ago