Which image shows the correct way of lining up vectors to add them

A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and

Airida [17]

Answer:

Explanation:

Given

$N_1=1 rev/s$

angular velocity $\omega =2\pi N_1=6.284 rad/s$

Combined moment of inertia of stool,student and bricks $=6\ kg.m^2$

Now student pull off his hands so as to increase its speed to suppose $N_2$ rev/s

$\omega _2=2\pi N_2$

After Pulling off hands so final moment of inertia is

$I_2=2\ kg-m^2$

Conserving angular momentum as no external torque is applied

$I_1\omega _1=I_2\omega _2$

$6\times 6.284=2\times \omega _2$

$\omega _2=18.85\ rad/s$

$N_2=3 rev/s$

7 0

3 years ago

A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin

N76 [4]

Answer: The total momentum before the docking maneuver is $mV_{1}+MV_{2}$ and after the docking maneuver is $(m+M) U$

Explanation:

Linear momentum $p$ (generally just called momentum) is defined as mass in motion and is given by the following equation:

$p=m.v$

Where $m$ is the mass of the object and $v$ its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision $p_{i}$ will be the same as the total momentum of these same two objects after the collision $p_{f}$ ".

$p_{i}=p_{f}$

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

$p_{i}=mV_{1}+MV_{2}$

Where:

$m$ is the mass of the vehicle

$V_{1}$ is the velocity of th vehicle

$M$ is the mass of the space station

$V_{2}$ is the velocity of the space station

And the final momentum is:

$p_{f}=(m+M)U$

Where:

$U$ is the velocity of the vehicle and space station docked

6 0

3 years ago

Read 2 more answers

An organism is made up of

lilavasa [31]

B.) A group of cells working together. If it were D.), the question would have had to be A cell is made up of...

8 0

3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

$\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C$

Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

$\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C$

6 0

3 years ago

What is the acceleration of a 10 kg mass pushed 5 n forceushed by a 5n(kg-m/s2) (use gresa method)

Pepsi [2]

In this problem,

Applied force(F) = 10 N

The object’s mass (m) is 5 kg.

Having said that,

An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.

i.e., Mass + Acceleration = Force (a)

F= m×a

Therefore,

A= F÷m

A= (10÷5) m/sec²

A= 2 m/sec²

Consequently, the object’s acceleration,

A=2 m/sec²

Concept of force and acceleration:

This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.

It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.

To know more about such problems, visit:

brainly.com/question/16743612

#SPJ4

6 0

1 year ago