What percent of air by volume is nitrogen

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

Read 2 more answers

A player holds two baseballs a height h above the ground. He throws one ball vertically upward at speed v0 and the other vertica

Degger [83]

Answer:

a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

v = √(v₀² + 2g h)

The times to get to the ground

ball 1

v = v₀ - g t₁

t₁ = $\frac{v_{o} - v }{ g}$

ball 2

v = -v₀ - g t₂

t₂ = - \frac{v_{o} + v }{ g}

From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

Δt = t₂ -t₁

Δt = $\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]$

Δt = 2 v₀ / g

6 0

3 years ago

Anyone know how to do this ?

soldi70 [24.7K]

Not really but I need points lol

3 0

3 years ago

Read 2 more answers

Gyroscopic wander can be divided in to two categories, one is DRIFT, what is the other?

Tju [1.3M]

Gyroscopic wander can be divided into two categories and these are:

Drift
Topple

<h3>What is gyroscopic wander?</h3>

Gyroscopic wander can be defined as a movement of the spin axis (axis of rotation) away from a specific fixed direction.

Based on scientific information and records, there are two main types of gyroscopic wander and these include the following:

Drift

Topple

Read more on gyroscopic wander here: brainly.com/question/12168194

#SPJ12

4 0

2 years ago

When does air become turbulent around a thrown ball?

kherson [118]

Answer: B (on the rough side)

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8 0

3 years ago

What percent of air by volume is nitrogen​

What percent of air by volume is nitrogen