Hartman value profile

In a very busy off-campus eatery one chef sends a 201 g broccoli-tomato-pickle-onion-mushroom pizza sliding down the counter fro

Murrr4er [49]

Answer:

The correct answer is:

(A) to the left

(B) at speed -0.8725 m/s

Explanation:

The given values are:

Plate 1:

Mass,

m₁ = 201 g

Velocity,

v₁ = +1.79 m/s

Plate 2:

Mass,

m₁ = 335 g

Velocity,

v₁ = -2.47 m/s

According to the conservation of momentum, we get

⇒ $m_1v_1+m_2v_2=(m_1+m_2)v$

then,

⇒ $v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

On substituting the values, we get

⇒ $=\frac{201\times 1.79+335\times (-2.47)}{201+335}$

⇒ $=\frac{359.79-827.45}{536}$

⇒ $=\frac{-467.66}{536}$

⇒ $=-0.8725$ (to the left)

3 0

3 years ago

Hii guys!! ~ i've comeback again!! could you please help me with an assignment?? i have a couple of i know questions!! thank you

Nimfa-mama [501]

Answer:

Speed = 10.24 m/s.

Explanation:

Given the following data;

Distance = 100m

Time = 9.77

To find her speed;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

$Speed = \frac{distance}{time}$

Substituting into the equation, we have;

$Speed = \frac{100}{9.77}$

Speed = 10.24 meter per seconds. 

5 0

3 years ago

How do land and sea breeze work?

Studentka2010 [4]

by the wind and air flow in the wind

7 0

4 years ago

A moving walkway at an airport has a speed v1 and a length L. A woman stands on the walkway as it moves from one end to the othe

bearhunter [10]

Answer:

t=L/ $V_1$

Explanation:

solution:

Let E be an observer, and B a second observer traveling with velocity $V_{BE}$ as measured by E. If E measures the velocity of an object A as $V_{AE}$ then B will measure A velocity as

$V_{AB}$ = $V_{AE}$ - $V_{BE}$

Applied here,

the walkway (W) and the man (M) are moving relative to Earth (E}, the velocity of the man relative to the moving walkway is

$V_{MW}$ = $V_{ME}$ - $V_{WE}$

$V_1=V_{WE}$ ,

$V_2=V_{MW}$

The time required for the woman, traveling at constant speed $V_1$ relative to the ground, to travel distance L relative to the ground is :

t=L/ $V_1$

3 0

4 years ago

A lump of clay whose rest mass is 4 kg is travelling at three-fifths the speed of light when it collides head-on with an identic

Harman [31]

Answer:

mass of the composite lump is 10 kg

Explanation:

given data

mass = 4 kg

to find out

mass of composite lump

solution

we know energy is conserved so

so m1 = m2 = m0 that is 4kg

and

E(1) release+ E(2) release = E(1,2) rest

so γ(1)m(1)c² + γ(2)m(2)c² = Mc² ..........................1

that why here

|v(1)| = |v(2)| = 3/5 c ......................2

and

γ = 1 / √(1 − v²/c²) .......................3

put here v = 3 and c is 5

γ = 1 /√(1 − 9/25)

γ = 5/4

so

γ(1) = γ(2) = γ = 5/4

so from equation 1

γ(1)m(1)c² + γ(2)m(2)c² = Mc²

M = 2γm0

M = 2(5/4 )(4)

M = 10 kg

so mass of the composite lump is 10 kg

7 0

4 years ago