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velikii [3]
3 years ago
13

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

ball bounces off with no loss of speed, the clay sticks. What is the change in momentum of the clay and ball, respectively, assuming that toward the wall is the positive direction?
Physics
2 answers:
shutvik [7]3 years ago
8 0

Answer:

Change in momentum of the wet lump of clay = -mv

Change in momentum of the bouncy rubber ball = -2mv

Explanation:

Momentum, p = mv

Change in momentum of the wet lump of clay

The initial velocity of the wet lump of clay of mass, m is v, so its initial momentum is p₁ = mv. Since it sticks to the wall, its final velocity is zero. So its final momentum is p₂ = m × 0 = 0. Its change in momentum Δp = p₂ - p₁ = 0 - mv = -mv

Change in momentum of the bouncy rubber ball

The initial velocity of the rubber ball of mas m is v, so its initial momentum is p₃ = mv. Since it bounces of the wall with no loss of speed and thus moves in the opposite direction to its initial direction, its final velocity is -v. So its final momentum is p₄ = m × -v = -mv. Its change in momentum Δp = p₄ - p₃ = -mv - mv = -2mv

nadezda [96]3 years ago
6 0

Your answer


<h2>-mv; -2mv</h2>

Good luck!

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A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

3 0
3 years ago
The pH of acid rain would be _____. greater than 7 less than 7 exactly 7 exactly 14
Nookie1986 [14]
Less than 7, Ph of 7 is neutral Acid Rain would be between 4 and 0. 
7 0
3 years ago
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A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

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8 0
3 years ago
On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

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Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
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Mashutka [201]

Answer:

b) The downward force of gravity

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\sum F=m.a

right after you throw the ball, only the force exerted by the gravity will affect the ball (neglecting air resistance):

m.g=m.a\\a=g

so the object will eventually be going down again.

8 0
3 years ago
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