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2 years ago

3

A ball weighing 10 kg traveling at 1.4 m/s hits and rebounds back off a wall and now travels 1.4 m/s in the opposite direction.

What is the work done on it?

1 answer:

Marizza181 [45]2 years ago

4 0

<em>NO work is done on the ball.</em>

Its momentum changes, but its kinetic energy doesn't.

<u>Before it hits the wall:</u>

Momentum = m v = (10 kg) x (1.4 m/s) = +14 kg-m/s

Kinetic energy = (1/2)(m)(v²) = (1/2)(10kg)(1.4)² = 9.8 Joules

<u>After it hits the wall:</u>

Momentum = m v = (10 kg) x (-1.4 m/s) = -14 kg-m/s

Kinetic energy = (1/2)(m)(v²) = (1/2)(10kg)(-1.4)² = 9.8 Joules

No change in energy ==> no work done on the ball.

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A point charge q1=−4.00nc is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nc is at the point x

Katen [24]

electric field between mid point of two charges is given by

$E = E_1 + E_2$

here we have

$E_1 = \frac{kq_1}{r^2}$

$E_1 = \frac{9*10^9 * 4 * 10^{-9}}{0.4^2}$

$E_1 = 225 N/c$

now similarly for other charge

$E_2 = \frac{kq_2}{r^2}$

$E_2 = \frac{9*10^9 * 6 * 10^{-9}}{0.4^2}$

$E_2 = 337.5 N/C$

so the net electric field will be given as

$E = 225 + 337.5$

$E = 565.5 N/C$

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2 years ago

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as your roller coaster climbs to the top of the steepest hill on its track when does the first car have the greatest potential e

pochemuha

...the potential energy that you build while going up the hill on the roller coaster could be let go as kinetic energy -- the energy of motion that takes you down the hill of the roller coaster.

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2 years ago

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

$T = 3183 N$

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2 years ago

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.

Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7° northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

$V_{y}$ = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

$V_{x}$ = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{13.89^{2} +9.89^{2} }$ = <em>17.05 mph This is your airspeed</em>

b) To get your direction, we use

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = 13.89/9.89 = 1.413

∅ = $tan^{-1}$ (1.413) = <em>54.7° northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

$V_{y}$ = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{4.11^{2} +9.89^{2} }$ = <em>10.71 mph This is your airspeed</em>

Your direction will be,

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = -4.11/9.89 = -0.416

∅ = $tan^{-1}$ (-0.416) =<em> -22.58° this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

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2 years ago

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

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2 years ago