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PolarNik [594]
3 years ago
5

Two balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.75 kg and an initial velocity of 5.

0 m/s to the right. The second ball has a mass of 0.40 kg and an initial velocity of 3.5 m/s to the left. What is the decrease in kinetic energy during the collision?
Physics
1 answer:
balandron [24]3 years ago
3 0

Explanation:

Momentum conservation

V=3.38

Delta kinetic energy =1/2m(vi^2-v^2)

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Classical mechanics is an extremely well tested model. Hundreds of years worth of experiments, as well as most feats of engineer
djyliett [7]

Answer:

The Answer is 0.019998c

Explanation:

Please see the attached Picture for the answer.

7 0
3 years ago
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
A car slows down uniformly from a speed of 22 m/s to rest in 4.0 seconds. How far did it travel in that time
krek1111 [17]

Answer:

\boxed{\sf Distance \ travelled = 44 \ m}

Given:

Initial speed (u) = 22 m/s

Final speed (v) = 0 m/s (Rest)

Time taken (t) = 4 seconds

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

\boxed{ \bold{s =  (\frac{v + u}{2} )t}}

By substituting value of v, u & t in the equation we get:

\sf \implies s = ( \frac{0 + 22}{2} ) \times 4 \\  \\  \sf \implies s =  \frac{22}{2}  \times 4 \\  \\  \sf \implies s = 11 \times 4 \\  \\  \sf \implies s = 44 \: m

\therefore

Distance travelled by car (s) = 44 m

4 0
4 years ago
suppose you are in a car driving down a road. a large truck passes your car. what does the car have a tendency to do?
balandron [24]

it moves toward the truck because increased air movement between the car and the truck decreases pressure.

Hope this helped :) <3

6 0
3 years ago
Read 2 more answers
Two blocks are connected by a very light string passing over a massless and frictionless pulley. Traveling at constant speed, th
Semenov [28]

Answer:

<h2>i. 9J and </h2><h2>ii.  -9J</h2>

Explanation:

Please see attached the diagram and also the FBD for your reference.

Step one:

given data

Weight w1=20N

Distance s1=75cm= 0.75m

Weight w2=12N

Distance s2=75cm =0.75m

Clearly, we are asked to find the work done by gravity and the tension in the string.

we know that work is defined as force times the distance traveled

W=F*D

and to know the force we need the acceleration, seeing that the system is stationary the acceleration due to gravity 9.81m/s^2 is not needed here

i . Work done by gravity = w2*s2=12*0.75=9J

ii. The tension is equal but opposite = -9J

4 0
3 years ago
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