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3 years ago

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While texting his girlfriend that he was running late for their date, Jordan rear-ended a car stopped at a red light. Since thi

s was clearly an inelastic collision, some of his car's kinetic energy was transformed to other types of energy. What types of energy are most likely to result from such a collision?

2 answers:

SSSSS [86.1K]3 years ago

5 0

Kinetic energy is the basic energy built in a moving vehicle or object. When the cars impact with another car or object the kinetic energy must be dispersed This is dependent of the speed of the object and the mass.

RideAnS [48]3 years ago

3 0

Heat energy and sound

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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3 years ago

The white ring seen here, within our solar system, is a(n) ________ belt.

Gelneren [198K]

Asteroid belt :-) hope it helped

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3 years ago

Read 2 more answers

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

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3 years ago

Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground

Readme [11.4K]

PE = (mass) (gravity) (height)

PE = (0.005 kg) (9.8 m/s²) (5 m)

<em>PE = 0.245 Joule</em>

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3 years ago

Can you help me please the question says

bearhunter [10]

Answer:

p= 4 m/v

Explanation:

v=l*w*h

v=(25)(2)(3)

v=150

p=600/150

p=4

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3 years ago