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2 years ago

7

A truck traveling at 60 km / h comes to a complete stop due to a traffic signal. Does this change in velocity violate the law of

conservation of energy? Explain

1 answer:

Furkat [3]2 years ago

4 0

Explanation:

No. Energy is still conserved. The kinetic energy of the truck is converted into heat.

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A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

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2 years ago

Why does a geostationary satellite must orbit around Earth's equator, rather than in some other orbit (such as around the poles)

LiRa [457]

Answer:

A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.

Explanation:

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1 year ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

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2 years ago

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

$\underline{\left \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}$

$\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}$

Learn more here:

brainly.com/question/14466080

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2 years ago

What is the main fuel consumed in the core of a red giant?<br> a. H<br> b. C<br> c. Fe<br> d. He

vladimir1956 [14]

<span>What is the main fuel consumed in the core of a red giant?
The </span><span>main fuel consumed in the core of a red giant is He or helium. The answer is letter D.</span>

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3 years ago