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3 years ago

True or false friction makes energy vanish

Physics

2 answers:

Alexxx [7]3 years ago

7 0

False, friction produces heat, kinetic energy has been changed to heat energy.

Kaylis [27]3 years ago

7 0

Answer:

False.

Explanation:

The first law of thermodynamics states that energy , in a closed system, its not created nor destroyed, remains constant. The universe its a closed system, so the total energy in the universe must remains constant. The energy doesn’t vanish due to friction, but its dissipated into the surroundings, in the form of heat.

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A vehicle with a mass of 2,553 kg accelerates at 10 m/s2. Find the force on the vehicle in Newtons.

Lerok [7]

Answer:

F = 25530 N

Explanation:

F(net) = ma

F(net) = 2553 kg•(10) m/s² = 25530 N

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3 years ago

Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may

Tpy6a [65]

Answer:

The answer is potential energy

Explanation:

The potential energy is the energy possessed by a body by virtue of it position

For example the water at the top of the dam is being held at a height h above the bottom of the dam

Then the potential energy

PE= weight of the water* the height

PE= m*g*h

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4 years ago

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave

Marina86 [1]

Answer:

$a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree$

Explanation:

Puck A's initial speed is $v_a_1=14.6m/s$ and move in a direction $\theta_b=28.0\textdegree$ after the collision.

# $P_1=P_2$ since there's no external force on the system( $P=mv).$

#The collision equation can be written as;

$m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2$

The kinetic energies before and after the collision are expressed as:

$K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2$

$0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2$

Let +x along A's initial direction and +y along A's final direction makes the angle $62\textdegree$

$\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j$

#Substitute in $v_{b2}^2=213.16-v_{a2}^2$ :

$\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\$

$v_{a2}=12.891\ m/s$

Hence, the the speed of puck A after the collision is 12.891 m/s

#. The velocity of A after the collision is;

$\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j$

Substitute $\dot v_{a2}$ into $\dot v_{b2}=(14.6i)-\dot v{a2}$ :

$\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j$

This is the velocity of puck B after the collision, it's speed is:

$v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s$

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

$\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree$

Hence, the direction of B's velocity after the collision is 62°

4 0

4 years ago

A sample of water in equilibrium with its vapor in a closed 0.6-L container has a vapor pressure of 23.8 torr at 25°C. The conta

Lyrx [107]

Answer:

Explanation:

Given data:

vapor pressure P_1 = 0.6 L

Volume = 2L

Temperature remain constant

No change in vapor pressure. This is because vapor pressure is calculated from the temperature of the liquid and not the volume of the container. Since the temperature is constant therefore vapor pressure remian constant..

8 0

3 years ago

The diameter of a spherical balloon that is being filled with air is increasing at the rate of 3 inches more than the time, t. W

Sever21 [200]

The description of the problem is a bit ambiguous as "<span>increasing at the rate of </span>3 inches more than the time" can be interpreted in many ways. Assuming the increase rate is the (3+t) per second and as it expands for 3 seconds, then the diameter of the sphere would be:
d= (3 + t)*t
d= 3t + t^2= 18

r= 1/2 d
r= 1/2*18= 9

volume= 4/3 * pi * r^3
volume= 4/3 * pi * 9^3
volume= 4/3 * pi * 729= 972 pi cubic inches

7 0

4 years ago

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