Do not have definite size and always take the shape of their contanier

When heat is converted into another form of energy, the total amount of energy is constant. Which law best illustrates this stat

deff fn [24]

The law of conservation of energy states that in a closed or isolated system, the amount of energy remains constant because energy can neither be created or destroyed. It can only be transferred from one form into another. This applies to all forms of energy.

8 0

3 years ago

Which of the following changes would decrease the coefficient of friction needed for this ride?

posledela

Friction occurs between two contacting surfaces. The coefficient of friction is very much dependent on the roughness of these surfaces. Some of the many ways in which the coefficient can be lessened or decreased are to lubricate the surface or make it shiny by eliminating the spikes which caused the roughness.

7 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago

A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

lisabon 2012 [21]

The centripetal acceleration a is 4.32 $\times$ 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m, Time = 100 s.

Computing the v = 0.4 m / 100 s

v = 4 $\times$ 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

centripetal acceleration a = v^2 / r

= (4 $\times$ 10^-3)^2 / 0.037

a = 4.32 $\times$ 10^-4 m/s^2.

6 0

3 years ago

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

Let's calculate

v = √(1.23² + 2 9.8 1.69)

v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

W = ΔK

-fr x = 0- ½ m v²

Newton's second law

N- W = 0

The equation for the friction is

fr = μ N

fr = μ m g

We replace

μ m g x = ½ m v²

x = v² / 2μ g

Let's calculate

x = 5.89² / (2 0.255 9.8)

x = 6.94 m

6 0

3 years ago

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