Weather stays the same all the time

Give main forms before and after for the following energy transformations: A gasoline engine.

klemol [59]

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➷ It would be chemical energy to kinetic energy. The chemical energy from the gasoline is being transferred to kinetic energy.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

4 years ago

A 74-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.56 m,

Reptile [31]

Answer:

Work done, W = 1.44 kJ

Explanation:

Given that,

Mass of boy, m = 74 kg

Initial speed of boy, u = 1.6 m/s

The boy then drops through a height of 1.56 m

Final speed of boy, v = 8.5 m/s

To find,

Non-conservative work was done on the boy.

Solution,

The work done by the non conservative forces is equal to the sum of total change in kinetic energy and total change in potential energy.

$W=\dfrac{1}{2}m(v^2-u^2)+(0-mgh)$

$W=\dfrac{1}{2}m(v^2-u^2)-mgh$

$W=\dfrac{1}{2}\times 74\times (8.5^2-1.6^2)-74\times 9.8\times 1.56$

W = 1447.21 Joules

or

W = 1.44 kJ

Therefore, the non conservative work done on the boy is 1.44 kJ.

4 0

3 years ago

A sound wave is often called a pressure wave because there are regions of high and low pressure established in them medium throu

Bingel [31]

Explanation:

A sound wave is often called a pressure wave because there are regions of high and low pressure established in them medium through which the sound wave travels. The regions of high pressure are known as <u>Compressions</u> and the regions of low pressure are known as <u>Rarefactions</u> . Sound waves are composed of compressions and rarefactions. Compressions are the parts where the molecules are congusted and pressed together. However in the rarefactions molecules are relax and have enough space for expansion. Sound waves are the logitudnal waves and always been defined as the motion of the medium particles parallel to the wave motion.

7 0

3 years ago

A wire with a circular cross section and a resistance R is lengthened to 9.66 times its original length by pulling it through a

damaskus [11]

Answer:

The resistance of the wire after it is stretched is 93.31R.

Explanation:

Resistance is the property of the material to oppose the current flow through it. It is given by the relation :

R = (ρl)/A

Here ρ is resistivity, l is length of wire and A is the area of the wire.

Let l₀, and A₀ are the original length and original circular cross section area of the wire. while l₁ and A₁ are the new length and new circular cross section area of the wire.

Volume of the original wire, V₀ = A₀ x l₀

Volume of the new wire, V₁ = A₁ x l₁

According to the problem. volume remain same. So,

V₀ = V₁

A₀ x l₀ = A₁ x l₁

It is given that l₁ = 9.66 x l₀. Substitute this value in the above equation;

A₀ x l₀ = A₁ x 9.66 x l₀

A₁ = A₀/9.66

Resistance of the original wire, R = (ρl₀)/A₀

Resistance of the new wire, R₁ = (ρl₁)/A₁

Substitute the value of l₁ and A₁ in the above equation.

R₁ = (ρ x l₀ x 9.66)/(A₀/9.66) = 93.31 x (ρl₀)/A₀

But (ρl₀)/A₀ = R. hence,

R₁ = 93.31 R

8 0

3 years ago

a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed

Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

$\Delta x \Delta p\geq \frac{h}{4\pi}$ (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}$

The minimum kinetic energy is calculated by using the following formula:

$k=\frac{(\Delta p)^2}{2m}$ (2)

m: mass of the proton = 1.67*10^{-27}kg

$k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J$

in eV you have:

$2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV$

The kinetic energy of the proton is 1.29eV

7 0

3 years ago