Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?

As mass increases what happens to the kinetic energy

AnnyKZ [126]

As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)

5 0

3 years ago

g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes ar

zhenek [66]

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

X = (246)(5.62 x 10⁻⁷ m)/2

Therefore,

5 0

3 years ago

A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work ( $W$ ) done on an object is equal to the change in its kinetic energy ( $\Delta K$ ):

$W=\Delta K = K_{f}-K_{i}$ (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation $K=\frac{I\omega^{2}}{2}$ for the kinetic energy so:

$\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}$ (2)

with $\omega_{i}$ the initial angular velocity, $\omega_{f}$ the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is $I=MR^{2}$ , using those on (2)

$\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2}$ (3)

By (3) on (1):

$W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}$

$W=19806.62\,J$

B) Average power is work done divided by the time interval:

$P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}$

$P=1320.44\,W$

NOTE: We use the relation $1rpm*\frac{2\pi}{60s}=\frac{rad}{s}$ to convert 280 rev/min(rpm) to 29.32 rad/s

4 0

4 years ago

A golfer strikes her tee shot with the following velocities, vx = 106 m/s & vy = 66.2 m/s. a. What is the velocity of flight

lara [203]

Answer:

a) V ≈ 125 m/s; b) Δt = 13.24 s; c) ΔS ≈ 1450 m

Explanation:

a) We have just to calculate the vector resultant.

V² = 106² + 66.2²

V² = 15618.44

V ≈ 125 m/s

b) The time of flight is equal to the time to reach the maximum height summed to the time to reach the land.

In vertical:

V = V₀ + a * t

V = 66.2 - g * t

0 = 66.2 - 9.8 * t

t ≈ 6.76 s

So: Δt = 13.24 s

c) In horizontal:

V = ΔS / Δt

106 = ΔS / 13.52 ⇒ ΔS = 106 * 13.52

ΔS = 106 * 13.52

ΔS = 1433,12

ΔS ≈ 1450 m

3 0

1 year ago

The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in

solmaris [256]

Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,

which is equivalent to an energy change of

Delta E = (0.00523 u)*(931.5MeV/1u)

Delta E = 4.87 MeV

Converting 4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule

5 0

4 years ago