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2 years ago

In chemistry, the law of conversation allows for energy allows for which of the following?

Physics

2 answers:

morpeh [17]2 years ago

8 0

B. Energy to be stored

IrinaK [193]2 years ago

6 0

I think the second choose, energy to be stored.

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A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

Bumek [7]

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s speed of ball in the moving frame (the truck)

U = -20 m/s speed of primed (rest) frame with respect to moving frame

So V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

3 0

3 years ago

A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths

rosijanka [135]

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m - 1/2)λ

d = (m - 1/2)(λ/2n)

refractive index of the thin film, n = 1.34

minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

= 99.07nm

3 0

3 years ago

A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha

Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

$N = \dfrac{m g}{cos\theta-\mu sin \theta}$

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

$\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}$

taking cos θ from nominator and denominator

$F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g$

$\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}$

now, inserting all the given values

$\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}$

μ = 0.6

7 0

3 years ago

A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input

KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( $\frac{1}{2fRC}$ )V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 / ( (120 * 1 * 62.2) )15V

= 2V

The peak-to-peak ripple voltage = 2V

3 0

3 years ago

Compared to the inertia of a 1-kilogram mass, the inertia of a 4-kilogram mass is

Mumz [18]

Explanation:

The inertia of a 4 kg mass is four times as great as a 1 kg mass.

5 0

3 years ago