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Montano1993 [528]

4 years ago

6

A rock is rolled in the sand. It starts at 5.0 m/s, moves in

1 answer:

Lemur [1.5K]4 years ago

3 0

Answer:

magnitude of the average acceleration is 4.16 m/s²

Explanation:

given data

initial velocity u = 5 m/s

distance s = 3 m

final velocity v = 0

to find out

magnitude of the average acceleration

solution

we will apply here linear motion equation that is

v²-u² = 2×a×s ..............1

put here all these value

v is final velocity and u is initial velocity and s is distance

0²-5² = 2×a×3

a = -4.16

so magnitude of the average acceleration is 4.16 m/s²

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The power of a purely resistive lead is always positive although the current and voltage are sometimes negative. explain

pickupchik [31]

Answer:

Current is in phase with voltage in a resistive circuit. Note that the wave form for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads.Explanation:

7 0

3 years ago

What relationship exists between the amplitude of a wave and amount of disturbance in the water?

guapka [62]

A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.

6 0

3 years ago

A lamp is labelled '230 V, 100 W'. How many joules of electrical energy is changed to thermal energy and light if the lamp is sw

viva [34]

Given;

V = 230V

Power, P = 100W

time, t = 2hrs = 7200s

from,

P = IV

and Energy, E= Pt

E = 100*7200

E = 720000 Joules

E = 720KJ

4 0

3 years ago

Read 2 more answers

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

The morning of a birthday party, a balloon is filled with 8.5 L of helium (He) when the temperature is 294 kelvin. The party sta

Phoenix [80]

Explanation:

the morning of the birthday party balloon filled with the 2.5 Litre of helium

temperature is 294kelvin

the party starts at the 4 p.m.

temperature rises 305 Kelvin.

the new volume = 4 litre.

At same temperature,

P

1

V

2

=P

2

V

2

(Boyle's law)

P

1

=10atm;P

2

=1atm

V

1

=4l=V

2

=8l

But while filling balloons from cylinder when pressure in cylinder becomes 1 atm then further filling is not possible (P

1

′

=9atm)

Let n be the number of balloons that can be filled.

∴P

1

′

V

1

=n(P

1

V

1

)

9×8=n(4×1)

n=

4

9×8

=18balloons

5 0

3 years ago