A 845kg dragster (very fast car) accelerates from 2m/s to 30m/s in 0.9s. Determine the average force exerted on the dragster.

You are traveling in a spaceship at a speed of 0.70c away from Earth. You send a laser beam toward the Earth traveling at veloci

Dovator [93]

The question is asking to describe and state and calculate what do the observer on the earth measure for the speed of the laser beam, and base on my research, the answer would be v = 1bc, I hope you are satisfied with my answer and feel free to ask for more

6 0

2 years ago

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

2 years ago

yesenia wants to bake cookies in her electric oven. the oven will transform electrical energy into ________ energy?

lana [24]

<span>The oven will transform electrical energy into heat energy</span>

5 0

3 years ago

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Light of wavelength 600nm illuminates a diffraction grating. The second-order maximum is at angle 39°. How many lines per millim

fomenos

Answer:

the number of lines is 526

Explanation:

The wavelength λ =600nm = 600 × 10⁻⁶ mm

The diffraction angle θ = 39°

Recall the expression for the relation between the wavelength, angle and central maxima distance.

relation between the wave length, angle and central maxima distance

d = nλ / sinθ

Here n = 2 for second order maxima and d is the distance

= 2(600 × 10⁻⁶) / sin 39°

= 1200 × 10⁻⁶ / 0.6293

= 1.9 × 10⁻³ mm

N = 1/d

= 1 / 1.9 × 10⁻³

= 526

The grating has a line density of 526 lines per millimeter

3 0

2 years ago

A certain atom has only three energy levels. From lowest to highest energy, these levels are denoted n = 1, n = 2, and n = 3. Wh

shtirl [24]

Answer:

D. 160 nm

Explanation:

The energy released from n = 3 to n = 1 must be equal to the sum of energies released from n = 3 to n = 2 and from n = 2 to n = 1. Therefore,

Energy of Photon from 3 to 1 = Energy of Photon from 3 to 2 + Energy of Photon from 2 to 1

$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}\\\\\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$

where,

λ = wavelength of photon released from 3 to 1 = ?

λ₁ = wavelength of photon released from 3 to 2 = 800 nm

λ₂ = wavelength of photon released from 2 to 1 = 200 nm

Therefore,

$\frac{1}{\lambda} = \frac{1}{800\ nm} + \frac{1}{200\ nm}\\\\\frac{1}{\lambda} = 0.00625 nm^{-1}\\\\\lambda = \frac{1}{0.00625 nm^{-1}}\\\\\lambda = 160 nm$

Therefore, the correct option is:

5 0

2 years ago