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3 years ago

How many millimeters are 120 meters ? Help ! Please <3

Physics

2 answers:

blsea [12.9K]3 years ago

8 0

Answer:

120,000

Explanation:

Millimeters to meters calculation-

Multiply by 1,000.

120 x 1,000 = 120,000.

This is the correct answer and formula.

Hope this helps!

natulia [17]3 years ago

4 0

The answer is 120000

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Which of the following is not an example of work?

barxatty [35]

<span>The following which is not an example of work is </span>C. holding a tray in the cafeteria line because <span>if force displaces an object it should work. I think it's clear and I am pretty sure this answer will help you.</span>

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3 years ago

For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll

Schach [20]

Answer:

1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

E = K T

The relationship of the speed of light

c =λ f = f / ν 1/λ= ν

The Planck equation is

E = h f

Let's start the transformations

c = f λ = f / ν

f = c ν

E = h f

E = h c ν

E = KT

h c ν = K T

T = h c ν / K =( h c / K) ν

Let's replace the constants

h = 6.63 10⁻³⁴ J s

c = 3 10⁸ m / s

K = 1.38 10⁻²³ J / K

v = 1 cm-1 (100 cm / 1 m) = 10² m-1

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

A = h c / K = 1,441 10⁻²

T = 1.44K

ν = 103 cm⁻¹ = 103 10² m

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

T = (h c / K) 1 /λ

T = 1,441 10⁻² 1 / 1,097 10⁷

T = 1.3 10⁻⁹ K

E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

E = KT

T = E/K

T = 1.6 10⁻¹⁹ /1.38 10⁻²³

T = 1.16 10⁴ K

3 0

3 years ago

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

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3 years ago

What measures the amount of displacement in a transverse wave

MAVERICK [17]

Unlike a longitudinal wave, a transverse wave moves about, perpendicular to the direction of propagation. The particles in a transverse wave do not travel along the direction of propagation, but only oscillate up and down on its equilibrium position. With this, the displacement can be determined by measuring (in the case of electronic waves, using an oscilloscope or spectrum analyzer) and setting the desired units to measure the wave in.

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3 years ago