All categories

3 years ago

Assuming the input of energy continues for another 2.5 seconds, where will the particle be?

1 answer:

3 0

Well i had the same question on my test, and when the graph in 2.5 seconds goes up from the equilibrium it reaches a positive maximum, so that would be your answer.

b. positive maximum

You might be interested in

Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int

neonofarm [45]

Answer:

The value is $V_n = 2.2498 \ m^3$

Explanation:

From the question we are told that

The volume of liquid nitrogen is $V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3$

The density of nitrogen at gaseous form is $\rho_n = 1.2929 \ kg/m^3$ = The dry air at sea level

Generally the density of nitrogen at liquid form is

$\rho _l = 808 \ kg/m^3$

And this is mathematically represented as

$\rho_l = \frac{m}{V_l }$

=> $m = \rho_l * V_l$

Now the density of gaseous nitrogen is

$\rho_n = \frac{m}{V_n }$

=> $m = \rho_n * V_n$

Given that the mass is constant

$\rho_n * V_n = \rho_l * V_l$

$1.2929* V_n = 808 * 3.6*10^{-3}$

=> $V_n = 2.2498 \ m^3$

3 0

3 years ago

The process of combining two small nuclei into one nucleus of larger mass is called _____.

fiasKO [112]

The process of splitting one large nucleus into
smaller ones is nuclear fission.

The process of combining two small nuclei into
one larger one is nuclear fusion.

3 0

3 years ago

Read 2 more answers

if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud

sveta [45]

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

(1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass .

On Earth, that mass weighs 4,900,000 newtons .

(about 1,102,300 pounds)

(about 551 tons)

3 0

4 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$