All categories

3 years ago

A benefit of using good human relations is that you'll

Physics

2 answers:

pantera1 [17]3 years ago

6 0

The Answer Is B adjust to new situations better.

OLEGan [10]3 years ago

5 0

Its either B or C but im leaning more towards B but i could be wrong

You might be interested in

A 8.0×104kgspaceship is at rest in deep space. Its thrusters provide a force of 1200kN. The spaceship fires its thrusters for 20

Sladkaya [172]

Answer:

Time = 40 s

It will take the spaceship 40s to coast that distance.

Explanation:

Using the impulse momentum equation;

Impulse = change in momentum

Ft = m(∆v) ........1

Given;

Force F = 1200kN

time of action t = 20s

Mass of spaceship m = 8.0×10^4 kg

∆v = change in velocity

Substituting the values into equation 1;

1200kN × 20s =8.0×10^4 kg × ∆v

∆v = 1200000×20/80000

∆v = 300 m/s

Since it was initially at rest, V0 = 0

Final velocity V2 = 300 + 0

V2 = 300m/s

Time To travel 12 km,

Time = distance/speed

Distance = 12 km = 12000m

Speed = 300m/s

Time = 12000/300

Time = 40 s

It will take the spaceship 40s to coast that distance.

4 0

3 years ago

Type of force that keep objects moving in a circle or arv

Svetlanka [38]

Answer: a net force

7 0

3 years ago

A proton is trapped in a one-dimensional well that is 200 pm wide. What is the ground state energy of the proton, if the potenti

Harrizon [31]

Answer:

E=1.50\times 10^{-18}J

Explanation:

Energy of the one dimensional infinite well,

$E=\frac{n^{2}h^{2} }{8mL^{2} }$

Given that, $L=200 pm\\L=200\times 10^{-12}m$

For the ground state n=1,

Therefore energy is,

$E=\frac{(6.626\times 10^{-34})^{2}}{8(9.1\times 10^{-31})(200\times 10^{-12}) ^{2} }\\E=1.50\times 10^{-18}J$

8 0

3 years ago

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

8 0

3 years ago

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that

Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

$g_e$ = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

$m=M-0.1M\\\Rightarrow m=0.9M$

$g_e=G\frac{M}{r^2}$

$g=G\frac{0.9M}{r^2}$

Dividing the equations we get

$\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2$

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0

4 years ago