Ask question

Ask question

All categories

3 years ago

6

Thermal escape of an atmosphere is most pronounced on worlds where the gravity is low and the temperature is high.

1 answer:

Alona [7]3 years ago

7 0

The answer is false

You might be interested in

What is the primary outgoing radiation put off by the sun?

ira [324]

I really don’t know but I think it’s D

7 0

3 years ago

A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$

$\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}$

$\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}$

$a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}$

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

4 0

2 years ago

A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec

AnnyKZ [126]

2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.

5 0

2 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

PIT_PIT [208]

Answer:

$9.96\cdot 10^{-10}J$

Explanation:

The capacitance of the parallel-plate capacitor is given by

$C=\epsilon_0 k \frac{A}{d}$

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

$A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2$ is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F$

Now we can calculate the energy of the capacitor, given by:

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

$U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J$

4 0

2 years ago