Can I see the work for number 55

Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and coll

gregori [183]

Answer:

a) t = 3.027 10⁻⁹ s , b) y = 2.25 10⁻² m

Explanation:

We can solve this problem using the kinematic relations

a) as on the x-axis there is no relationship

vₓ = x / t

t = x / vₓ

We reduce the magnitudes to the SI system

x = 5.6 cm (1m / 100 vm) = 0.056 m

we calculate

t = 0.056 / 1.85 10⁷

t = 3.027 10⁻⁹ s

b) the time is the same for the two movements, on the y axis

y = v₀t + ½ a t²

as the beam leaves horizontal there is no initial vertical velocity

y = ½ a t²

let's calculate

y = ½ 5.45 10¹⁵ (3.027 10⁻⁹)²

y = 2.25 10⁻² m

6 0

3 years ago

Consider a gate whose Shape is a parabolic arch of height 3m and base width 3m. If the origin of the coordinate system is at the

vitfil [10]

Answer:

An arch in a memorial park, having a parabolic shape, has a height of 25 feet and a base width of 30 feet.

6 0

3 years ago

Define e constitutional principle of limited government

Sonbull [250]

In a limited government, the power of government to intervene in the exercise of civil liberties is restricted by law, usually in a written constitution. It is a principle of classical liberalism, free market libertarianism, and some tendencies of liberalism and conservatism in the United States

3 0

3 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$