All categories

3 years ago

HELP ME PLZ I NEED TO PASS

Physics

2 answers:

Varvara68 [4.7K]3 years ago

6 0

Answer:

50 newtons

Explanation:

subtract directions

ipn [44]3 years ago

5 0

Answer:

D. 50 N

Explanation:

The 2 forces of the cart will counteract each other because they are being pushed in opposite directions. The 150 N will counteract the 150 N on the other side. But since there's an extra 50 N on the left, it will push 50 N to the right. The 2 150 N will do nothing because they are pushing in opposite directions. For example, if two people are pushing something in the opposite direction with the exact same force, the pushed object won't move.

You might be interested in

The main difference between speed and velocity involves

DiKsa [7]

A. Direction

Speed is just distance divided by time, but velocity is displacement divided by time and displacement has direction. Speed will always be positive, but velocity can be either positive or negative.

8 0

3 years ago

Read 2 more answers

What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

6 0

3 years ago

During a total lunar eclipse, the moon

Anit [1.1K]

A, C, and D all happen at different stages
of a total lunar eclipse.

I'll describe the stages of the eclipse, but before I do, I just need
to clarify: The Earth doesn't have an umbra or a penumbra, but
its shadow does.

-- the eclipse begins when the first edge of the moon
   moves into the penumbra of Earth's shadow; ( C )
   this part of the moon grows steadily.

-- After a while, the first edge of the moon begins to move
   into the umbra of Earth's shadow ( A ), and gets very dark.

-- The total phase of the eclipse begins when the ENTIRE
    moon is in the umbra of Earth's shadow.

Then everything happens in reverse.

-- Eventually, the leading edge of the moon moves out
   of the shadow's umbra, into the penumbra. This part
   steadily grows.

-- After a while, none of the moon is in the umbra, and
   the whole thing is in the penumbra. The moon is
   fully illuminated, but not quite as bright as it should be.

-- Soon, the leading edge of the moon leaves the penumbra
of Earth's shadow, and gets brighter. This portion of the moon
    steadily grows, until ...

-- the moon completely leaves the penumbra, all of it is as bright
as it's supposed to be. The eclipse is completely over. ( B )

==> The whole process lasts several hours.

==> Everybody on the night side of the Earth sees the same thing
   at the same time. It doesn't matter WHERE you are on the night
   side ... if you can see the moon in the sky, you see the present
   phase of the eclipse.

==> The lunar eclipse can only happen at the Full Moon. In fact, the
   mid-point of the total phase is the exact moment of Full Moon.

8 0

3 years ago

Read 2 more answers

An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th

AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$ Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0

3 years ago

Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul

timofeeve [1]

Answer:

a) $K_{max}$ = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

$K_{max}$ = h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

E = 2,486 10⁻¹⁸ J

Let's reduce to eV

E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

E = 15.5 eV

Now we can calculate the kinetic energy

$K_{max}$ = h c / f - fi

$K_{max}$ = 15.5 -13.6

$K_{max}$ = 1.9 eV

b) Extra energy = 10.2 eV

The total kinetic energy of electrons is

Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

ΔE = E photon - Total kinetic energy electron

ΔE = 15.5 - 12.1

ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

$E_{n}$ = 13.606 / n2

n = √ 13.606 / En

n = √ (13606 / 3.4)

n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

8 0

3 years ago