If the period of the pendulum is 2s, what is the frequency?

(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from

victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c) $a_{satellite}= 3.13 m/sec^2$

d) $a_{earth}=2.32\times10^{-22} m/s^2$

Explanation:

The force experienced by the satelite is giveb by

$F= \frac{Gm_{satellite}m_{earth}}{r^2} \\$

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

$F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}$

⇒F= 1396.52 N

now,

$a_{satellite}= \frac{F}{m_{satellite}}$

$a_{satellite}= \frac{1396.52}{445}$

$a_{satellite}= 3.13 m/sec^2$

also,

$a_{earth}= \frac{F}{m_{earth}}$

$a_{earth}= \frac{1396.52}{(6\times10^24)}$

$a_{earth}=2.32\times10^{-22} m/s^2$

3 0

3 years ago

A flask contains equal masses of f2 and cl2 with a total pressure of 2.00 atm at 298k. What is the partial pressure of cl2 in th

Sindrei [870]

Answer:

Partial Pressure of F₂ = 1.30 atm

Partial pressure of Cl₂ = 0.70 atm

Explanation:

Partial pressure for gases are given by Daltons law.

Total pressure of a gas mixture = sum of the partial pressures of individual gases

Pt = P(f₂) + P(cl₂)

Partial pressure = mole fraction × total pressure

Let the mass of each gas present be m

Number of moles of F₂ = m/38 (molar mass of fluorine = 38 g/Lol

Number of moles of Cl₂ = m/71 (molar mass of Cl₂)

Mole fraction of F₂ = (m/38)/((m/38) + (m/71)) = 0.65

Mole fraction of Cl₂ = (m/71)/((m/38) + (m/71)) = 0.35 or just 1 - 0.65 = 0.35

Partial Pressure of F₂ = 0.65 × 2 = 1.30 atm

Partial pressure of Cl₂ = 0.35 × 2 = 0.70 atm

7 0

3 years ago

I would give brainliest if you have me!

Ket [755]

She can first measure the mass on the scale, then measure the cm^3 by putting water in the cylinder and measuring the original water level minus the water level after you put the rock in. The take the measurement from the scale (g) and divide it by the measurement in the graduated cylinder (c^3).

5 0

3 years ago

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and di

lozanna [386]

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

B = μ₀ N/L I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

B = (μ₀ I / L) N

the amount between paracentesis constant, in the case of 4 loop the field is worth

B = cte 4

N B

4 4 cte

3 3 cte

2 2 cte

1 1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

6 0

3 years ago