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iragen [17]
3 years ago
15

Мета: :

Physics
1 answer:
Ierofanga [76]3 years ago
3 0
I can’t answer this
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In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of
nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

F_g = F_c

\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}

now we can say that kinetic energy of satellite is  given as

KE = \frac{1}{2}M_s v^2

KE = \frac{GM_sM_p}{2r}

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

U = -\frac{GM_sM_p}{r}

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0
3 years ago
A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

7 0
3 years ago
A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height
krok68 [10]

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

8 0
3 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
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