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Vlada [557]
3 years ago
9

Define orbital velocity

Physics
1 answer:
sveta [45]3 years ago
8 0

the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.

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Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com
Bas_tet [7]

Answer:

Failure rate   = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at =  140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷  (number of breakdowns)    ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be  = (Total up time) ÷ (number of breakdowns)      .......2

=  \frac{2000}{2}   =  1000  

so here gap between occurrences is

gap between occurrences=  140 - 20

gap between occurrences = 120 hour

and

MTBF  will be

MTBF = 1000 - 120

MTBF = 880 hours  

and

Failure rate (FR)  will be

Failure rate (FR) =  1 ÷ MTBF    ................3

Failure rate (FR) = R÷T     ......................4

as here R is the number of failures and T is total time

so Failure rate (FR)  = 20%

4 0
3 years ago
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

8 0
3 years ago
To ya is modeling the discovery of electromagnetic induction. Which procedures should she use
Vedmedyk [2.9K]

Answer:

moving a magnet into a coil of wire in a closed circuit.

Ed 2020

8 0
3 years ago
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
Vaselesa [24]

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

3 0
3 years ago
A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

8 0
3 years ago
Read 2 more answers
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