What is the weight of a 10-kilogram object on Earth?

maria [59]

it will be very hot so that causes nobody to walk on venus because its the hottest outer planet.

hope I helped :)

3 0

2 years ago

An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an

ANEK [815]

Answer:

a)

$z=561.7$

b)

$z=214.1$

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

a)

f = frequency = 55.7 Hz

Impedance is given as

$z=\frac{1}{2\pi fC} - 2\pi fL$

$z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)$

$z=561.7$

b)

f = frequency = 11000 Hz

Impedance is given as

$z= - \frac{1}{2\pi fC} + 2\pi fL$

$z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)$

$z=214.1$

4 0

2 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

2 years ago

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab (dielectric c

adoni [48]

Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

$C = \frac{\epsilon*A}{d}$

Where ε, is the dielectric constant of the material that fills the space between plates.
When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
At the same time, the capacitance of a capacitor, by definition, is as follows:

$C =\frac{Q}{V}$

If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
So, the change in the charge of the positive plate is +2.5 nC.

3 0

2 years ago

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would pro

Gnesinka [82]

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of <u>Eddington luminosity</u>

<h3>What are stars?</h3>

Stars are a fixed luminous point in the sky which is a large and remote incandescent body

So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

Learn more about stars:

brainly.com/question/13018254

#SPJ1

4 0

1 year ago