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joja [24]
3 years ago
10

Where is the sun located in the galaxy?

Physics
2 answers:
Anon25 [30]3 years ago
8 0
I believe it is with an outer arm
GarryVolchara [31]3 years ago
6 0
I think it's 3. within an outer arm
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If you had started with a larger mass, how would the half-life change?
iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0
3 years ago
You are in your car driving on a highway at 23 m/s when you glance in the passenger-side mirror (a convex mirror with radius of
Irina-Kira [14]

Answer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}

\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}

Solve s'

\frac{1}{s'}  = \frac{2}{1.50 m} - \frac{1}{2.0 m}

\frac{1}{s'} =  1.833 m

s' = - 0.545 m

To determine the speed of the trick to the highway

\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}

\frac{ds}{dt} = 26.88 m/s

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

8 0
4 years ago
2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
AnnyKZ [126]

Answer:

b. negative

Explanation:

neutrons have a negative charge and protons have a proton has a positive charge

3 0
3 years ago
Read 2 more answers
Consider a uniformly wound solenoid having N=210 turns, length l=0.18 m, and cross-sectional area A = 4.00 cm2. Assume l is much
strojnjashka [21]

Answer:

Explanation:

Number of turns

N = 210turns

Length of solenoid

l = 0.18m

Cross sectional area

A = 4cm² = 4 × 10^-4m²

A. Inductance L?

Inductance can be determined using

L = N²μA/l

Where

μ is a constant of permeability of the core

μ = 4π × 10^-7 Tm/A

A is cross sectional area

l is length of coil

L is inductance

Therefore

L = N²μA / l

L=210² × 4π × 10^-7 × 4 × 10^-4 / 0.18

L = 1.23 × 10^-4 H

L = 0.123 mH

B. Self induce EMF ε?

EMF is given as

ε = -Ldi/dt

Since rate of decrease of current is 120 A/s

Then, di/dt = —120A/s, since the current is decreasing

Then,

ε = -Ldi/dt

ε = - 1.23 × 10^-4 × -120

ε = 0.01478 V

ε ≈ 0.015 V

4 0
3 years ago
What is happening to you as you walk across the carpet
ira [324]
You are attracting electricity<span />
6 0
3 years ago
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