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3 years ago

Problem 5 please! I need an equation to solve

Physics

2 answers:

Maru [420]3 years ago

3 0

N(t) = N(0) x [1/2]^(t/th)
N(t) = 12.5
N(0) = 100
t = 10
th = half life ... your answer

ziro4ka [17]3 years ago

3 0

You could try the equation (12.5/100) = (1/2)^x .

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

3 years ago

Calculate the magnitude of the electric field at the location of given that the square is 6.05 cm on a side.

Serggg [28]

The term “electric field” refers to the physical field that surrounds electrically charged particles and acts to either attract or repel all other charged particles in the field (also known as an E-field).

It can also refer to the physical field surrounding a system of charged particles. Electric fields are composed of electric charges and time-varying electric currents.

Both electric and magnetic fields are manifestations of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature.

Electrical technology makes use of electric fields, which are significant in many branches of physics.

For instance, in atomic physics and chemistry, the electric field acts as an attractive force to hold atoms’ atomic nuclei and electrons together. It is also the force that causes atoms to chemically link together to form molecules.

To know more about electric field visit:

brainly.com/question/14811118

#SPJ4

4 0

1 year ago

after singing for a few minutes the soloist rejoins the choir a second choir, consist of 90 additional singers then joins in eac

vredina [299]

If the soloist produces "x" decibels and the 10-person choir produces "y" decibels, combined they will produce "x+y" decibels.

The second choir has 90 additional singers, we base our description on the first choir. If a 10-person choir produces "x+y" decibels, then the 90 person choir produces 10 (x+y) decibels.

7 0

3 years ago

Which of the rolling statement is not true about proper tire care? A. If you see a weaver bar across the width of the tread. B.

skad [1K]

The correct answer should be <span>A. If you see a wear bar across the width of the tread, it's time to change the tire

This is an incorrect sentence because you should change a tire before you see the wear bar. At the time you see the bar, you should've already changed it. </span>

7 0

3 years ago

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

4 0

3 years ago