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2 years ago

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What happens when you decrease the thrust on your scooter? A. You stop B. Nothing happens C. You fall over D. You speed up Reset

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1 answer:

mars1129 [50]2 years ago

4 0

Answer:

D. You speed up

Explanation:

hope it helps

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If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i

slavikrds [6]

Answer:

$3t_{1/2}$

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

$N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}$ (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em> $t_{1/2}$ <em>: is the half-lifes.</em>

By solving equation (1) for t we have:

$\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}$

<u>Having that:</u>

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

$t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}$

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0

2 years ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0

2 years ago

Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener

Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 $\frac{m^{3} }{kg}$

% Error = 20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 $\frac{m^{3} }{kg}$

% Error = - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 $\frac{J}{kg k}$

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ $\frac{V}{m}$ = $\frac{R T}{P}$

⇒ Here $\frac{V}{m}$ = Specific volume = v

⇒ v = $\frac{R T}{P}$

Put all the values in the above formula we get

⇒ v = $\frac{323}{10^{6} }$ ×81.49

⇒ v = 0.02632 $\frac{m^{3} }{kg}$

This is the specific volume by ideal gas equation.

Actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.02632 - 0.021796 = 0.004524 $\frac{m^{3} }{kg}$

% Error = $\frac{0.004524}{0.021796}$ × 100

% Error = 20.75 %

(b). From the generalized compressibility chart the value of specific volume

$\frac{V}{m}$ = v = 0.0142 $\frac{m^{3} }{kg}$

The actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.0142 - 0.021796 = $\frac{m^{3} }{kg}$

% Error = $\frac{- 0.0076}{0.021796}$ × 100

% Error = - 34.85 %

3 0

2 years ago

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere

Nataliya [291]

Answer:

The correct answer is D $I_{30}$ / $I_{50}$ = 1.5

Explanation:

In this exercise the moment of inertia equation should be used

I = ∫ r² dm

In addition to the parallel axis theorem

I = $I_{cm}$ + M D²

Where $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

L = 100.00 cm (1m / 100 cm) = 1.0000 m

$x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

λ = M / L = dm / dx

dm = λ dx

Let's replace in the moment of inertia equation

I = ∫ x² ( λ dx)

We integrate

I = λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]

I = λ/3 L³/4

I = 1/12 λ L³

Let's replace the linear density with its value

I = 1/12 (M/L) L³

I = 1/12 M L²

Let's calculate with the given values

I = 1/12 M 1²

I = 1/12 M

This point is the center of mass of the rod

Icm = I = 1/12 M = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

D = $x_{cm}$ - x

D = 0.50 - 0.30

D = 0.20 m

Let's calculate

$I_{30}$ = $I_{cm}$ + M D²

$I_{30}$ = 1/12 M + M 0.202

$I_{30}$ = M (1/12 + 0.04)

$I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

$I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

$I_{30}$ / $I_{50}$ = 1.48

The correct answer is d 1.5

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2 years ago

A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t

7nadin3 [17]

I don’t know what the angle is in your diagram so I used the angle from the vertical.

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2 years ago