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3 years ago

Variable.

Physics

1 answer:

zloy xaker [14]3 years ago

4 0

Answer:

Independent

Dependent

Mass

Explanation:

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A silver bar 0.125 meter long is subjected to a temperature change from 200 C to 100 C . What will be the length of the bar afte

dimulka [17.4K]

\Delta L= \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.

8 0

3 years ago

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A uniformly charged thin ring has radius 15.0 cm and total charge 20.0 nC . An electron is placed on the ring's axis a distance

Ivanshal [37]

Answer:

b) 1.67×10^7 m/s

Explanation:

The solution is attached in the attachment section

8 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

7 0

3 years ago

A wheelbarrow is a complex machine that combines which simply marchines?

koban [17]

It's most likely the combination of a bucket and the wheel.

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3 years ago

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