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A student made a model of an onion skin cell she viewed using a microscope. The scale is 1:500.The students cell model has a len

r-ruslan [8.4K]

The original will be 15\500 to find the real length

6 0

3 years ago

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An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

5 0

3 years ago

An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.

podryga [215]

Answer:

$9.38\times 10^7 m/s$

Explanation:

We are given that

Potential ,V=25 kV= $25\times 10^3 V$

Distance,r =1 cm= $\frac{1}{100}=0.01 m$

1 m=100 cm

Mass of electron, m= $9.1\times 10^{-31} kg$

Charge, q= $1.6\times 10^{-19} C$

We have to find the final velocity of the electron.

Speed of electron, $v=\sqrt{\frac{2qV}{m}}$

Using the formula

$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}$

v= $9.38\times 10^7 m/s$

Hence, the final velocity of the electron= $9.38\times 10^7 m/s$

3 0

4 years ago

What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your

densk [106]

Answer:

The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

Explanation:

Given that,

Third charge $q_{3}=49.0\ nC$

Distance $x_{3}=-1.085\ m$

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

$F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}$

$F_{13}=-17268.3\times10^{-9}\ N$

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

$F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}$

Put the value into the formula

$F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}$

$F_{23}=11800.2\times10^{-9}\ N$

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

$F_{net}=F_{13}+F_{23}$

$F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}$

$F_{net}=-0.0000054681\ N$

$F_{net}=-5.468\times10^{-6}\ N$

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

5 0

3 years ago

How is sound produced by using pipel wind instruments

AlekseyPX

The air inside the pipes of a wind instrument vibrates. ... The result is longitudinal standing waves in the air column inside the pipe. The ends, whether open or closed, create nodes or antinodes in these standing waves.

5 0

3 years ago

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