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4 years ago

The cells lie odjacent to the sieve tubes

Physics

1 answer:

nignag [31]4 years ago

4 0

Answer:

<h2>Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell. </h2>

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Consider two transformers. Transformer A has 300 turns on its primary and 500 turns on its secondary. Transformer B has 250 turn

rusak2 [61]

Answer:

Secondary voltage on second transformer is 200 volt.

Explanation:

It is given two transformer

Let us consider first transformer.

Number of turns in primary $N_p=300$

Numb er of turns in secondary $N_s=500$

Now consider second transformer

Number of turns in primary $N_p=250$

Number of turns in secondary $N_s=1000$

Now it is given that same voltage of 50 volt is applied to primary of both the transformer.

For second transformer

$\frac{N_p}{N_s}=\frac{V_p}{V_s}$

$\frac{250}{1000}=\frac{50}{V_s}$

$V_s=200volt$

So secondary voltage on second transformer is 200 volt

4 0

4 years ago

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s

Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

$I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}$

where I is moment of inertia

$w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm$

7 0

3 years ago

A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does t

34kurt

Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

$\Delta T=T_{f}-T_{i}$

$\Delta T=21-2.0$

$\Delta T=19^{\circ}C$

We need to calculate the carrier lengthen

Using formula of length

$\Delta L=\alpha_{steel}\times L_{0}\times\Delta T$

Put the value into the formula

$\Delta L=1.2\times10^{-5}\times370\times19$

$\Delta L=0.08436\ m$

Hence, The carrier lengthen is 0.08436 m.

8 0

3 years ago

A 1.45 kg falcon catches a 0.515 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is

True [87]

Answer:

Their velocity after the impact is 20.85 m/s.

Explanation:

Given that,

Mass of falcon, $m_1=1.45\ kg$

Mass of dove, $m_2=0.515\ kg$

Initial speed of the falcon, $u_1=26.5\ m/s$

Initial speed of the dove, $u_2=4.95\ m/s$

We need to find the final velocity after the impact. When the falcon catches the dove, it will becomes the case of inelastic collision. The conservation of momentum will be :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{1.45\times 26.5+0.515\times 4.95}{(1.45+0.515)}\\\\V=20.85\ m/s$

So, their velocity after the impact is 20.85 m/s.

8 0

3 years ago

A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top

raketka [301]

when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
 $E=K = 100 J$
where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.

3 0

4 years ago

The cells lie odjacent to the sieve tubes​

The cells lie odjacent to the sieve tubes