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3 years ago

What type of plant tissue is NOT part of the vascular bundle?

Physics

2 answers:

mr_godi [17]3 years ago

7 0

Answer:

cortex

Explanation:

Cambium

The vascular cambium produces secondary xylem on the inside of the ring, and secondary phloem on the outside, pushing the primary xylem and phloem apart.

Phloem and Xylem

vascular bundles, each containing both xylem and phloem.

Cortex

The region of cells in the stem is called the cortex

jek_recluse [69]3 years ago

3 0

Answer:

Cortex

Explanation:

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Cho hai điện tích q1=q2=8.10^-7 C đặt cách nhau 5cm. Xác định cường độ điện trường tại điểm:

Aleksandr [31]

Answer: b

Explanation:

5 0

3 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

5 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

The instantaneous velocity of an object is the blank of the object with a blank

777dan777 [17]

The instantaneous velocity of the object is its speed and direction at that instant.

7 0

3 years ago

Read 2 more answers

It takes you 9.8 min to walk with an average velocity of 1.9 m/s to the north from the bus stop to the museum entrance.

Dahasolnce [82]

60 x 9.8 = 588
588 x 1.9 = 1117.2

5 0

3 years ago