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2 years ago

Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?

Physics

1 answer:

Maslowich2 years ago

7 0

The available options are:

Mint is a dicot.

Mint is a monocot.

Mint is an angiosperm.

Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

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Briefly describe the formation of the planets from the solar nebula

saul85 [17]

<span>The formation of the Solar System began 4.6 billion years ago with the gravitational collapse of a small part of a giant molecular cloud. Most of the collapsing mass collected in the center, forming the Sun, while the rest flattened into a proto-planetary disk out of which the planets, moons, asteroids, and other small Solar System bodies formed.

Hope it helped
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6 0

3 years ago

Calculat the acceleration of a person at latitude<br> 40degreesowing to the rotation of the earth.

beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

acceleration of person = g - ω²R ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × $10^{-5}$ rad/s and R is earth radius that is 6.37 × $10^{6}$ m

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × $10^{-5}$ )² × 6.37 × $10^{6}$

acceleration of person = 9.77 m/s²

6 0

3 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago

A bullet is fired horizontally from a 17.7 m high cliff at a speed of 482.0. What is the distance from the cliff that the bullet

Marysya12 [62]

Answer:

Horizontal distance=?m

Explanation:

Horizontal velocity,u=482ms⁻¹

Height of the cliff=17.7m

Horizontal distance,R=?

R=v×√2h/g

5 0

3 years ago

Can someone help with my physics homework? please

Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

$W = F\cdot \Delta s$ (1)

Where:

$F$ - Force applied by the cart, measured in newtons.

$\Delta s$ - Distance travelled by the car, measured in meters.

$W$ - Work done on the cart, measured in joules.

If we know that $F = 888\,N$ and $\Delta s = 22\,m$ , then the work done on the cart is:

$W =(888\,N)\cdot (22\,m)$

$W = 19536\,J$

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

d) If we know that $F = 1111\,N$ and $\Delta s = 22\,m$ , then the work on the cart is:

$W = (1111\,N)\cdot (22\,m)$

$W = 24442\,N$

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0

3 years ago