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3 years ago

8

if bananas are curved because they grow towards the sun ☀️so say if a was born towards the sun would i have been a curved baby o

r what

2 answers:

Ivenika [448]3 years ago

3 0

ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ

Tomtit [17]3 years ago

3 0

Answer:

It's because of the sun! Bananas are curved so they can retrieve sunlight. Bananas go through a process called 'negative geotropism'. ... So, bananas do not grow directly towards the sun rays but grow upwards to break through the canopy.

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A train slows from 60m/s to 20m/s in 50s

Dovator [93]

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

hope this helps! <3

7 0

1 year ago

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Calculate the net force and the acceleration on the block

Darina [25.2K]

Answer:

Net Force = 10N

Acceleration = 2m/s^2

Explanation:

calculate the net force and the acceleration on the block

Net force on the block F = mass * acceleration

Net force acting in the positive direction = 4N + 6N = 10N

Mass = 5kg

According to newton's second law;

a = F/m

a = 10N/5

a = 2m/s^2

hence the acceleration on the block is 2m/s^2

7 0

3 years ago

A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti

Pavel [41]

If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

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3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago

What is meant by a quark’s “flavor”?

Viktor [21]

Answer:

b

Explanation:

because a quark have many kinds of flavor

7 0

3 years ago

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