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3 years ago

Why are bananas curved?

Physics

2 answers:

Vladimir [108]3 years ago

5 0

Answer:

Bananas are curved because they grow towards the sun

Explanation:

Bananas are curved so they can retrieve sunlight. Bananas go through a process called 'negative geotropism'. What it means is that bananas grow away from the ground, instead of growing towards it, hence the 'negative' geotropism.

fgiga [73]3 years ago

5 0

Answer:

Because they grow towards the sun and the sun is what makes things grow like water to Beneath each flower petal, a row of tiny banana fruits start to grow. Once they're much bigger in size, the fruit goes through a process called negative geotropism. Which basically means instead of continuously grow towards the ground, they start to turn towards the sun, in order to retrieve light.

Explanation:

You might be interested in

A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km

Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

$v_f=v_o+at$

Since the car starts at rest, you can write the following equation:

$20=0+a(5) \\\\a=20\div 5=4 m/s^2$

Now that you have the acceleration, you can do this:

$d=v_o+\dfrac{1}{2}at^2$

Once again, there is no initial velocity:

$d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m$

Hope this helps!

8 0

3 years ago

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0

3 years ago

A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is

Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

= √(T/A×ρ)

= √[2.96×10^4/(4.49×10^-3×7860)]

= 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0

3 years ago

A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}$

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.

3 0

3 years ago

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

6 0

3 years ago