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A hummingbird flies 1.2 m along a straight path at a height of 3.4 m above the ground. Upon spotting a flower below, the humming

Viefleur [7K]

Answer:1.84

Explanation:

Given

Humming bird travels 1.2 m along a straight line at height of 3.4 m

Now bird drops 1.4 m to hover in front of the flower

So bird net displacement is

displacement $=\sqrt{1.2^2+1.4^2}$

displacement $=\sqrt{3.4}$

=1.84

7 0

3 years ago

Think about the way your body works when you eat an ice cream cone. Provide an explanation of how organs are working together as

AlexFokin [52]

Answer:

Explanation:

BAHAHAHAH karely omg HAHAHA un idke

8 0

3 years ago

Testing Plant Growth with Fertilizer

kykrilka [37]

b) by the end of the fifth day, group b will be more than one cm taller than group a. eliminate

Explanation:

The correct prediction based on the data shown is that by the end of the fifth day, group B plants will be more than 1cm taller than group A plants.

let us examine the table closely;

Day Group A (Water Only) Plant Height (cm) Group B (Water and Fertilizer)

1 2.0 2.0

2 2.2 2.3

3 2.3 2.8

4 2.5 3.2

5 2.6 3.8

We can see that by the end of the fifth day, the height differences between the two plants, 3.8 - 2.6 = 1.2cm, this is greater than 1cm. This claim from the experiment is very correct.

Other predictions are false.

learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

5 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$