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3 years ago

How do birds manage to fly without flapping their wings?

1 answer:

3 0

It's called gliding.
So, they will get up high and collect their balance and be nice and steady. Then, they can spread their wings out and glide. The wind will carry them from their.
How do we manage to stand without constantly walking? It's the same exact thing.

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Facts about river deltas

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3 years ago

Write the following as powers of ten with one figure before the decimal point:

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3 years ago

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mihalych1998 [28]

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3 years ago

Read 2 more answers

Two equal resistors are connected in series with a 1.50V battery. In order to keep the current at 0.030 A, the resistors much ea

AlexFokin [52]

Answer: 25 Ohms

Explanation:

From this question, the following parameters are given:

Voltage V = 1.5 v

Current I = 0.03A

From Ohm's law;

V = IR

Where R = resultant resistance of the two resistors.

Substitute V and I into the formula and make resultant R the subject of formula.

1.5 = 0.03 × R

R = 1.5/0.03

R = 50 Ohms

From the question, it is given that Thr two equal resistors are connected in series.

R = R1 + R2

But R1 = R2

50 = 2R1

R1 = 50/2

R1 = 25

R1 = R2 = 25 Ohms

Therefore, the resistors must each have a value of 25 Ohms

4 0

4 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$

$p_e=1.35\times 4.61\times \sin28^{\circ}$

$p_e=2.9218\ kg.m.s^{-1}$

6 0

3 years ago