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tankabanditka [31]

3 years ago

10

Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? W

hat is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?

1 answer:

Akimi4 [234]3 years ago

8 0

Answer:

Same direction to produce maximum magnitude and opposite direction to produce minimum magnitude

Explanation:

Let a be the angle between vectors A and B. Generally when we add A to B, we can split A into 2 sub vectors, 1 parallel to B and the other perpendicular to B.

Also let A and B be the magnitude of vector A and B, respectively.

We have the parallel component after addition be

Acos(a) + B

And the perpendicular component after addition be

Asin(a)

The magnitude of the resulting vector would be

$\sqrt{(Acos(a) + B)^2 + (Asin(a))^2}$

$= \sqrt{A^2cos^2a + B^2 + 2ABcos(a) + A^2sin^2a}$

$= \sqrt{A^2(cos^2a + sin^2a) + B^2 + 2ABcos(a)}$

$= \sqrt{A^2 + B^2 + 2ABcos(a)}$

As A and B are fixed, the equation above is maximum when cos(a) = 1, meaning a = 0 degree and vector A and B are in the same direction, and minimum with cos(a) = -1, meaning a = 180 degree and vector A and B are in opposite direction.

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kupik [55]

Answer:

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Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

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$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$

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A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

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$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

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$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

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$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

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2 years ago

How dose an atom change if all of its electrons are removed

cluponka [151]

Answer:

The atom becomes a positively charged ion.

Explanation:

The building blocks of an atom are protons, neutrons, and electrons.
The protons and neutrons present in the core of the atom are called nucleus.
The electrons are scattered in an ordered way around the nucleus.
The protons are positively charged and the electrons are negatively charged particles. The neutrons do not possess any charges.
Binding energy is supplied to the atom to remove an electron.
It is possible to remove the electrons of the lighter elements.
When an electron is removed from the hydrogen atom. It becomes positively charged ion or simply proton.
When all of the electrons are removed from the helium atom, it becomes a positively charged α particle.
It is practically very difficult to remove all of the electrons from the heavier elements.
When all of the electrons are removed from an atom it becomes an unstable positively charged ion.

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