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Montano1993 [528]

3 years ago

13

A vector A~ has a negative x component 2.89 units in length and a positive y component 3.17 units in length. The vector A~ can b

e written in unit vector notation: a1ˆı + a2ˆ. Find the value of (a) a1 (units) (b) a2 (units) (c) What is the magnitude of A~ (units)? (d) What is the direction of A~ (degrees)? 1

1 answer:

Fantom [35]3 years ago

5 0

Answer:

a1 = -2.89 units

a2 = 3.17 units

magnitude = 4.2896 units

direction = 132.35 degrees

Explanation:

We can write the vector in component form as: <-2.89, 3.17>

That means that its i and j components are:

a1 = -2.89 units

a2 = 3.17 units

The magnitude of the vector is given by:

$\sqrt{(-2.89)^2 +3.17^2} \approx 4.2896$

The vectors direction can be found from the tangent function, ad noticing that the vector must reside on the Second Quadrant:

$arctan(-3.17/2.89) + 180^o\approx 132.35^o$

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3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

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3 years ago

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enyata [817]

Answer:

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Explanation:

<u>Given:</u>

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where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

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3 years ago

In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?

Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

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The nature of charges refers to the properties of charges.

There are two types of charges:

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The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

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2 years ago