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3 years ago

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Vector e is 0.111m long in a 90 deg direction. Vector f is 0.233 m long in 400 deg direction. Find the direction of their vector

sum

1 answer:

-BARSIC- [3]3 years ago

7 0

Answer:

50.6

Explanation:

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When you jump off the earth, your momentum changes, but the Earth does not move. 1)If momentum is always conserved, why do we no

Mazyrski [523]

This question is off-base and misleading from the beginning.

When you jump off the Earth, your momentum changes, <em>and the Earth moves away from you with an equal change of momentum in the opposite direction</em>.

1). Momentum is conserved when you jump. But we don't feel the Earth moving. Since the Earth's mass is a bazillion times greater than YOUR mass, the speed with which the Earth moves away from you is only one bazillionth of your speed. That way, the product of (mass) x (speed) is the SAME for you and for the Earth, and momentum is conserved.

2). <em>Of course !</em> If everyone jumped at the same time, the Earth's momentum would change. In answer-(1), I explained that the Earth's momentum changes whenever <em>ONE PERSON</em> jumps. So 7 billion people all jumping at the same time would certainly make it change.

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3 years ago

A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the

REY [17]

Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

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3 years ago

Everything we know about stars comes from studying their _____.

denis-greek [22]

Light / Radiation. The light radiation defines its color. Thermal radiation for its temperature and the overall appearance of the wavelengths the star emits gives off the info from where astronomers and scientist are able to build up knowledge on the age and current state of the stars

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3 years ago

Read 2 more answers

Explain why an equation may be homogenous with respect to its unit but still be incorrect

natima [27]

When both sides of an equation give the same units, same numerical values, and same concept we refer to the equation as being correct. ... Removing constants from correct equations make them homogeneous but incorrect.

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3 years ago

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago