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3 years ago

9

Mathew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the bal

ls journey will gravity be the greatest force acting on the ball

1 answer:

Cloud [144]3 years ago

8 0

When It begins to drop because that when gravity will have its strongest pull on the object.

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Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

The force of replusion between two like charged particles will increase if

Alex_Xolod [135]

Answer:

The distance of separation is decreased

Explanation:

From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.

7 0

3 years ago

Please help. Questions are in the image.

yuradex [85]

Speed: measure of how fast an object travels
Velocity: measure of how fast, and in what direction, an object travels
Distance: how far an object has traveled on some path
Position: where an object is located in some reference system
Displacement: the difference between an objects starting position and it’s ending position

8 0

3 years ago

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

grandymaker [24]

P1V1/T1 = P2V2/T2

(1.6atm)(168cm^3)/(255K) = (1.3atm)V2/(285K)

Final volume = 231cm^3

6 0

4 years ago

Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket's engine is to be fired

ValentinkaMS [17]

Answer:

$\frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=1096.633$

Explanation:

$m_i$ = Initial mass of rocket

$m_f$ = Final mass of rocket

u = Initial velocity

$v_r$ = Relative velocity

v = Velocity

From the rocket equation

$v=u+v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v}}=e^1\\\Rightarrow \frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=2.7182$

when $v=7v_r$

$v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{7v_r}{v_r}}=e^7\\\Rightarrow \frac{m_i}{m_f}=1096.633$

$\frac{m_i}{m_f}=1096.633$

6 0

3 years ago