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3 years ago

11

A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h

eight is the ball originally dropped?

1 answer:

dlinn [17]3 years ago

8 0

Answer:

How long does the ball fall is t_2 = 13.66 (s).

From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

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What happens to rocks as water combines

insens350 [35]

The answers is
D. The acid creates cracks in the rocks, which
allow air to circulate through the rock,
causing it to weather

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3 years ago

A student pulls out his or her chair in order to sit down. The student pulls the chair 0.75 m with a force of 20 N. How much wor

IRINA_888 [86]

Work = Force x displacement
Force= 20N
Displacement= 0.75m

W= 20(0.75)
W= 15 Nm

3 0

3 years ago

In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing t

Stells [14]

Answer:

Explanation:

Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.

Hence;

1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.

2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.

3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.

4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.

4 0

3 years ago

Let's say you are on the third story of an apartment building that has a swimming pool. For some insane reason you think it is a

labwork [276]

Answer:

To make it into the pool you must run and jump at

$\displaystyle v_o=x.\sqrt{\frac{g}{2y}}$

Explanation:

Horizontal Launch

When an object is thrown with a specified initial speed in the horizontal direction, it describes a curved path that finishes when it hits the ground level after traveling certain horizontal distance x and a vertical height y from the launching point. The horizontal speed is always constant and the vertical speed increases due to the effect of gravity. It can be found that the horizontal distance reached by the object when launched at an initial speed in a given time t is

$x=v_o.t$

And the vertical distance is

$\displaystyle y=\frac{g.t^2}{2}$

If t is the total flight time, then x and y are maximum and we can find a relation between them. Solving for t in the first equation

$\displaystyle t=\frac{x}{v_o}$

Substituting in the second equation

$\displaystyle y=\frac{g}{2}\left ( \frac{x}{v_o}\right )^2$

Rearranging

$\displaystyle \left ( \frac{v_o}{x}\right )^2=\frac{g}{2y}$

Solving for $v_o$

$\displaystyle v_o=x.\sqrt{\frac{g}{2y}}$

There are many applications for the horizontal launch. One common situation is when someone wants to drop something on certain terrain at a specific approximate point when traveling in a plane at a given height. Once the object is left fall, it has the same speed as the plane, so the plane speed can be estimated to make the best possible launch, or given that speed, we can know in advance where the object will reach ground level

4 0

3 years ago

Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the

Brrunno [24]

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

$137.89848\times 10^{-6}\ J$

$111.20842\times 10^{-6}\ J$

2.98273 V

Explanation:

$C_1=25\ mu F$

$C_2=31\ mu F$

V = Voltage = 6 V

Equivalent capacitance is given by

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

Equivalent capacitance is $13.83928\times 10^{-6}\ F$

Energy stored is given by

$E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J$

Total energy stored is $249.10704\times 10^{-6}\ J$

Charge is given by

$Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C$

Voltage is given by

$V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V$

$E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J$

Energy strored in C1 is $137.89848\times 10^{-6}\ J$

$V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V$

$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

So, the energy is equivalent

Equivalent capacitance

$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V

8 0

4 years ago