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3 years ago

Can two negatively charged balloons move apart without ever touching why or why not

2 answers:

Nutka1998 [239]3 years ago

8 0

Yes, because electric charges have electric field surrounding them that allow them to exert forces on other objects without touching them.

Sloan [31]3 years ago

4 0

Your answer would indeed be, yes, because electric charges have electric field surrounding them that allow them to exert forces on other objects without touching them.

just took the test and got it right! Hope this helped!

You might be interested in

The boy on the tower throws a ball 20m downrange. What is his pitching speed?

san4es73 [151]

Answer:

The pitching speed of the ball is 19.7 m/s

Explanation:

Here, we can use the third equation of motion, $v^{2} = u^{2} - 2as$

whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled

Here, the initial velocity of the the ball is given as zero and the acceleration due to gravity is 9.8 , the distance 's' is given as 20 m

Using the equation, $v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s$

Hence, the pitching speed of the ball is 19.7 m/s

5 0

2 years ago

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

How many zebras automatically survive the first interaction with the lions in Generation 1?

lesya692 [45]

I think the answer is c. but I think it depends on how many zebras you have

5 0

3 years ago

A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2

MArishka [77]

From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
= 2×9.81×0.43
= 8.4366
u = √8.4366
=2.905 m/s
Hence the initial velocity is 2.905 m/s
Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
= 2.905/9.81
= 0.296 seconds

3 0

3 years ago

What is the equation for an inelastic collision

abruzzese [7]

M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.

7 0

2 years ago