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4 years ago

Two obiect accumulated a charge of

Physics

1 answer:

tamaranim1 [39]4 years ago

3 0

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

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

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

$q_1=4.5\mu C=4.5\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2.8\mu C=2.8\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

$F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N$

So, the closest answer is

A) 181.24 N

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Lauryn serves a volleyball with a mass of 2.1 kg. The ball leaves her hand with a speed of 30 m/s.

Naya [18.7K]

Hi! Let's see.

<h3><u>TOPIC: Kinetic Energy.</u></h3>

Explanation:

Data:

Mass(m) = 2.1 kilograms
Speed(v) = 30 m/s
Kinetic Energy(Ek) = ?

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Use the kinetic energy formula:

$\boxed{\boxed{\bold{E_{k}=\frac{m*(v)^{2}}{2}}}}$

Replace by the data:

$\boxed{E_{k}=\frac{2.1kg*(30\frac{m}{s})^{2}}{2}}}}}$

We multiply speed squared:

$\boxed{E_{k}=\frac{2.1kg*900\frac{m}{s}}{2}}}$

Multiply in the numerator, without forgetting the units:

$\boxed{E_{k}=\frac{1890J}{2}}}}$

Both numerator and denominador operations are divided:

$\boxed{\bold{E_{k}=945\ J}}$

Answer: The kinetic energy is <u>945 Joules.</u>

Cordially Alejanndraax. Greetings!

8 0

3 years ago

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0

3 years ago

Who first proposed that each layer of rock represented a specific interval of geologic time?

RoseWind [281]

Hutton, a Scottish geologist.

4 0

3 years ago

A bungee jumper has a mass of 60kg and uses a 25m long bungee cord (unstretched length) with an elastic coefficient of 800N/m. a

KonstantinChe [14]

Answer:

a ) 2.68 m / s

b ) 1.47 m

Explanation:

The jumper will go down with acceleration as long as net force on it becomes zero . Net force of (mg - kx ) will act on it where kx is the restoring force acting in upward direction.

At the time of equilibrium

mg - kx = 0

x = mg / k

= (60 x 9.8 ) / 800

= 0.735 m

At this moment , let its velocity be equal to V

Applying conservation of energy

kinetic energy of jumper + elastic energy of cord = loss of potential energy of the jumper

1/2 m V² + 1/2 k x² = mg x

.5 x 60 x V² + .5 x 800 x .735 x .735 = 60 x 9.8 x .735

30 V² + 216.09 = 432.18

V = 2.68 m / s

b ) At lowest point , kinetic energy is zero and loss of potential energy will be equal to stored elastic energy.

1/2 k x² = mgx

x = 2 m g / k

= (2 x 60 x 9.8) / 800

= 1.47 m

3 0

3 years ago

Read 2 more answers

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

3 0

4 years ago