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3 years ago

HELP ME

Physics

2 answers:

STatiana [176]3 years ago

5 0

Answer:

Explanation:

Electromagnetic and gravitational forces have equations that clearly describe them.

Electromagnetic Fe = K q₁ q₂ / r²

Gravitational Fg = G m₁ m₂ / r²

Analyzing these expressions a bit we can see some things in which they are EQUAL:

• They depend on the inverse of the squared distance

• They are long range, they become zero In the infinite

• They are proportional to a system constant, in one case the charge and in the other the mass

We can also observe some DIFFERENCES:

• Gravitational always attractive, the electric can be attractive or repulsive depending on the charge

• When we calculate their magnitudes the electric force in 10 40 times greater than the gravitational

* The electric force can be zero if the charge is zero, the gravitational force can never be zero

Katarina [22]3 years ago

3 0

Explanation:

alike

no contact with the object itself

they can attract objects

they're between 2 objects

difference

source of foundations

type of forces

gravity is only for huge objects

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Arisa [49]

Answer:

Explanation:

a = gsinθ

a = 9.8sin30

a = 4.9 m/s²

v² = u² + 2as

v² = 0² + 2(4.9)(10)

v² = 98

v = √98 = 9.8994949...

v = 9.9 m/s

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3 years ago

What does the vibrating object do to the air particles around it ?

dimulka [17.4K]

Answer:

When an object vibrates, it causes the molecules in the air around it to move. These molecules collide with nearby molecules, forcing them to vibrate as well. As a result, they collide with more surrounding air molecules.

Explanation:

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3 years ago

A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected

vekshin1

Answer:

(A) $9.14\times 10^{-9}sec$

(B) $6.20\times 10^{-3}A$

Explanation:

We have given inductance $L=5.41\mu H=5.41\times 10^{-6}H$

Resistance $R=0.949kohm=0.949\times 10^3ohm$

Time constant of RL circuit is equal to $\tau =\frac{L}{R}$

$\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec$

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

$i=i_0(1-e^{\frac{-t}{\tau }})$

According to question

$0.799i_0=i_0(1-e^{\frac{-t}{\tau }})$

$e^{\frac{-t}{\tau }}=0.201$

${\frac{-t}{\tau }}=ln0.201$

${\frac{-t}{5.7\times 10^{-9} }}=-1.6044$

$t=9.14\times 10^{-9}sec$

(b) Current at $t=\tau sec$

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})$

$i=6.20\times 10^{-3}A$

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3 years ago

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Shkiper50 [21]

The average adult in the us spends 24 hours watching televistion each week

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4 years ago

Read 2 more answers

Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 m

skad [1K]

Answer: $a= -1029.97 m/s^{2}$

Explanation:

We know Haley is driving with an initial velocity of $V_{o}=73mi/h$ and then she has to change to a velocity of $V_{f}=55mi/h$ in a distance $d=0.50 mi$ , and we also know we are dealing with constant acceleration $a$ .