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Rzqust [24]
3 years ago
8

Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the charge of object 1 is doubled AND the charge

of object 2 is tripled, then the new electrostatic force will be ____ units
Physics
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

432 units

Explanation:

Let the charges be q and Q separated by a distance r. The electrostatic force , F = kqQ/r² = 72 units. If q = 2q and Q = 3Q, then the new electrostatic force is

F = k × 2q × 3Q/r² = 6kqQ/r² = 6 × 72 = 432 units

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When you lift an object, you add the energy of lifting to the object.<br> Tor F
PtichkaEL [24]

It's true, when we lift an object we add energy to it.

because, when we lift an object by applying force , the object attains a height and hence the energy gets stored in it, in the form gravitational potential energy .

6 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
3 years ago
A gas in a sealed container has a pressure of 50 kPa at 27°C. What will the pressure of the gas be if the temperature rises to 8
alexgriva [62]

Answer:

the final pressure of the gas is 60 kPa.

Explanation:

Given;

initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa

initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k

final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K

Let the final pressure of the gas = P₂

Apply pressure law;

\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} = \frac{50,000 \times 360}{300}  = 60,000 \ Pa = 60 \ kPa

Therefore, the final pressure of the gas is 60 kPa.

4 0
3 years ago
A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0
3 years ago
Why don’t trade winds blow straight toward the equator<br> Apex
rosijanka [135]

Trade winds near equator blows in curve path instead of straight path. This is because of earth rotation. This effect of earth rotation that cause wind to move in curve motion is called Coriolis effect. These kind of wind blows at the northeast of the North hemisphere and southeast of the South hemisphere. The trade wind are warm and it blows due to rising of hot air from equator.

6 0
3 years ago
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