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Leokris [45]
3 years ago
7

Each of two identical objects carries a net charge. The objects are made from conducting material. One of them is attracted to a

positively charged ebonite rod, and the other is repelled by the rod. After the objects are touched together, it is found that they are each repelled by the rod. What can be concluded about the initial charges on the objects?
a) initially, both objects are positive, with both charges having the same magnitude.
b) Initially both objects are negative, with both charges having the same magnitude.
c) Initially one object is positive and one is negative, with the positive charge having a greater magnitude than the negative charge.
d) Initially one object is positive and one is negative, with the negative charge having a greater magnitude than the positive charge.
Physics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

It can be concluded that:

c) Initially one object is positive and one is negative, with the positive charge having a greater magnitude than the negative charge.

Explanation:

There is the property of charges by virtue of which like charges repel each other and unlike charges attract each other.

Given that,

Initially, one of the object is attracted to positively charged ebonite rod, therefore, it must have negative charge initially and the other is repelled by the rod, therefore, it must have positive charge initially.

When the two objects touched together, there is a redistribution of charges on them, such that, the net charge on both the objects distributes equally on both.

Since, it is found that they are each repelled by the rod after the objects touched each other, it means they both acquire positive charges.

Both the objects acquire positive charge implies that the net charge on the two objects, i.e., the sum of charges on the two objects, must be positive before they were touched which means that the object with the positive charge had charge greater in magnitude.

So the correct conclusion is:

c) Initially one object is positive and one is negative, with the positive charge having a greater magnitude than the negative charge.

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8 0
3 years ago
shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

8 0
3 years ago
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

1000\times\dfrac{0.10400}{0.02}=\rho_{m}

\rho_{m}=5200\ kg/m^3

Hence, The density of the metal is 5200 kg/m³.

6 0
3 years ago
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in
Maksim231197 [3]

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

                            v =  (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

                            mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

                            mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

                            unit (v) = vec (v) / mag (v)

                            unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

                            unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

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3 years ago
How is the concept of wind best described?
konstantin123 [22]
I believe that it is the first one just a guess tho. So don't trust me, just in case
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