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scoray [572]
3 years ago
13

You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative char

ge, while the other sphere has no net charge.
(a) If the spheres are close together but do not touch, will they
(i) attract each other,
(ii) repel each other, or
(iii) exert no force on each other
Physics
1 answer:
kkurt [141]3 years ago
8 0

Answer:attract each other

Explanation:

When two-sphere, one with a negative charge and another neutral is brought close together but do not touch then they try to attract each other.

This because of the polarization of the neutral sphere as it is placed in the vicinity of a negatively charged sphere. The negatively charged sphere will induce the positive charge in the neutral sphere and they will attract each other according to Columb law.

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A superhero flies 285 m from the top of a tall building at an angle of 35◦ below the horizontal.
iren2701 [21]
287.14108 it simple prantherogram theorem 
 
7 0
3 years ago
A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area
Airida [17]
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
p_1 = p_2
which becomes
\frac{F_1}{A_1}= \frac{F_2}{A_2}
where
F_1=3600 N is the force on the first piston
A_1=0.030 m^2 is the area of the first piston
F_2 is the force on the second piston
A_2=0.015 m^2 is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
F_2=F_1  \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N
7 0
3 years ago
Read 2 more answers
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
A flat uniform circular disk (radius = 2.30 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the
Vadim26 [7]

The resulting angular speed = 0.6 rad / s.

<u>Explanation:</u>

Here there is no external torque acting on the system thus we can apply the law of conservation of angular  momentum  

Angular momentum of the man = Iω

Where I = Inertia of the man about the axis of rotation

or         I = M r 2

            I  = 50 * 1.25*1.25 = 78.125

w = Angular velocity of the man, that can be calculated as follows

Tangential velocity of man = v = 2m/s  

So time taken to describe this circle is t = (2*pi* r) / v

Now angle described in 1 revolution θ = 2*pi radians

This angle is subtended in time t = (2*pi* r) / v

Thus angular speed = w = θ/t = 2*pi* ( v/ 2π r) = v/r = 2.70 / 1.25 = 2.16 rad/s

So angular momentum of man = Iw = 78.125 * 2.16 = 168.75.

To conserve the angular momentum before and after,

Angular momentum of disk = angular momentum of the man  

           i.e.             Iw of disk = 168.75

                                disk of I = (disk of M*R^2) / 2

                                              = (1.00 * 102 * 2.30 * 2.30) / 2

                                              = 269.79

                 Thus 269.79 of disk of w = 168.75

      Resulting angular speed of disk = 168.75 / 269.79 = 0.6 ras / s

7 0
3 years ago
You push against a steamer trunk with a force of 800 n at an angle alpha with the horizontal . the trunk is on a flat floor and
galina1969 [7]
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
6 0
3 years ago
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