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avanturin [10]
1 year ago
13

An inflated rubber balloon is rubbed with a wool cloth until an excess of 1.00 × 107 electrons is on the balloon. What is the ma

gnitude of the charge on the balloon?
Physics
1 answer:
REY [17]1 year ago
8 0

The magnitude of the charge on the balloon is 1.6 x 10⁻¹² C.

<h3>What is the magnitude of the charge on the ball?</h3>

The magnitude of the charge on the ball is calculated by determining the total charge equivalent to the given number of electrons.

The charge of one electron = 1.6 x 10⁻¹⁹ Coulombs

Now, we are going to estimated the total charge of 1 x 10⁷ electrons.

1 electron =  1.6 x 10⁻¹⁹ C

1 x 10⁷ electrons = ?

= (1 x 10⁷ electrons x 1.6 x 10⁻¹⁹ C) / (1 electron)

= 1.6 x 10⁻¹² C

Thus, the total charge of 1 x 10⁷ electrons is obtained by multiplying the magnitude of charge of one electron to the entire given electrons.

Learn more about charge of electron here: brainly.com/question/9317875

#SPJ1

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Why do sex cells contain only half the number of chromosomes needed for offspring? Explain
Bogdan [553]

Answer:

None, egg cells don't have chromosomes. No, sex cells do have chromosomes. Meiosis reduces chromosome number so that sex cells (eggs and sperm) have a half set of chromosomes–one homolog of each pair. This is the haploid number.

3 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
When magma cools quickly, what kind of texture or
Ksivusya [100]

Answer:

a

Explanation:

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6 0
3 years ago
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

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The rebirth of science and learning during the fifteenth century is termed the
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