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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?

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

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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

lora16 [44]

Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 50.4 m/s

Final speed of the ball, v = -37 m/s (As it returns)

(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

$J=m(v-u)$

$J=0.06\times (-37-(50.4))$

J = -5.24 kg-m/s

(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 0.06\times ((-37)^2-(50.4)^2)$

W = -35.1348 Joules

Hence, this is the required solution.

6 0

3 years ago

The temperature of a sample of water increases from 20°c to 46.6°c as it absorbs 5650 calories of heat what is the mass of the s

kompoz [17]

M=energy transferred/ (temperature change*specific heat) M= 5650/(26.6*1.0) M=212g

4 0

3 years ago

Which statement about cellulose is true?

zlopas [31]

Is there any answers? Or is it asking you to choose?

6 0

3 years ago

Read 2 more answers

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

1.

109.6 cm , - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

2 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago