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Andrej [43]
3 years ago
10

Which sentence uses the correct adverb to make a comparison?

Physics
1 answer:
Arturiano [62]3 years ago
8 0
Option C would be the right one
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A small piece of aluminum (atomic number 13) contains 10^15 atoms (The atomic number is the number of protons; it determines the
Otrada [13]

Answer:

3 micro coulombs = 3 * 10E-6 coulombs     charge of aluminum

13 * 10E15 * 1.6 * 10E-19 = 2.08 E-3 Coulombs - charge of atoms in Al

3 * 10E-6 / 2.08 * E-3 = 1.44 * E-3 = .00144 = .144 %

.00144 of the original electrons would have to be lost

8 0
3 years ago
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is 2.0 m from th
ASHA 777 [7]

Answer:

13 m

Explanation:

It is given that :

I got a haircut  sitting at a place having two parallel mirrors at a distance = 6.5 m apart

My head is at a distance of 2 m from the nearer mirror.

Now the light from the back of my head must go to (6.5 - 2) = 4.5 m to the back mirror.

Then it must go to 6.5 m to the front mirror and 2 m from the front mirror to my eyes.

So in order to see the back of my head, it will be = 6.5 + 2 + 4.5 = 13 m away.

4 0
3 years ago
1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.
Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

Hence, this is the required solution.

8 0
4 years ago
A boat displaces a volume V of water as it floats on a fresh-water lake. When the boat moves into the ocean, what volume of salt
Mamont248 [21]

Answer:

Option (b)

Explanation:

let the weight of boat is W.  In equilibrium condition, the weight of boat is equal to the buoyant force acting on the boat.

The buoyant force acting on the boat is equal to the weight of water displaced by the boat.

In case of fresh water:

Weight of the boat = weight of fresh water displaced by the boat

W = Volume of fresh water displaced x density of fresh water x g

W = V x 1 x g

W = V x g ....... (1)

In case of salt water:

Let the volume of salt water displaced is V'.

Weight of the boat = weight of salt water displaced by the boat

W = Volume of salt water displaced x density of salt water x g

W = V' x 1.02 x g    ..... (2)

Equate equation (1) and equation (2), we get

V x g = V' x 1.02 x g

V' = 0.98 V

Thus, option (b) is true.

7 0
4 years ago
An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze(the
Amanda [17]

Answer:

a)  E = k Ze (1- r³ / R³)  1/r², b) E=0, c)   E = -6.62 10¹⁰  N / C

Explanation:

a) For this we can use the law of Gauus

         Ф = E- dA = q_{int} / ε₀

where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product

       E A =q_{int} / ε₀

the area of ​​a sphere  

      A = 4π r²

      E 4π r² = q_{int} / ε₀

      E = 1 / 4πε₀   q_{int} / r²

       k = 1 /4π ε₀

       E = k q_{int} / r²       (1)

       

let's analyze the charge inside the gaussian sphere,

let's use the concept of density for electrons, since they indicate that the charge is evenly distributed

     ρ = Q / V

where the volume of the sphere is

    V = 4/3 πr³

     Qe = ρ V

     Qe = ρ 4 / 3π r³

the density of the electrons is

     ρ = Ze 3 / (4π R³)

where R is the atomic radius

we substitute

       Qe = Ze r³/ R³

for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.

The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)

     q_{int} = q_proton + Q_electron

     q_{int} = + Ze - Qe

     q_{int} = + Ze - Ze r³ / R³

     q_{int} = Ze (1- r³ / R³)

   

  we substitute in equation 1

     E = k Ze (1- r³ / R³)  1/r²

b) on the surface of the atom r = R

therefore the electric field is zero

      E = 0

c) Calculate the electric field for the Uranium for

       r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m

     

       E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³   1/ (5 10⁻¹¹)²

       E = -6.62 10¹⁰  N / C

6 0
4 years ago
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