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My name is Ann [436]
3 years ago
6

The unit of length most suitable for measuring the thickness of a cell phone is a ( megameter, meter, millimeter, nanometer ) Th

e unit of length most suitable for measuring the height of a backyard tree is a ( megameter, meter, millimeter, nanometer ).
Physics
1 answer:
puteri [66]3 years ago
5 0

Answer: 1.millimeter 2.meter

Explanation:

a phones thickness can be measured with such a small measurement and a tree is not big enough for a megameter but not small enough for a millimeter so it'd be a meter

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Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
3 years ago
If a wave is traveling at a certain speed and it's frequency is doubled,what happens to the wavelength of that water​
Elena L [17]

Answer:

If the frequency is doubled the wavelength is only half as long.If you put your fingertip in a pool of water and repeatedly move it up and down,you will create circular water waves.

Hope this helps!!!

4 0
3 years ago
When a force of 20.0N is applied to a spring, it elongates 0.20m. Determine the period of oscillation of a 4.0kg object suspende
choli [55]
So what we can do is apply the<span> Hooke's law wich states that
F = -kx ( P.S the -ve sign means opposite in direction ) 
Also we will need to determine the spring's constant with the formula:
k = F / x 
Where F = the force ( = 20 N ) 
x = the displacement of the end of the spring from it's position ( = 0.20 m ) 
k = the spring's constant ( = unknown ) 
So this would be: k = 20 / 0.20 = 100 N/m 
The period of oscillation of 4 kg : T = 2 * pi * square root m / k 
T = 2 * pi * square root 4 / 100 
T = 1.256 seconds
Hope it helps</span>
7 0
3 years ago
The boiling point of a substance is _72 degree Celsius. This temperature will be equivalent to Kelvin scale is-------.
Vlada [557]

Answer:

345 K

Explanation:

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object.

Generally, it is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

<u>Given the following data;</u>

  • Boiling point = 72°C

<em>To convert the temperature in degree Celsius to Kelvin, we would use the following mathematical expression;</em>

Kelvin = 273 + °C

Substituting into the formula, we have;

Kelvin = 273 + 72

<em>Kelvin = 345 K</em>

<em>Therefore, the temperature of 72°C will be equivalent to 345 K on the Kelvin scale.</em>

3 0
3 years ago
A thin, horizontal copper rod is 1.0792 m long and has a mass of 53.1794 g. The acceleration of gravity is 9.8 m/s 2 . What is t
horsena [70]

Answer:

0.1675 A

Explanation:

Since BILsin\theta=mg then making I the subject I=\frac {mg}{BLsin\theta} where L is the length of the rod, in this case given as 1.0792 m, B is magnetic field which is given as 2.88578 T, m is the mass which is  53.1794 g which is equivalent to 0.0531794 Kg . For minimum current, sin\theta=1.

Substituting the given values then

I=\frac {0.0531794\times 9.81}{1.0792\times 2.88578}=0.167512525 A\approx 0.1675 A

3 0
3 years ago
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