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3 years ago

An instrument for observing very small objects is

1 answer:

6 0

Microscope
A microscope shines a light through a slide with the sample of the object on and you look through a lens that allows you to magnify it by a very significant amount depending on what kind of microscope you have (a powerful electron microscope can magnify to the atomic level).

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In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the

AleksandrR [38]

Answer:

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

Explanation:

Electric field due to plates

Ef = V/d

Ef = 2033 V / (2.0 * 10^-2 m )

Ef = 101650 V/m

So, we can write

Ef * q = m*g

q = m*g / E f

The mass can be equal using the density and the volume so:

m = ρ * v

The volume can be find as:

v = 2.298 x 10 ⁻ ¹⁶ m³

q = ρ * v * g / Ef

q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

5 0

2 years ago

2) A car travels 5 miles north, and then 10 miles south. What is the person's DISTANCE and

Liono4ka [1.6K]

Answer:

distance traveled is 15 mi

displacement is 5 mi

Explanation:

Distance takes time into account and adds up all the tiny displacements during the entire period of the trip.

Displacement ignores time and looks only at the change in position from the starting point to the ending point.

6 0

2 years ago

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

Law Incorporation [45]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is $r=6940 km=6.94 \cdot 10^6 m$ , the gravitational force is $F=9.21 \cdot 10^4 N$ and the mass of the Earth is $m_1=5.98 \cdot 10^{24} kg$ , therefore we can rearrange the previous equation to find m2, the mass of the telescope:
$m_2 = \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg$

6 0

3 years ago

Read 2 more answers

An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int

LiRa [457]

12 V is the f.e.m. $\epsilon$ of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
$\epsilon - Ir$
this is equal to the voltage drop on the resistance of the motor R:
$RI$
so we can write:
$\epsilon - Ir = RI$
and using $r=0.0305~\Omega$ and $R=0.055~\Omega$ we can find the current I:
$I= \frac{\epsilon}{R+r}=140~A$

8 0

2 years ago

What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?

alex41 [277]

Using K.E=1/2MV^2
answer is 125joules

5 0

3 years ago