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Brrunno [24]
3 years ago
6

What is The most basic unit of an Element on the Periodic Table

Physics
1 answer:
lesya [120]3 years ago
8 0

Answer:

atom(which contains protons, neutrons and electrons)

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A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a right turn and travels 1.33 km east before maki
charle [14.2K]

Answer:

AD\approx1.7582\ km

Explanation:

Follow the schematic in which point A is the warehouse and point D is the destination.

Now we observe the triangle constructed \Delta ADB:

here:

AB\perp BD

AB=2.6-1.45

AB=1.15\ km

&

BD=1.33\ km

As we know that displacement is the shortest distance between two points.

<u>Using Pythagoras theorem:</u>

AD=\sqrt{AB^2+BD^2}

AD=\sqrt{(1.15)^2+(1.33)^2}

AD\approx1.7582\ km

3 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
4 years ago
A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t
zubka84 [21]

Answer:

a) Q1=Q2=480μC   V1=240V   V2=60V

b) Q1=96μC   Q2=384μC   V1=V2=48V

c) Q1=Q2=0C    V1=V2=0V

Explanation:

Let C1 = 2μC  and  C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C

Q_{T} = V_{T}*C_{T} = 480\mu C

V1 = \frac{Q1}{C1}=240V

V2 = \frac{Q2}{C2}=60V

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

V1=V2   So,  \frac{Q1}{C1} = \frac{Q2}{C2}

Total charge is the same calculated for part (a), so:

\frac{Qt - Q2}{C1} = \frac{Q2}{C2}   Solving for Q2:

Q2 = 384μC    Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C  V1=V2=0V

3 0
3 years ago
How far would a spacecraft moving in a circular orbit 500 km above Pluto's surface travel during that time?
Ilia_Sergeevich [38]

Answer:

The speed of the spacecraft should be 719.35m/s

Explanation:

if the spacecraft is orbiting the planet with a circular orbit, the gravitational force must act as a centripetal force. This means:

F_G=F_c\\\frac{GmM}{d^2}=m\frac{v^2}{d}

In this case, the pluto's mass M is 1.3099·10^22 kg. The radius of the planet R is 1188.3Km and G is the gravitational constant. Therefore:

\displaystyle\frac{G M}{d^2}=\frac{v^2}{d}\\v=\sqrt{\frac{GM}{d} } =\sqrt{\frac{GM}{(R + 500km)} } =719.35m/s

7 0
3 years ago
Can a observation be accurate but not precise?
laila [671]
Sure 
"Mount Everest is 29,000 feet tall" or in scientific terms 2.9E4 
This statement/observation is accurate to 2 decimal places. It is precise only to (perhaps) the nearest thousand feet. 
More precise 
"Mount Everest is 29,029 feet tall"
3 0
4 years ago
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