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Juliette [100K]
4 years ago
11

Which component of the atom has the least mass?

Physics
2 answers:
mrs_skeptik [129]4 years ago
6 0

B) Electrons have the least amount of mass

34kurt4 years ago
5 0

B is the correct answer i hope this helps.

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1: Solve for acceleration. Write the units of your answer next to the number
aleksandrvk [35]
1. 7.1111m/s Recurring
Solution a = v/t basically divide the velocity by the time taken
2. 9.375m/s
Same solution as before
5 0
3 years ago
A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then s
yaroslaw [1]

Question:

A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then sealed and the air above the liquid is evacuated. The cork:

A. sinks slightly  

B. rises slightly  

C. floats at the same height  

D. bobs up and down about its old position

Answer:

The correct answer is C)  floats at the same height  

Explanation:

The liquid is incompressible because its density very high and leaves no room for further compaction whether or not there is atmospheric pressure. So when you put a cork on the liquid, pressure or no pressure, there is no displacement hence it floats on the same height regardless of the absence of air.

Cheers!

6 0
3 years ago
A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at
natka813 [3]
First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

YOUR ANSWER IS C.
7 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
Read 2 more answers
Equation of uniformly accelerated motion​
Romashka [77]

This means acceleration a is constant.

Let

a) vo be the initial speed, at t=0

b) v be the final speed after time t

c) d distance travelled in time t

Then we have:

a) v=vo+a×t

b) v²=vo²+2×a×d (Galilei's equation)

c) d=vo×t+a×t²/2

d) average speed vm=(vo+v)/2

3 0
4 years ago
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