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Shtirlitz [24]
3 years ago
7

Paragraph/Comprehension type questions.

Physics
1 answer:
MatroZZZ [7]3 years ago
4 0

Answer:

1>500gf

1>300gf

its answer

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A constant friction force of 25 Nacts on a 65 kg skier for 20
Radda [10]

Answer:

The skier's change in velocity is 7.69 meters per second.

Explanation:

The Newton's second law tells force is equal to the change on the linear momentum of a body:

\sum\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}

If we approximate the differential \frac{d\overrightarrow{p}}{dt} to \frac{\Delta\overrightarrow{p}}{\Delta t}:

\sum\overrightarrow{F}=\frac{\Delta\overrightarrow{p}}{\Delta t}

Using that linear momentum is mass times velocity:

\sum\overrightarrow{F}=\frac{m\Delta\overrightarrow{v}}{\Delta t}

Solving for \Delta\overrightarrow{v}:

\Delta\overrightarrow{v}=\frac{\Delta t\sum\overrightarrow{F}}{m}=\frac{(20\,s)(25\,N)}{65\,kg}

\Delta\overrightarrow{v}=7.69\,\frac{m}{s}

3 0
3 years ago
The distance versus time graph for Object A and Object B are shown.
lianna [129]
Both move with constant speed
5 0
3 years ago
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Which kind of star is most likely to be found in the halo?
san4es73 [151]
<span>The most likely type of star to be found in the halo are stars classified as M stars. These stars absorb red light, have temperatures under 3000K, have an average mass of .3 times the mass of the sun, have an average radius of .4 times the radius of the sun, and have .04 times the luminosity of the sun.</span>
6 0
4 years ago
How does a rubber rod become negatively charged through friction?
stira [4]
I think it is c I'm only in 7th grade but I'm pretty sure that the answer is c
5 0
4 years ago
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The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t
Radda [10]

Answer:

a)F = 6816.5680 N

b) F' = 0 N

c)F''=28195.5N

Explanation:

Magnitude of charges M=1.6\muC

Distance d=2.6

Generally the equation for  Net Force is mathematically given by

For First Drawing

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

F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}

F = 6816.5680 N

For second Drawing

F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}  

F' = 0 N

For Third Drawing

F'' =\frac{ k q^2}{d^2} * \sqrt (2)

F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}

F''=28195.5N

4 0
3 years ago
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