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Eduardwww [97]
2 years ago
7

E 3.6 What force is needed to give a mass of 20 kg an acceleration of 5 m/s??

Physics
2 answers:
nadya68 [22]2 years ago
8 0

Explanation:

We know that,

F=ma

F=20×5

F=100 N

luda_lava [24]2 years ago
6 0

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
  • F=20×5
  • F=100N

Hence, the needed force is 100N.

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During wich two time intervals does the particle undergo equal displacement
sweet-ann [11.9K]
Basically, you want to take the integral of each interval and compare them. The two intervals with the same integral represent equal displacement of the particle. And since delta(x) is always 2, all you have to do is average the initial and final velocities of each interval and multiply by two to find total displacement.

Hope it helped.

Edit to show calculations:

2 * [ (0 + 10)/2 ] = 10 for interval AB
2 * [ (7 + 3)/2 ] = 10 for interval DE
8 0
3 years ago
What are the folds called inside the mitochondria?
max2010maxim [7]
The inner membrane has many overlapping folds called cristae. Inside the inner membrane there is the mitochondrial matrix, it contains enzymes that are used in creating ATP. 
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7 0
3 years ago
Read 2 more answers
If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?
adelina 88 [10]

Answer:

150 inches (12.5 ft)

Explanation:

The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.

The following is the equation one needs to solve:

10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\

therefore solving for the distance "x" gives as the answer (in inches):

10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in

which can also be given in feet as: 12.5 ft

3 0
3 years ago
In which city did the wright brothers take off on their very first flight
Nezavi [6.7K]
<h2>Answer: Kitty Hawk, North Carolina </h2>

The Wright brothers, Wilbur and Orville, were pioneers of aviation, since they flew in a device heavier than air, which was inconceivable at that time.

Their first successful flight was on December 17th, 1903 in Kitty Hawk, North Carolina, which lasted only 12 seconds in which their plane (the Flyer I, with 341 kg, 6.4 m long and a wingspan of 12.3 m) traveled 37 m without touching the ground. This was achieved through the help of an external catapult that "threw" them into the air.

It should be noted that the Wright brothers only studied until high school, however, their passion for solving the problem of the human inability to fly, their perseverance and experience acquired over the years in their bicycle company, led them to reach that goal. An achievement that marked the beginning of the  aviation era.

7 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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