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san4es73 [151]
3 years ago
11

Which table is it I’ll mark brainlest

Physics
1 answer:
alexira [117]3 years ago
5 0

Answer:

we need the graph to answer the question.

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A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
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2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
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А man runs 20 km west is hows. He then runs 20km 60° north or west 2.5 hours. what's the average velocity the Runnner?​
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Explanation:

\large{ \star \fcolorbox{magenta}{purple}{ \fcolorbox{magenta}{blue}{ \fcolorbox{magenta}{pink}{ \fcolorbox{magenta}{red}{ \sf{ añswër}}}}}} \star

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
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The resistance is 27.5 ohms
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4 years ago
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Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act
adoni [48]

Answer:

Juno scientific payload includes:

  • A gravity/radio science system (Gravity Science)
  • A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
  • A vector magnetometer (MAG)
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  • An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

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3 years ago
A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0
3 years ago
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