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kotykmax [81]
3 years ago
8

What are the units of acceleration?

Physics
1 answer:
Marina86 [1]3 years ago
5 0

Acceleration is any change in speed or direction of motion.

The dimension of speed is [length/time],
so a change is [length/time²].

Popular units include [meter/second²] and [feet/second²] .
________________________

Direction almost always boils down to an angle, (which technically
has no dimensions), so a change in direction is  [angle/time] .

Popular units include [radian/second] and [degree/second] .

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How will frost on the wings of an airplane affect takeoff performance?
san4es73 [151]

Frost will disturb the smooth flow of air over the wing, unpleasantly distressing its lifting competence. In other words, this spoils the even flow of air over the wings, by this means decreasing lifting capability. Also, frost may avoid the airplane from becoming flying at normal departure speed.

8 0
3 years ago
What item can be arranged to transform electrical energy to mechanical energy
meriva

Answer:

battery iron wire

Explanation:

7 0
2 years ago
Calculate V1 and V2 ( V = voltage )
Kitty [74]

FORMULA:

  • V = IR, where V = P.D; I = Current; R = Resistance.

ANSWER:

Total equivalent resistance for circuit:

R(eq) = R1 + R2 [It is in series]

  • 330Ω + 470Ω
  • 800Ω

Now, Current passing through whole circuit:

I = V/R

  • 16/800
  • 1/50 ampere

We know that, In series combination current passing through whole circuit is same.

So, V¹ = IR¹

V¹ = 1/50 × 330

  • V¹ = 6.6 volt

And V² = IR²

V² = 1/50 × 470

  • V² = 9.4 volt
7 0
3 years ago
Read 2 more answers
A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000
allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

  • <em>Mass of the car, m = 1700 kg</em>
  • <em>Velocity of the car, v = 21 m/s</em>
  • <em>Time of motion, t = 10 s</em>
  • <em>Additional energy provided by the engine, E₁ = 22,000 J</em>
  • <em>Energy used in overcoming friction, E₂ = 3,666.67 J</em>

The change in the energy applied to the car is calculated as;

\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta  E = 18,333.33 \ J

The final speed of the car is calculated as follows;

\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2  = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

4 0
2 years ago
2 kg<br> Use 10 m/s2 for g
Iteru [2.4K]

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

6 0
1 year ago
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