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leonid [27]
3 years ago
12

The speed at which an object falls and the acceleration at which it falls are

Physics
1 answer:
allochka39001 [22]3 years ago
6 0

No, the speed at which an object falls is not equal to the acceleration at which it falls.

Answer:

Option B

Explanation:

Speed is defined as how fast an object can cover a specific distance and in what time it covers. So it is measured as the ratio of distance covered to the time taken to cover that distance. While acceleration is the rate of change of velocity. Moreover, speed is a scalar quantity and acceleration is a vector quantity. So most of the times, the direction will play an important role in the varying values of speed and acceleration. Also, acceleration of an object will depend upon the force and mass of the object. Thus, speed and acceleration will not attain same value always.

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How is resonance used in musical instruments?
9966 [12]

Answer:

The answer is B: to Amplify the sound

7 0
3 years ago
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A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t
diamong [38]

Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

5 0
2 years ago
Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates  to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m=1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}=7*10^{14} m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

X=u[2y/a]^{1/2}

=4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}

=3*10^{-2}= 3 cm

5 0
3 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?​
astra-53 [7]

Given:-

  • Speed of the unicycle = 20 m/s
  • Time taken = 15 s

To Find: Distance travelled by the unicycle.

We know,

s = vt

where,

  • s = Distance travelled,
  • v = Speed &
  • t = Time taken.

Therefore,

s = (20 m/s)(15 s)

→ s = (20 m)(15)

→ s = 300 m (Ans.)

8 0
3 years ago
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