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Vikki [24]
3 years ago
12

An object falls freely from rest the total distance covered by it in 2s will be​

Physics
1 answer:
tresset_1 [31]3 years ago
5 0

Answer:

Distance, S = 19.6 meters

Explanation:

<u>Given the following data;</u>

Time = 2 seconds

We know that acceleration due to gravity is equal to 9.8 m/s².

Also, the initial velocity of the object is equal to zero because it's starting from rest.

To find the total distance covered by the object, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Where;

  • S represents the displacement or height measured in meters.
  • u represents the initial velocity measured in meters per seconds.
  • t represents the time measured in seconds.
  • a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 0*2 + \frac {1}{2}*(9.8)*2^{2}

S = 0 + 4.9*4

S = 4.9*4

<em>Distance, S = 19.6 meters</em>

<em>Therefore, the total distance covered by the object is 19.6 meters.</em>

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Answer:

The liquid is vaprizing at point C

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3 years ago
A football quarterback throws a football for a long pass. While in the motion of throwing, the quarterback moves the ball , star
sergejj [24]

This question is incomplete, the complete question is;

A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;

a) F_throw = 8.083 N

b) F_throw = 9.181 N

c) F_throw = 2.284 N

d) F_throw = 16.014 N

e) None of these is correct

Answer:

the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

Explanation:

Given that;

m = 0.408 kg

d = 1.909 m

u = 0 { from rest}

t = 0.439 s

Now using Kinetic equation

d = ut + 1/2 at²

we substitute

1.909 = (0 × 0.439) + 1/2 a(0.439)²

1.909 = 0 + 0.09636a

1.909 = 0.09636a

a = 1.909 / 0.09636

a = 19.8111 m/s²

Now force applied will be;

F = ma

we substitute

F = 0.408 ×  19.8111

F = 8.0828 ≈ 8.083 N

Therefore the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

5 0
3 years ago
In a lab, the mass of object A is 2.5 kg. Object A weighs:
dangina [55]

Answer:

25N

Explanation:

Assuming the lab is on earth:

w = mg = 2.5 (9.81) = 25N

8 0
3 years ago
Pure water has a pH of 7. Pure water _______. A. is a neutral substance B. could be either an acid or a base C. is a base D. is
malfutka [58]

A. is a neutral substance.

Only neutral substances (like pure water) can have a pH of 7.

5 0
3 years ago
Read 2 more answers
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina
Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

                                 =7200\times\dfrac{2\pi}{60}

                                 =754\ rad/s

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

    using equation of angular motion

 \omega_f - \omega_i = \alpha t

 754 - 0 =190\times t

    t = 4 s

rotation of wheel in 4 s

\theta =\omega_i t+  \dfrac{1}{2}\alpha t

\theta = \dfrac{1}{2}\alpha t^2

\theta = \dfrac{1}{2}\times 190 \times 4^2

 θ = 1520 rad

 \theta = \dfrac{1520}{2\pi}

 \theta =242\ rev

now, revolution of the disk in next 6 s

angular velocity is constant

\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}

754 = \dfrac{\theta_f-1520}{10-4}

θ_f = 6044 rad

θ_f = \dfrac{6044}{2\pi}

revolution of the computer hard disk

θ_f =  962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0
3 years ago
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