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Bogdan [553]
3 years ago
7

An initially stationary 4.3 kg object accelerates horizontally and uniformly to a speed of 11 m/s in 3.4 s. (a) In that 3.4 s in

terval, how much work is done on the object by the force accelerating it
Physics
2 answers:
Nataliya [291]3 years ago
8 0

Answer:

The work done on the object by the force accelerating it is 520.31 J

Explanation:

Given;

mass of an object = 4.3 kg

horizontal velocity of the object, v = 11 m/s

time of acceleration, t = 3.4 s

work done is given as the product of force and distance

Work done = Fd

horizontal distance traveled by the object within 3.4 s, is calculated as follows;

X = Vt + ¹/₂gt², gravity has little or no influence on horizontal displacement, thus g = 0

X = Vt

X = 11*3.4 = 37.4 m

Force on the object, F = ma = m(v/t) = 4.3(11/3.4) = 13.912 N

work done = Fd = 13.912 x 37.4 = 520.31 J

Therefore, the work done on the object by the force accelerating it is 520.31 J

kompoz [17]3 years ago
3 0

Answer: W=260.174J

Explanation: since the object is stationary, it initial velocity U = 0

final velocity V = 11 m/s, time t = 3.4s, distance S = ?, acceleration a = ? work done W = ? force F = ?

W = FS

a = v-u/t = 11-0/3.4 = 3.2353m/s^2

to calculate the distance, let look at one of the equations of motion

V^2=U^2+2as hence s = V^2-U^2/2a = 11*11/2*3.235 = 18.7017m

But force F = MA (mass*acceleration)

F= 4.3*3.2353 = 13.91179N

therefore work done W = 13.91179*18.7017 = 260.174J

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Which is an example of something heated by conduction?
babunello [35]

Answer:

A waffle iron heated by coils

Explanation:

A waffle iron heated by coils - conduction

Food heated in a microwave oven - radiation

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5 0
4 years ago
Read 2 more answers
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
3 years ago
uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th
adoni [48]

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

7 0
3 years ago
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha
sineoko [7]

Answer:

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

T=2\pi \sqrt{\frac{I}{Mgd}} (1)

Where:

  • I is the moment of inertia
  • M is the mass of the pendulum
  • d is the distance from the center of mass to the pivot
  • g is the gravity

Let's solve the equation (1) for I

T=2\pi \sqrt{\frac{I}{Mgd}}

I=Mgd(\frac{T}{2\pi})^{2}

Before find I, we need to remember that

T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s

Now, the moment of inertia will be:

I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}  

Therefore, the moment of inertia is:

I=0.37 \: kgm^{2}

I hope it helps you!

7 0
3 years ago
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