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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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A friend insists that electric current is the same as electrical energy. Work with a partner to develop a rebuttal
kotykmax [81]

Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”

Explanation: Hope it Helps have a blessed day

4 0
3 years ago
When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do
sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, \theta=0

And so, the work done is

W=(75)(5)(cos 0)=375 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

6 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
3 years ago
Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force
xeze [42]

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
7 0
3 years ago
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