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ANEK [815]
4 years ago
9

A woman steps in front of a child to keep him from running off. which term best describes this example? negative work positive w

ork negative force positive force
Physics
2 answers:
vladimir1956 [14]4 years ago
5 0

Answer:

Negative force.

Explanation:

Here the woman is applying a force to stop the motion of the child so the nature of force is negative. after applying the force child stops without any displacement means displacement is zero.

Since we know that

    WORK= FORCE * DISPLACEMENT

    WORK= FORCE* 0 (ZERO)

    WORK= ZERO.

means no work is done .only negative force would be the answer.

lana [24]4 years ago
4 0

Answer:

negative force

Explanation:

Here woman is applying the force to stop the child or to prevent him from running.

Here we know that the force applied by the woman is opposite to the motion of child but here the child is not displaced by this applied force but the child is only prevented his running.

So here the work done by the woman is zero as there is no displacement by this force.

But here we can say that the the force of woman is against the motion of child so here this force is opposite force and hence it is termed as negative force here.

so correct answer will be

negative force

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AlekseyPX

Answer:

Oceanic crust moves easly. Continental crust does not move

8 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
A point charge that is exactly q =3 mu or micro CC is at the origin. In this problem, assume that the Coulomb constant k = 8.99
Simora [160]

Answer:

(a) V1 = 8990.00 V

V2 = 8960.13 V

Explanation:

Parameters given:

q =3 mC

k = 8.99 * 10⁹ Nm²/C²

x1 = 3 m

x2 = 3.01 m

Electric potential is given as:

V = kq/r

Where

k = Coulombs constant

q = charge

r = distance

Potential at x1 is:

V1 = (8.99 * 10⁹ * 0.000003)/(3)

V1 = 8990.00V

Potential at x2 is:

V2 = (8.99 * 10⁹ * 0.003)/(3.01)

V2 = 8960.13 V

7 0
3 years ago
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DaniilM [7]
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3 years ago
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valentinak56 [21]
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C. Protostar
During this stage, the young Star is still gathering its mass from the residual of its parent  molecular cloud

hope this helps
8 0
4 years ago
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