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2 years ago

A force (5ỉ - ;)N moves an object from the point P (1,3) m to the point Q (3, 8) m.

Physics

1 answer:

natita [175]2 years ago

3 0

Answer:

b) 5 J

Explanation:

Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)

To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:

Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j

The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:

Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J

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245289457u2895785/+55891235723895718957=

Bingel [31]

why Is there just a random "u" in the middle of the equation awa

thequestion is impossible to do btw

Imao the u in the middle doesn't makes sense

if you edit the question I'll help you w it)

have a good day bro cya

6 0

2 years ago

Read 2 more answers

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

2 years ago

when an object falls to the ground, only the object moves down but the earth's motion is not noticeable.why?

vlada-n [284]

Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.

7 0

2 years ago

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

8 0

3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

2 years ago