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Oliga [24]
3 years ago
14

A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is

decreasing at a rate of 3.1m/s2.
What is the student's mass?
What is the net force on the student?
What value does the scale read?
Physics
1 answer:
jasenka [17]3 years ago
3 0

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

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Bumek [7]

The correct answer to the question is 130.4 N.

CALCULATION;

The mass of the bullet is given as m = 28 gram = 0. 028 kg.

The initial velocity of the bullet u = 55 m/s

The final velocity of the bullet  v = 18 m/s.

The distance covered by the bullet through the sand bag s = 29 cm.

                                                                                                   = 0.29 m

Let the acceleration of the bullet is a .

From equation of kinematics, we know that-

                                         v^2-u^2=\ 2as

                                        ⇒ a=\ \frac{v^2-u^2}{2s}

                                               =\ \frac{(18)^2-(55)^2}{2\times 0.29}\ m/s^2

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The negative sign is used due to the fact that force is opposing in nature. Its velocity is decreasing with time.

From Newton's second law of motion, we know that net force on a body is equal to the product of mass with acceleration.

Mathematically F = ma.

Hence, the frictional force exerted on the bullet is calculated as -

                                              F = m × a

                                                 = 0.028 × (-4656.897) N

                                                 = -130.4 N                     [ANS]

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Answer:

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Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

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Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

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The electric field 4cm from the wire is

É = 10nc { }^{ - 1}

Given:

charge is distributed uniformly along a long straight wire.

electric field 2cm from the wire is 20n/c

To find:

electric field 4cm from the wire

what is electric field?

Electric field is the region around charge particle or charged body in which if another charge is placed it experiences electrostatic force.An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

E   ∝ \frac{1}{r}

\frac{E}{É}  =  \frac{r2}{r1}

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É =  \frac{40}{4}

É = 10nc {}^{ - 1}

thus the electric field 4cm from the wire is

É = 10nc { }^{ - 1}

learn more about electrostatic force from here: brainly.com/question/28166144

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Answer:

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Now, work done by a force 'F' causing a displacement 'S' such that the angle between the force and displacement is \theta is given as:

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Work=(100\ N)(2\ m)(\cos(60))\\\\Work=(200\times 0.5)\ \textrm{N-m}\\\\Work=100\ J

Therefore, the work done by the force is 100 J.

8 0
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