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3 years ago

What happens to the direction of an object as an unbalanced force acts on it? Give an example

Physics

2 answers:

xz_007 [3.2K]3 years ago

8 0

Hello :)

Answer:

When an unbalanced force acts on an object, the side with the greater force's direction makes the object move along its direction. For example, when two cars collide on a head-on collision, both vehicles will move in the direction that the vehicle with the greater force is moving in.

Pie3 years ago

5 0

Answer:

When an unbalanced force acts on a body the side with the greater force's dircetion makes the object move along its direction

Also to find the net force acting on the bofldy you can subtract the two force acting on the body

In case of balanced force the net force will always be 0

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Hyaline, elastic, & fibrocartilage!

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3 years ago

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A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

Solution:

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

Find the work done:

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

Find the time taken

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

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3 years ago

A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m

BaLLatris [955]

Answer: a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass

 m = son's mass = M/3

 V = father's initial speed

 v = son's initial speed  (1/2)MV^2 = (1/2)*(1/2)*m v^2

 M*V^2 = (1/2)(M/3)v^2

 V^2/v^2 = 1/4

 V = v/2  Second equation:

 (1/2)M*(V + 1.4)^2 = (1/2)m*v^2

 = (1/2)*(M/3)*(3V)^2

 cancel out the M's and (1/2)'s

 (V + 1.4)^2 = 3V^2

 V^2 + 2.8V + 1.96 = 3V^2

 V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

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3 years ago

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A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the cu

Effectus [21]

Answer:

Compared with the current in the first coil, the current in the second coil is unchanged.

Explanation:

All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit.

The steady state of current in the LR circuit is:

I= V/R (1 - e^-Rt/L)

Where I= current

R= Resistance

V= Voltage

Where R/L is the time constant.

For a conducting wire, it has a very small resistance. The time constant will be in microseconds. The current will be in a steady state after few second. The current is independent on the inductance and dependent on the resistance. The length of wire and the resistance here are the same. Therefore, the current remains unchanged.

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3 years ago

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Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is

raketka [301]

R = rate of flow = 0.370 L/s H = height = 2.9 m T= time = 3.9 s V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2 G value = 9.8 m/s2 Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N Wa = weight of accumulated water after 3.9 s Fi = force of impact of water on the bucket S = reading on the scale = Wa + Wb + Fi mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg Therefore, Wa = 1.443 x 9.8 = 14.1414 N Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt) = 0.37 x 7.539 = 2.78943 N S = 14.1414 + 6.762 + 2.78943 = 23.692 N

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3 years ago