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3 years ago

A person speed walking down a street covers 20m in 5s. What is the speed of the individual? *

Physics

1 answer:

omeli [17]3 years ago

8 0

Answer:

4 m/s.

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

Speed is defined as the rate of change of distance moved with time. Mathematically, it is expressed as:

Speed (S) = Distance (d) /time (t)

S = d/t

With the above formula, we can easily calculate the speed of the individual as follow:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

S = d/t

S = 20/5

S = 4 m/s

Therefore, the speed of the individual is 4 m/s

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3 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

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Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

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3 years ago

A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at

attashe74 [19]

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

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=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6

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u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

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