Brittleness, malleability, ductility, and viscosity fall into the

A person walks 45 m East and then walks 39 m at an angle 40◦ North of East. What is the magnitude of the total displacement? Ans

vredina [299]

Answer:

r = 78.95 m

Explanation:

given,

Person Walk in east, r₁ = 45 m

Then he walk, r₂= 39 m in the direction of 40◦ North of East.

writing the total displacement in x- direction

r₁ₓ = 45 m

r₂ₓ = 39 cos 40°

r₂ₓ = 29.87 m

total displacement in x-direction

rₓ = r₁ₓ + r₂ₓ

rₓ = 45 + 29.87

rₓ = 74.87 m

displacement in y-direction

r_{1y} = 0 m

r_{2y} = 39 sin 40°

r_{2y} = 25.07 m

total displacement in y- direction

r_y = 25.07 m

using Pythagoras theorem for the magnitude calculation

$r = \sqrt{r_x^2 + r_y^2}$

$r = \sqrt{74.87^2 +25.07^2}$

r = 78.95 m

The magnitude of total displacement is equal to 78.95 m.

8 0

3 years ago

A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, an

Leokris [45]

Answer:

(a) 0.833 j

(b) 2.497 j

(c) 4.1625 j

(d) 4.995 watt

Explanation:

We have given force F = 5 N

Mass of the body m = 15 kg

So acceleration $a=\frac{F}{m}=\frac{5}{15}=0.333m/sec^2$

As the body starts from rest so initial velocity u = 0 m/sec

(a) From second equation of motion $s=ut+\frac{1}{2}at^2$

For t = 1 sec

$s=0\times 1+\frac{1}{2}\times 0.333\times 1^2=0.1666m$

We know that work done W =force × distance = 5×0.1666 =0.833 j

(b) For t = 2 sec

$s=0\times 2+\frac{1}{2}\times 0.333\times 2^2=0.666m$

We know that work done W =force × distance = 5×0.666 =3.33 j

So work done in second second = 3.33-0.833 = 2.497 j

(c) For t = 3 sec

$s=0\times 3+\frac{1}{2}\times 0.333\times 3^2=1.4985m$

We know that work done W =force × distance = 5×1.4985 =7.4925 j

So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j

(d) Velocity at the end of third second v = u+at

So v = 0+0.333×3 = 0.999 m /sec

We know that power P = force × velocity

So power = 5× 0.999 = 4.995 watt

4 0

4 years ago

a stationary police officer directs radio waves emitted by a radar gun at a vehicle toward the officer. Compared to the emitted

Arisa [49]

Answer:

D) Higher Frequency

Explanation:

Higher frequency because it says a vehicle “towards” the officer

3 0

3 years ago

What type of relationship does this graph show?

Schach [20]

Negative because the graph goes down

8 0

4 years ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

Natasha_Volkova [10]

Answer

given,

distance of first satellite = 48,000 Km

distance of second satellite = 64,000 Km

orbital period = 6.39 day

Using equation of time period

$T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_{pluto}}}$

now, from the above equation we can say that only variable is Time period and r is the radii of orbit.

from the first satellite

$\dfrac{T_{charon}}{r^{3/2}_{charon}}=\dfrac{T_{sat1}}{r^{3/2}_{sat1}}$

$T_{sat1}=\dfrac{T_{charon}\ r^{3/2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (48000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=24.5\ days$

for second satellite

$T_{sat2}=\dfrac{T_{charon}\ r^{3/2}_{sat2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (64000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=37.7\ days$

7 0

3 years ago