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3 years ago

A lady walks 10 m to the north, then she turns and continues walking 30 m due east.

Physics

1 answer:

ANEK [815]3 years ago

4 0

Answer:

The distance covered is 40 m and the displacement is 31,6m.

Explanation:

The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .

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Write two examples of machine which changes the direction of force

Ostrovityanka [42]

Answer:

Washign machine and swingset

Explanation:

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3 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago

HELPPPPP it’s due in 10 minutes

iris [78.8K]

Answer:

A.c

Explanation:

The chromosphere is above the photosphere, the visible "surface" of the Sun. It lies below the solar corona, the Sun's upper atmosphere, which extends many thousands of kilometers above the chromosphere into space. The plasma (electrically charged gas) in the chromosphere has a very low density.

In basic terms it is the 2nd one out from the core.

7 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

In what year did the Organization of Petroleum Exporting Countries (OPEC) refuse to sell oil to the U.S. due to political differ

Sedaia [141]

A) 1973 I believe it had something to do with the Arab wars happening at the time

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3 years ago