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3 years ago

Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side.

1 answer:

8 0

Although a zig zag pattern going up a mountain means you walk further, the incline of the slope is a lot less so you don't have to work as hard.

You might be interested in

Is it possible that a speed of 254 and a speed of 100 could be the same speed?

EastWind [94]

Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.

4 0

3 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant fig

Igoryamba

Answer:

$A=6.36\times 10^5\ m^2$

Explanation:

It is given that,

Radius of the circle, $r=4.5\times 10^4\ cm=4.5\times 10^2\ m$

The area of the circle is given by :

$A=\pi r^2$

$A=\pi (4.5\times 10^2\ m)^2$

$A=636172.51\ m^2$

$A=6.36\times 10^5\ m^2$

As there is no uncertainty given in the radius of the circle. So, the area of the circle is $6.36\times 10^5\ m^2$ . Hence, this is the required solution.

5 0

3 years ago

When an object stops moving, changes speed, or changes direction, how does a scientists describe that condition?

soldi70 [24.7K]

The answer is A. unbalanced forces

5 0

3 years ago

average speed = distance covered divided by travel time. do some algebra and multiply both sides of this relation by travel time

mars1129 [50]

Based on the scenario,
The thing that the result say about distance covered is :
it tells you that the distance covered is equal to the average speed times the travel time

hope this helps

6 0

3 years ago