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3 years ago

13

A skier, starting from rest, slides down an icy frictionless 80o incline whose vertical height is 105 m. How fast are they going

when the reach the bottom?

1 answer:

leonid [27]3 years ago

4 0

Answer:

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1. Which of the following explains why a bicycle uses the pedals as a lever to move the heavy rider along the sidewalk?

olga_2 [115]

<span>1. Which of the following explains why a bicycle uses the pedals as a lever to move the heavy rider along the sidewalk?
</span>
C. reduces the resistance force
The mechanical advantage of a compound machine is related to the mechanical advantage of all the simple machines involved.
True

5 0

3 years ago

Read 2 more answers

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

3 years ago

The property of water that contributes to its ability to stick to certain surfaces is called

Gekata [30.6K]

Answer: Adhesion

Explanation: Adhesion is the tendency of dissimilar particles or surfaces to cling to one another (cohesion refers to the tendency of similar or identical particles/surfaces to cling to one another). The forces that cause adhesion and cohesion can be divided into several types. This allows Particles in things like water to stick to surfaces

3 0

3 years ago

__________, the best forecasting methods to use are time series models such as moving average. ANSWER Unselected When the future

DochEvi [55]

Answer:

Greetings!

The correct answer is the last one, "When the future level of some variable is seen as a function other than time".

Explanation:

Along other forecasting methods, the moving average finds utility in cases when seasonality is a factor. This term refers to a set of variables unknown to (or uncontrolled by) the observer that influence the series model in some way.

By applying a moving average to a time series one can mitigate such irregular effects.

I hope this helps!

8 0

3 years ago

An 18 gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm,1.02 mm, carries a constant current o

larisa86 [58]

Answer:

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

$J = \frac {I}{A}$

<em>where I: electric current and A: cross-sectional area of the copper wire</em>

<u>The cross-sectional area of the copper wire can be calculated by:</u>

$A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2}$

<u>Substituting the calculated area in the equation (1) we have:</u>

$J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}}$

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

$v = \frac {J}{n q}$

<em>where n: the free-electron density, q: module of the charge of the electron </em>

$v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)}$

$v = 1.50 \cdot 10^{-04} \frac {m}{s}$

So, the drift speed is 1.50x10⁻⁴ m/s.

Have a nice day!

4 0

4 years ago

Read 2 more answers