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3 years ago

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A graph with resistance in ohms on the x-axis from 0 to 200 and current in amps on the y-axis from 0 to 5. The graph is labeled

as current as a function of resistance at 50 V. Points are plotted at (10, 5), (20, 2.5), (100, .5) and (200, .25). There is an inverse curve through the points. Use the graph to answer the questions. Based on the graph, what happens to current as resistance increases? This type of graph shows which type of relationship between the two variables? What would be a reasonable estimate for current at a resistance of 50 Ω?

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

1. decreases

2. inverse proportionality

3. 1 A

Explanation

Just did it

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A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

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2 years ago

2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin

ycow [4]

Answer:

$\frac{0.065}{r}$

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

$v=\sqrt{\mu gr}....................(1)$

where $\mu$ is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = $9.81m/s^2$

r = ?

$\mu=?$

In order to solve for $\mu$ , we can simply make it the subject of formula from equation (1) as follows;

$v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)$

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

$\mu=\frac{0.8^2}{9.81r}$

$\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}$

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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

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strojnjashka [21]

I don't understand the language.....

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