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3 years ago

An acrobat on skis starts from rest 50.0 m above the

Physics

1 answer:

Brums [2.3K]3 years ago

4 0

The initial speed of the skier is 14 m/s.

The maximum height attained by the skier is 15 m.

<h3>Initial speed of the skier</h3>

The initial speed of the skier is determined by using the following kinematic equation.

$K.E = mgh\\\\\frac{1}{2} mv^2 = mgh\\\\\frac{1}{2} v^2 = gh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 10} \\\\v = 14 \ m/s$

<h3>Maximum height attained</h3>

The maximum height attained by the skier is calculated as follows;

$H = \frac{v^2 sin^2\theta}{2g} \\\\H = \frac{14^2 \times (sin45)^2}{2 \times 9.8} \\\\H = 5 \ m$

Thus, the maximum height attained by the skier is (10 + 5) = 15 m.

Learn more about maximum height of projectile here: brainly.com/question/12870645

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At what point will the electric field of a charged object be strongest?

puteri [66]

Answer:

The answer is C because they have to be close to be able to interact

7 0

2 years ago

A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$ , we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒ $n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

7 0

3 years ago

Read 2 more answers

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$