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mash [69]
2 years ago
8

A water balloon is hovering directly above the line join points ANB which are 4.6 km apart if the angles of elevation to the bal

loon from point a to B or 28.8° and 52.2 respectively find the altitude of the balloon
Physics
1 answer:
Anna35 [415]2 years ago
7 0

Answer:

Drawing the triangle:

H / x = tan 52.2 = 1.29

H / (4.6 - x) = tan 28.8 = .550

H = 1.29 x

H = .55 * 4.6 - .55 x

1.84 x = 2.53        combining equations

x = 1.38

4.6 - 1.38 = 3.22

Total base of triangle = 1.38 + 3.22 = 4.6

H / x = tan 52,2 = 1.29

H = 1.29 * 1.38 = 1.78 height of triangle

Check:

1.78 / 3.22 = tan 28.9    

This agrees with the given value of 28.8

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On a roller coaster, riders can experience a force of up to 4 g. What is the maximum acceleration of the roller coaster?
yKpoI14uk [10]

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7 0
3 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
velikii [3]

Answer:

In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .

7 0
3 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
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