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3 years ago

A scientist heats a flexible container full of neon gas. What will most likely happen to the container as the gas absorbs heat?

Physics

2 answers:

AnnZ [28]3 years ago

6 0

The container will burn up. it depends if it is glass or plastic

Volgvan3 years ago

3 0

It will most likely get hotter and hotter and eventually exsplod from all the presure.

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A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

Musya8 [376]

1) $0.61 m/s^2$

2) 13.9 s

Explanation:

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

$g=\frac{GM}{r^2}$ (1)

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

$r=4R$

where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

$g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2$

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

$g=0.61 m/s^2$ is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

$T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s$

4 0

3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago

What is power?

Xelga [282]

Answer:

b- how much work can be done in a given time

7 0

3 years ago

Read 2 more answers

A car which is traveling at a velocity of 1.6 m/s undergoes an acceleration of 9.2 m/s over a distance of 540 m. How fast is it

mixas84 [53]

Answer:

Vf = 99.7 m/s

Explanation:

In order to find the final velocity of the car, we will use the third equation of motion. The third equation of motion is given as follows:

$2as = V_{f}^{2} - V_{i}^{2}$

where,

a = acceleration of the car = 9.2 m/s²

s = distance traveled by the car = 540 m

Vf = Final Speed of the car = ?

Vi = Initial Speed of the car 1.6 m/s

$2(9.2\ m/s^{2})(540\ m) = (V_{f})^{2}-(1.6\ m/s)^{2}\\V_{f}^{2} = 9936\ m^{2}/s^{2} + 2.56\ m^{2}/s^{2}\\\\V_{f} = \sqrt{9938.56\ m^{2}/s^{2}}$

8 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago