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3 years ago

What is an inertial frame of reference? does a perfect inertial frame of reference exist, and if it does, give me an example?

Physics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

Within the realm of Newtonian mechanics, an internal frame of reference or internal reference frame, is one in which newton's law of motions is valid

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Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

What is a cheeseburger without cheese

NISA [10]

Answer:

It is a Hamburger

3 0

3 years ago

Read 2 more answers

An Astronaut on the Moon drops a rock straight downward from a height of 1.25 meters. If the acceleration of gravity on the Moon

beks73 [17]

Answer:

Since the astronaut drops the rock, the initial velocity of the rock is 0 m/s

<u>We are given:</u>

initial velocity (u) = 0 m/s

final velocity (v) = v m/s

acceleration (a) = 1.62 m/s/s

height (h) = 1.25 m

<u>Solving for v:</u>

From the third equation of motion:

v²-u² = 2ah

replacing the variables

v² - (0)² =2 (1.62)(1.25)

v² = 1.62 * 2.5

v² = 4 (approx)

v = √4

v = 2 m/s

The speed of the rock just before it lands is 2 m/s

8 0

4 years ago

Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

4 years ago

What is the speed of a wave with a wavelength of 2 m and a frequency of 9 Hz?

ale4655 [162]

Answer:

The speed of a wave would be 18 with a wavelength of 2 m and a frequency of 9 Hz.

7 0

3 years ago