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3 years ago

14

An object was dropped from the plane with 1000 km above the ground with a mass of 300 kg. Find the acceleration of the object ea

ch second until it reaches the ground.

1 answer:

Oxana [17]3 years ago

8 0

Answer:

$9.8m/s^2$

Explanation:

The acceleration due to gravity on earth is $9.8m/s^2$ regardless of the mass of the object. This means, in the absence of air resistance, both a 3000 kg object and a 300 kg will accelerate at the same rate of $9.8m/s^2$ towards earth.

The reason the acceleration due to gravity is the same for all masses is that by grace of nature, the inertial mass $m$ of the object from $F= ma$ equals the gravitational mass $m$ from $F =G \dfrac{mM}{R^2}$ , resulting

$ma=G\dfrac{mM}{R^2}$

$\boxed{a=G\dfrac{M}{R^2}}$

which is the acceleration that only depends on the mass of earth and its radius, and not on the mass of the object.

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3 years ago

Which is an example of a covalent bond?

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3 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

or

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

or

$f=\frac{33.84}{0.935}$

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2 years ago

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

vivado [14]

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

$a_{y}$ = acceleration along vertical direction = - 9.8 m/s²

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

$y = v_{oy}t + (0.5)a_{y}t^{2}$

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

$a_{x}$ = acceleration along vertical direction = 0 m/s²

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

$x = v_{ox}t + (0.5)a_{x}t^{2}$

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m

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2 years ago

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2 years ago