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4 years ago

A piece of metal has attained a velocity of 107.8 m/sec after fallinf for 10 seconds what is its initial velocity

Physics

1 answer:

soldi70 [24.7K]4 years ago

4 0

Answer:

7.8 m/s

Explanation:

Here object is falling with a gravitational acceleration there for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation

V = U + at

V - Final velocity

U - Initial velocity

a - acceleration

t - time

V=U+at

107.8=U + 10×10

= 7.8 m/s

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A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$

(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

Therefore, the frequency of $i^{th}$ normal mode is $f_i=\frac{v}{\lambda_i}$

6 0

3 years ago

Which can be observed both on Earth and in an accelerating ship in space that is free from the effect of any gravitational field

Reil [10]

Answer:

apples falling from trees
people's feet touching the ground
sky divers moving toward the ground
balls bending downward after being thrown

Explanation:

When a space ship is accelerating in space, there is a force known as Inertia that kicks in. Inertia will mirror the effects of gravity on the ship even if there is no gravitational field effect such that anything that would happen where there is gravity, would continue to happen.

This means that apples will fall from trees, people's feet will touch the ground, sky divers will be pulled downwards and balls will bend downwards when thrown as well. These are the same effects expected on earth where gravity pulls things towards the earth's core.

8 0

3 years ago

Read 2 more answers

Today's topic

babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

A body is said to be in non-uniform motion if it covers unequal distances in equal interval of time or vice-versa

4 0

2 years ago

Select all of the answers that apply.

Kruka [31]

The answer is letter A. meteorite bombardment.
During the Earth's earliest beginning, it went through a period of catastrophic and intense formation. By 3-8- 4.1 billion years ago, Earth's atmosphere was never the same as today. This is because of its formation during the pre-Cambrian period whereby t<span>he Earth formed under so much heat and pressure that it formed as a molten planet.
</span>

Earth was bombarded continuously by the remnants of the dust and debris — like asteroids, meteors and comets — until it formed into a solid sphere, pulled into orbit around the sun and began to cool down during the Hellish period (4.5 to 3.8 billion years ago).

<span> </span>.To know more of this topic, see attached file.

Download docx

6 0

3 years ago

Read 2 more answers

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

3 0

4 years ago