Match the following

You are loading a toy dart gun, which has two settings, the more powerful with the spring compressed twice as far as the lower s

Amiraneli [1.4K]

Answer:

option A

Explanation:

given,

compression for lower setting = x

work done to compress in lower setting = 5 J

compression in higher setting, x' = 2 x

work done in higher setting = ?

Work done in compression of spring at lower setting

$W = \dfrac{1}{2}kx^2$

$5 = \dfrac{1}{2}kx^2$ ............(1)

Work done in compression of spring at higher setting

$W' = \dfrac{1}{2}kx'^2$

$W'= \dfrac{1}{2}k(2x)^2$

$W'= 4\times \dfrac{1}{2}kx^2$

$W'= 4\times W$

from equation (1)

$W'= 4\times 5$

W' = 20 J

Work take for the higher setting is equal to 20 J

Hence, the correct answer is option A

4 0

3 years ago

What phases is shown in each of the diagrams above?

taurus [48]

Answer:
A. Interphase
B. Cytokinesis
C. Metaphase
D. Anaphase

Correct me if I was wrong :)

7 0

4 years ago

Which situation describes a non-contact force changing the speed of an object?(1 point)

SashulF [63]

A ball accelerating as it rolls down a hill

5 0

4 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl

spin [16.1K]

Answer:

L=0.654 m

Explanation:

Concepts and Principles

1- The speed of sound in air is expressed as a function of the temperature of air as follows:

v=(331 m/s)√(1+T_C/273°C) (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.

Standing Wave Patterns in Pipes:

A pipe open at both ends can have standing wave patterns with resonant frequencies:

f=v/λ=nv/2L n=1,2,3.........

where v is the speed of sound in air.

Given Data

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C

The flute is open at both ends.

Required Data

We are asked to determine the length of the tube.

Solution

The speed of sound in air at temperature T = 20°C is found from Equation (1):

v=(331 m/s)√(1+T_C/273°C)

=342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):

f=v/2L

Solve for L:

L=v/2f_1

L=0.654 m

7 0

3 years ago