Ask question

Ask question

All categories

1 year ago

8

Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the s

eparation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4

1 answer:

Nimfa-mama [501]1 year ago

4 0

Answer:

D) F

Explanation:

Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :

$F=\frac{GMm}{r^{2} }$ ......(1)

Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.

Since, New mass of ball A = 2M

New mass of ball b = 2m

Distance between the two balls = 2r

Substitute these values in equation (1).

$F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }$

$F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }$

Using equation (1) in the above equation.

F₁ = F

You might be interested in

A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

$f=\frac{v}{\lambda}$

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

$\lambda=\frac{v}{f}$

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

3 0

2 years ago

Vader's light saber is red, while Obi-Wan's light saber is blue, meaning that Obi-Wan's light saber is emitting _____ compared t

Bingel [31]

I assume it woukd be higher energy light waves. when fire is at its hottest state its blue because its burning off so much.

4 0

2 years ago

Read 2 more answers

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

azamat

Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

6 0

2 years ago

Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc

Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 / - 52.5 - 0 = 1 / f

f = -52.5 cm

= -.525 m

Power P = 1 / f = - 1 / .525

= - 1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = - 52.5 cm

1 / v - 1 / u = 1 / f

1 / v + 1 / 17.2 = - 1 / 52.5

1 / v = - 1 / 17.2 - 1 / 52.5

= - .05813 - .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0

2 years ago

The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is

Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

W = ΔEm

Where the work is

W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

Em₀ = K = ½ m v²

Final. When the system is at rest

$Em_{f}$ = $K_{e1}b +K_{e2}$ = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

∑ F = F - fr

Axis y

N- W = 0

N = W

The friction force has the equation

fr = μ N

fr = μ W

The job

W = (F – μ W) x

We substitute in the equation

(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values and solve

½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

x² 30 + 14.4 x - 1,125 = 0

x² + 0.48 x - 0.0375 = 0

We solve the second degree equation

x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

x = [-0.48 ± 0.617] / 2

x₁ = 0.0685 m

x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0

2 years ago