Ask question

Ask question

All categories

Norma-Jean [14]

3 years ago

15

If the distance between objects is the same for each pair, which of the following have the greatest gravitational force between

them?
A--a 65-kg object and a 65-kg object

B--a 65-kg object and a 75-kg object

C--a 60-kg object and a 70-kg object

D--a 50-kg object and an 80-kg object

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

B

Explanation:

You might be interested in

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$

Next, you calculate the wavelength of the sound waves by using the following formula:

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

vs: speed of sound = 343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

$\lambda=\frac{343m/s}{400Hz}=0.8575m$

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

$\Delta d=3.827m-2.10m=1.727m$

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

$\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2$

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0

3 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

8 0

3 years ago

¿Qué tiempo realiza un avión de la Cd. de México a Cancún, si la distancia entre ambos es de 1608.9Km y el avión viaja a una rap

Leya [2.2K]

I can help you with that if you translate to English

3 0

3 years ago

What is the difference between reflection and refraction? What changes and what does not change.

wolverine [178]

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

8 0

2 years ago