Ask question

Ask question

All categories

mr Goodwill [35]

3 years ago

7

Two objects of equal mass are a fixed distance apart. If the mass of each object could be tripled, the gravitational force betwe

en the objects would....

1 answer:

antiseptic1488 [7]3 years ago

5 0

gravitational force between two objects is given as

F = G m₁ m₂/r²

where m₁ = mass of first object , m₂ = mass of second object , r = distance between the two objects .

Initial case :

m₁ = m₂ = m

gravitational force between the objects is given as

F = G m²/r²

Final Case :

m₁ = m₂ = 3 m

new gravitational force between the objects is given as

F' = G (3m)²/r²

F' = 9 G m²/r²

F' = 9 F

hence the gravitational force between the two objects becomes 9 times.

You might be interested in

I need the answer to number #23 please will give BRAINLIEST

Murrr4er [49]

Yeah lowkey i think it’s B

3 0

3 years ago

A man of mass 85 kg runs up a flight of stairs of height 4.6 m in a time period

seraphim [82]

Explanation:

a) Power = work / time = force × distance / time

P = Fd/t

P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)

P ≈ 319 W

b) P = Fd/t

0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)

m = 47.6 kg

7 0

3 years ago

What is imperceptible human motion?

castortr0y [4]

Unable to be noticed or not felt because of feeling slight

3 0

3 years ago

Because the earth's orbit is slightly elliptical, the earth actually gets closer to the sun during part of the year. When the ea

BabaBlast [244]

Closer to the sun . . . orbital speed is faster.

Farther from the sun . . . orbital speed is slower.

Flag answer: Answer 13 Answer 13

5 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago