Ask question

Ask question

All categories

vampirchik [111]

3 years ago

8

A computer mouse (.234 kg) and a full coffee mug (.647 g) are sitting 42.0 cm apart on a desk. Find the gravitational force betw

een these objects.

1 answer:

shutvik [7]3 years ago

7 0

Answer:

euphoria

Explanation:

because euphoria

You might be interested in

A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a

soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g= $\frac{46}{1000} kg=0.046 kg$

$1kg=1000 g$

Constant force=F=25.6 N/m

Initial displacement= $A_1=0.296 m$

Final displacement= $A_2=0.12 m$

Time=t=4.55 s

Damping force= $F_x=-bv_x$

We have to find the magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)$

For maximum displacement $cos(w't+\phi)=1$

Therefore , $x=A_2$

Substitute the values

$A_2=A_1e^{-\frac{-b}{2m}t}$

$e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}$

$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

$\frac{2\times 0.046}{4.55}=0.9b$

$b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s$

Hence,the magnitude of damping constant b=0.022kg/s

3 0

3 years ago

Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Tr

BARSIC [14]

Answer:

39.40 MeV

Explanation:

<u>Determine the minimum possible Kinetic energy </u>

width of region = 5 fm

From Heisenberg's uncertainty relation below

ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV

when we apply this values using the relativistic energy-momentum relation

E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,

Also in a nuclear confinement ( E, P >> mc )

while The large value will portray a Non-relativistic limit as calculated below

K = h^2 / 2ma^2 = 1.52 GeV

7 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

beks73 [17]

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

$q_1' = 2q_1\\q_2' = 2q_2$

While the distance does not change, so the new force will be

$F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F$

so, the force will quadruple.

4 0

3 years ago

Please help thank you

Valentin [98]

Diagram A is the correct diagram

5 0

3 years ago

A lemming running 2.22m/s runs off a horizontal cliff. It lands in the water 7.45m from the base of the cliff. How much time was

Gelneren [198K]

3,56.........................

6 0

3 years ago