All categories

3 years ago

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?

1 answer:

5 0

$m_1=0,5kg \\ m_2=0,1kg \\ F=1,5*10^{-10}N \\ G=6,67*10^{-11}m^3/kg*s^2 \\ \\ r=? \\ \\ \\ F= \frac{Gm_1m_2}{r^2 } \\ \\ Fr^2=Gm_1m_2 \\ \\r^2=\frac{Gm_1m_2}{F} \\ \\ r= \sqrt{\frac{Gm_1m_2}{F} } = 
$

$=\sqrt{ \frac{6,67*10^{-11} \frac{m^3}{kg*s^2}*0,5kg*0,1kg }{1,5*10^{-10}N} } = \sqrt{ \frac{6,67*10^{-11} \frac{m^3}{kg*s^2}*0,05kg^2 }{1,5*10^{-10} \frac{kg*m}{s^2} } } = \\ \\ =\sqrt{ \frac{6,67 *10^{-11}*0,05 }{1,5*10^{-10}}m^2 } \approx\boxed{0,15m} \\ \\ \\ \\ \\ \\ "Non\ nobis,\ Domine,\ non\ nobis,\ sed\ Nomini\ tuo\ da\ gloriam." \\ \\ \\ \\ \underline{\underline{Regards\ M.Y.}}$

You might be interested in

Jerehuynho :D hhhghghghhg

Viktor [21]

Huh ?!?!?? What does that mean? Can I have brainliest I need one more for my next stage/level

7 0

3 years ago

Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object

vivado [14]

Answer:

C. The initial momentum should be equal to the final momentum due to the conservation of momentum.

$P_{initial} = mv_0\\P_{final} = (M+m)v_1\\v_1 = \frac{m}{M+m}v_0$

Since m/(M+m) < 1, v_1 > v_0.

Explanation:

Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.

Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.

Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.

4 0

3 years ago

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the d

frez [133]

The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

$\beta=10log(\frac{I}{I_{0}} )dB$ ,

where $I_{0}$ is a reference intensity. for sound waves, $I_{0}$ is taken to be $10^{-12} W/m^{2}$ . Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. $I=10I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. $I=100I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels ( $\Delta \beta_{2},\Delta \beta_{4}$ and $\Delta \beta_{8}$ ) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) $I=10I_{0}$

$\beta=10log(\frac{10I_{0}}{I_{0}} )$

$\beta=10log(10 )$

β = 10

The sound Intensity level with intensity 10x is 10dB.

B) $I=100I_{0}$

$\beta=10log(\frac{100I_{0}}{I_{0}} )$

$\beta=10log(100)$

β = 20

With intensity 100x, level is 20dB.

C) To calculate the change, take the f to be the factor of increase:

For $\Delta \beta_{2}$ :

$I=2I_{0}$

$\beta=10log(\frac{2I_{0}}{I_{0}} )$

$\beta=10log(2)$

β = 3

For $\Delta \beta_{4}$ :

$I=4I_{0}$

$\beta=10log(\frac{4I_{0}}{I_{0}} )$

$\beta=10log(4)$

β = 6

For $\Delta \beta_{8}$ :

$I=8I_{0}$

$\beta=10log(\frac{8I_{0}}{I_{0}} )$

β = 9

Change is

$\Delta \beta_{2},\Delta \beta_{4}$ , $\Delta \beta_{8}$ = 3,6,9 dB

6 0

3 years ago

a cement block accidentally falls from rest from the ledge of a 80.6-m-high building. When the block is 10.8 m above the ground,

fomenos

Answer:

0.229 seconds

Explanation:

Given:

y₀ = 80.6 m

v₀ = 0 m/s

a = -9.8 m/s²

We need to find the difference in times when y = 10.8 m and y = 2.10 m.

When y = 10.8 m:

y = y₀ + v₀ t + ½ at²

10.8 = 80.6 + (0) t + ½ (-9.8) t²

10.8 = 80.6 − 4.9 t²

4.9 t² = 69.8

t = 3.774

When y = 2.10 m:

y = y₀ + v₀ t + ½ at²

2.10 = 80.6 + (0) t + ½ (-9.8) t²

2.10 = 80.6 − 4.9 t²

4.9 t² = 78.5

t = 4.003

The difference is:

4.003 − 3.774 = 0.229

The man has 0.229 seconds to get out of the way.

5 0

4 years ago

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of fre- quency 110.0 MHz in opposite directi

Shkiper50 [21]

To solve the problem it is necessary to take into account the concepts related to frequency depending on the wavelength and the speed of light.

By definition we know that the frequency is equivalent to,

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

where,

c= Speed of light

$\lambda = Wavelength$

While the wavelength is equal to,

$\lambda = \frac{2L}{n}$

Where,

L = Length

n = Number of antinodes/nodes

PART A) For the first part we have that our wavelength is 110MHz, therefore

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

$\lambda = \frac{3*10^8}{11*10^6}$

$\lambda = 1.36m$

Therefore the distance between the nodal planes is 1.36m

PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,

$\lambda'=\frac{2L}{n}$

$L = \frac{\lambda n}{2}$

$L = \frac{8*2.72}{2}$

$L = 10.90m$

Therefore the length of the cavity is 10.90m

4 0

3 years ago