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3 years ago

The strength of the gravitational pull between two object's is determined by?

Physics

2 answers:

aniked [119]3 years ago

6 0

Answer:

The distance and the mass

Explanation:

According to Newton's law, force can be determined by F = G * M * m / R^2

iogann1982 [59]3 years ago

4 0

Mass and distance are the two factors

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A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. an echo is heard 2.80 s later. how far away is the ref

Nostrana [21]

The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
$S=2L$
The speed of the sound wave is: $v=343 m/s$ , and since it is moving by uniform motion, we can find the distance covered by the wave using
$S=vt=(343 m/s)(2.80 s)=960 m$
And we said this corresponds to twice the distance between us and the reflecting object, so:
$L= \frac{S}{2} = \frac{960 m}{2} =480 m$
so, the object is 480 meters away.

3 0

3 years ago

A cube of water 10 cm on a side is placed in a microwave beam having Ea = 11 kV/m‘ The microwaves illuminate one faceofthe cube,

Shtirlitz [24]

Answer:

The time is 133.5 sec.

Explanation:

Given that,

One side of cube = 10 cm

Intensity of electric field = 11 kV/m

Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.

We need to calculate the rate of energy transfer from the beam to the cube

Using formula of rate of energy

$P=(0.80)IA$

$P=0.80\times\dfrac{c\mu_{0}E^2}{2}\times A$

Put the value into the formula

$P=0.80\times\dfrac{3\times10^{8}\times8.85\times10^{-12}\times(1.1\times10^{4})^2}{2}\times(10\times10^{-2})^2$

$P=1285.02\ W$

We need to calculate the amount of heat

Using formula of heat

$E =mc\Delta T$

$E =\rho Vc\Delta T$

Put the value into the formula

$E=1000\times(0.10)^3\times4186\times41$

$E=171626\ J$

We need to calculate the time

Using formula of time

$t=\dfrac{E}{P}$

Put the value into the formula

$t=\dfrac{171626}{1285.02}$

$t=133.5\ sec$

Hence, The time is 133.5 sec.

3 0

3 years ago

Learning Task 2: Write the words that can be associated with the ''Music During Classical Era''. You may ask help from the membe

Natasha_Volkova [10]

Answer:

classical

concert

exposition

orchesta

devolopment

recapitulation

sympathy

sonata

sinfonia

soloist

Explanation:

5 0

3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0

1 year ago

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

DerKrebs [107]

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

5 0

3 years ago