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3 years ago

12

The law of universal gravitation offers a mathematical explanation for the attraction between the moon and Earth.

1 answer:

grandymaker [24]3 years ago

5 0

Answer:

True.

Explanation:

The given statement is true that the law of universal gravitation offers a mathematical explanation for the attraction between the moon and Earth.

According to this law, the forces between two masses can be calculated by the product of their masses and divided by the square of the distance between them. Mathematically, it can be written as :

$F=\dfrac{Gm_1m_2}{r^2}$

G is universal gravitational constant

Hence, the given statement is true.

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A man, facing a high wall, notices that an echo is heard 4.0 seconds after he makes a sharp sound. After walking 200m directly t

cestrela7 [59]

Answer:2.4

Explanation:basically 2.4 = 2.500 kg and then we add 3.4 seconds in all of them 4.0 + 3.4 = 2.5 ok

4 0

3 years ago

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise

Viefleur [7K]

Answer:

The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

$\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}$

$\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}$

$f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)$

$\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)$

$\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

$v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

Substitute the given values in the formula,

$v_{s}=343+\frac{262}{271}(343-0)$

$v_{s}=343+0.966(343)$

$v_{s}=343-331.33$

$v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Therefore, The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

4 0

3 years ago

a 7.5 kg bowling ball is in front of a large, compressed spring. The spring has a constant of 500 N/m, and 225N of force was req

Vikentia [17]

F=kx
X=F/k=225/500=.45 meters

6 0

3 years ago

in california the pacific plate slides past the north american plate. If the pcific plate is moving at a speed of 5 centimeters

frozen [14]

Answer:

6.34×10⁹ seconds

Explanation:

Applying,

V = d/t............. Equation 1

Where V = speed plate, d = distance covered by the plate, t = time it takes the plate to travel

make t the subject of the equation

t = d/V............ Equation 2

From the question,

Given: d = 5 cm/yr = (5×3.154×10⁻⁹) = 1.577×10⁻⁸ m/s, d = 100 meter

Substitute these values into equation 2

t = 100/1.577×10⁻⁸

t = 6.34×10⁹ seconds

Hence the time it takes the the plate to travel is 6.34×10⁹ seconds

3 0

3 years ago

What is the magnetic field inside a coil with the following conditions: 228 number of turns, 0.952 A of current and a length of

sdas [7]

Answer:

The magnetic field inside a coil is 0.0119 T.

Explanation:

Given that,

Number of turns = 228

Current = 0.952 A

Length = 2.29 cm

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu_{0}NI}{l}$

Where, B = magnetic field

l = length

I = current

N = number of turns

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times228\times0.952}{2.29\times10^{-2}}$

$B=0.0119\ T$

Hence, The magnetic field inside a coil is 0.0119 T.

8 0

4 years ago