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3 years ago

Newton's law of universal gravitation works well in ordinary situations on earth, but it does not work well

Physics

2 answers:

IRINA_888 [86]3 years ago

6 0

Answer:

For large masses,It does not work well

Explanation:

Newton's law of universal gravitation :

In the universe every particle attract the other particle by a force and this force is directly proportional to product of their masses and inversely proportional to distance between these two masses.This law given as follows

$F=G\dfrac{m_1m_2}{r^2}$

Where G is the constant.

Th one limitation of this law that the masses should not be too large.So this law does not work well for large bodies.

USPshnik [31]3 years ago

3 0

In large stars because they are too massive

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.

Solving for the velocity $v$ , we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} }$

and we put in the numerical values

$m_a = 0.045kg$ ,

$m_b = 1.40kg,$

$\Delta H = 0.4m,$

$g= 9.8m/s^2$

and simplify to get:

$\boxed{ v= 15.9m/s}$

The arrow was moving at 15.9 m/s

6 0

3 years ago

An 81.5-kg man stands on a horizontal surface.

OLga [1]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

$P=\dfrac{F}{A}$

$P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2$

P=181.11 x 10² N/m²

7 0

3 years ago

What units are used to measure mass and weight?

horrorfan [7]

Mass- Mass is measured in kilograms (kg).

Weight- Weight is measured in Newton’s.

4 0

2 years ago

Read 2 more answers

A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

Mice21 [21]

Answer:

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

(9 - 25) = 22t + 0.5(-9.8)t²

- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

t = 5.055 s = 5.06 s

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals

slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t = 7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0

2 years ago