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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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HELP ASAP TIMED TEST

balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

$U_1 = mgH$

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

$\displaystyle U_2=\frac{1}{2}mv^2$

Since the energy is conserved, U1=U2

$\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]$

For the speed to be double, we need to drop the snowball from a height H', and:

$\displaystyle mgH'=\frac{1}{2}m(2v)^2$

Operating:

$\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]$

Dividing [2] by [1]

$\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}$

Simplifying:

$\displaystyle \frac{H'}{H}=4$

Thus:

H' = 4H

Correct choice: b 4H

4 0

3 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago

a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis

Len [333]

Answer:

solution,
Given
load =400N
ld=0.2m
ed=0.6m
effort =150N

Explanation:

efficiency =output work/input work ×100%

l×ld/e×ed×100%

400×0.2/150×0.6×100%

80/90×100%

88.89%ans

7 0

3 years ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

Zinaida [17]

1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s

2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

5 0

3 years ago

Read 2 more answers

Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a

alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x $10^{6}$ J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0

3 years ago