The actuon of a lever is affected by the?

Compared to its weight on Earth, a 5kg object on the moon will weigh

shutvik [7]

Answer:

8.1 N/49 N=0.1653 which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 $\frac{m}{s^2}$ , the weight of the 5 kg object will be: $weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N$

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

$weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N$

Therefore the object's weight on the Moon compared to that on Earth will be:

$8.1N/49N=0.1653$

That is, 16.53% of the weight the object has on Earth.

5 0

3 years ago

Ok I have no clue for this one I’m not sure what to make out of this one please help

MatroZZZ [7]

Helium, Neon, and Xenon are all part of the same column on the Periodic Table. Such a column is referred to as a Group, because they have the same number of valence electrons in their outermost shell. Hope this helps!

5 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:

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