What is the weight of a person with a massa of 80kg

The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub

Fofino [41]

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

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

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

$F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}$

F' = F

7 0

3 years ago

What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?

ss7ja [257]

a+b= ?
3i +3j + (3i -3j) = ?
3i + 3j + 3i -3j =?
= 6i + 0j

3 0

3 years ago

Read 2 more answers

When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav

GuDViN [60]

I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.

4 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

How do you determine true displacement value of an object that has moved?

WITCHER [35]

Diceplacement is the distance an object has traveled in a certain direction
for example, if you were to walk North for 20m, then east for 40m, the distance you have traveled is 60m however your displacement is the distance between your starting position and your end position;
sqrt(20^2+40^2) = 44.7m
and because displacement is a vector, there needs to be a direction;
sin(theta)=40/44.7
theta=63.4 degrees East of North

therefore the true displacement is 44.7m at 63.4 degrees East of North

3 0

3 years ago