Answer:
4 times the mass of Earth
Explanation:
= Mass of Earth
= Mass of the other planet
r = Radius of Earth
2r = Radius of the other planet
m = Mass of object
The force of gravity on an object on Earth is
The force of gravity on an object on the other planet is
As the forces are equal
So, the other planet would have 4 times the mass of Earth
<h3>
Answer: A) 3 pronged cord</h3>
Explanation:
One prong is known as the hot wire, where the electricity comes in, while the second prong is the neutral wire and electricity leaves. Recall that a circuit must be made for electrons to flow. This means that electrons enters your house through the transmission wires and then the electrons leave along the same lines they entered (just along a different wire).
This set up describes a 2 pronged cord. Adding a third prong allows for a grounding to happen. In the event of an electrical surge, when too many electrons are flowing, disaster is likely to happen without some kind of safety features. The grounding prong allows electrons to seek the ground. You can think of it like a drain in a bathtub when too much water gets in the tub.
10.3
Explanation:
Step 1:
The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g
Step 2:
It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.
Step 3:
Since the pressure pushing up mercury and water is the same
* 
* g = 
* 
* g
= 
= (13.6 g/cm * 76 cm)/1 g/cm = 1033.6 cm
Step 4:
Now we need to express the answer in meters.
1 m = 100 cm.
1033.6 cm = 10.336 m
This can be rounded off to 10.3 m
Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
