Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.
Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.
Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.
Answer:
If it had more or less mass, the atmosphere would be very different with either too much ammonia and methane or too little oxygen and water
Explanation:
An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:
DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation
according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero
solving for x we get
x = 0.200 m
b)electric field at half way mean x =0.25
E = 3.84*10^{-4} N/C
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
