Answer:
D is not the a vector quantities
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.
Answer:
If one of the parents is white and the other is brown, their offspring will be either white or brown with equal probabilities. Rabbits in this population mate randomly; thus, the probability of mating two white rabbits is the same as the probability of mating between two brown rabbits.
Explanation:
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
The formula is P = E/t, where P means power in watts, E means energy j , and t means time in seconds. This formula states that power is the consumption of energy per unit of time.
P = 15 M / 10*60
M = mega = 10⁶
15 *10⁶ / 600
= 25000 watt
