Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Answer:
Force
Explanation:
The Force You Put Behind Said Apple Is Greater Than That Of Gravitational Pull
Example : The More Power You Put behind a basketball the higher the basketball will go
C. is the answer because acceleration is the change in velocity in time while velocity is speed with a direction
Answer:
Explanation:
A ) angular velocity ω = 2π / T
= 2 x 3.14 / 60
= .10467 rad / s
linear velocity v = ω R
= .10467 x 50
= 5.23 m / s
centripetal force = m v² / R
= mg v² / gR
= 834 x 5.23² / 9.8 x 50
= 46.55 N
B )
apparent weight
= mg - centripetal force
= 834 - 46.55
= 787.45 N
C ) apparent weight
= mg + centripetal force
= 834 + 46.55
= 880.55 N.
D )
For apparent weight to be zero
centripetal force = mg
mg = mv² / R
v² = gR
= 9.8 x 50
= 490
v = 22.13 m /s
time period of revolution
= 2π R /v
2 x 3.14 x 50 / 22.13
= 14.19 s
Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
