Answer:
453 gm
Explanation:
<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
Answer:
a. 2.1 s
b.0.48 Hz
c. A=24cm
d. 72cm/s
Explanation:
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?
What are the period,
period is the time taken for a wave particle to make one complete oscillation
a) 31 / 15 = 2.066 seconds
= 2.1 s
(b) frequency
: this the number of oscillation made in one seconds.
it is also the inverse of the period.
= oscillations / time
= 15/31= 0.48 Hz
(c) amplitude
: maximum displacement from the origin
amplitude = 1/2 of the difference of oscillation marks
= 1/2(57-10) = 47/2cm
23.5cm
A=24cm
(d) maximum speed of the glider?
V=ωA
angular frequency *Amplitude
V=a*pi*f*amplitude
2π x frequency x amplitude = maximum speed
= 2π x .48 x 24
=72.38 cm/s
72cm/s
Answer:
upthrust or BUOYANT FORCE =Vdg
volume=LWH
upthrust=(4cm×5cm×2cm)×1g/cm²×g
upthrust=40cm³×1g/cm³×g
upthrust=40gf or 0.04kg×10m/s²=0.4N
weight of the displaced liquid is upthrust.
so mass=40g or 0.04kg
upthrust=40gf or 0.4Nand mass of the displaced liquid=40g or 0.04kg
please mark brainliest, hope it helped
Answer:
B
Explanation:
Acceleration = rate of change in velocity
a = (21-12)/(0.45) = 20cm/s^2
Answer:
the net force would be 3N in the upward direction since the two forces acting on the left and right of the object cancel out.
Explanation:
