Answer:
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
Explanation:
Knowing that the density of lead is 
and the volume, we can calculate the true weight of the piece of lead:
When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.
When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.
Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.
Now that we have the two measurements, we can calculate the difference:
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s
Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m
Answer:
a. 3/4λ
d. 1/4λ
Explanation:
When the wavelength of the sound waves is λ and the two waves are having same frequency the waves are said to be out of phase if their phase difference is in the multiples of 
or 180°.
When the two waves are out of phase then their opposite maxima coincide at the same time resulting in the minimum amplitude of the resulting wave throughout.
- As we observe from the schematic that the a wave has sinusoidal pattern of variation and we get a maxima after each
of the distance.
- Here we have two speakers out of phase therefore on shifting one of the speakers by the odd multiples of we have the maxima or the extreme amplitudes.
So, we must place the microphone at 3/4λ and 1/4λ to pickup the loudest sound.
Answer:
a) Focal length of the lens is 8 cm which is a convex lens
b) 6 cm
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Explanation:
u = Object distance = 4 cm
v = Image distance = -8 cm
f = Focal length
Lens Equation
a) Focal length of the lens is 8 cm which is a convex lens
Magnification
b) Height of image is 2×3 = 6 cm
Since magnification is positive the image upright
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
