Answer:
Explanation:
gauge pressure due to a liquid column of density d and height h is given by the following expression .
P = hdg
The pressure depends upon height of liquid column and not on the cross sectional area .
In first cylinder .
gauge pressure = .40 atm
hdg = .40 atm
cross sectional area of cylinder = π r²
The radius of second cylinder is twice of the first , cross sectional area will be 4 times .
The volume remains the same when the liquid is poured into second cylinder
volume = cross sectional area x height .
As cross sectional area of second cylinder is 4 times , height of liquid column in second cylinder = h / 4 .
gauge pressure in second cylinder = h / 4 x d x g = hdg / 4
.40 / 4 = .10 atm
gauge pressure in second cylinder = .10 atm.
Answer:
449.824 kg
Explanation:
F = Gm₁m₂/r²
Above equation can be used to find the gravitational force between two objects.
Here F is the Gravitational force. m₁ , m₂ are the masses of the objects and r is the distance between the objects. Let's take m₁ as the mass of the moon. Then m₂ is the mass of the satellite.
We are given that F = 324 N , m₁ = 7.3 * 10²²kg and r = 2.6 * 10 ⁶m.
324 = 6.67×10⁻¹¹ × m₁ × 7.3×10²² / (2.6 × 10⁶)²
m₁ = 449.824 kg
Answer:
<em>The divers pull in their limbs and curl up their bodies because</em> doing so decreases their moment of inertia and increases their angular velocity
Explanation:
<em>The conservation of angular momentum</em> states that in a rotational system, the initial angular momentum is equal to the final angular momentum if no torque acts on it.
Angular momentum is equal to the product of the moment of inertia about its axis and the angular velocity
Angular momentum = Iω
where I = moment of inertial = mass x
I = m
angular momentum = mω
since angular momentum is constant, one can see that decreasing the radius of rotation about the body by curling in the limb will cause the moment of inertia to decrease and the angular velocity to increase.
NB: mass of the body is constant.
Answer:
magnitude of the induced emf in the coil is 0.0153 V
Explanation:
Given data
no of turns = 20
area = 0.0015 m²
magnitude B1 = 4.91 T/s
magnitude B2 = 5.42 T/s
to find out
the magnitude of the induced emf in the coil
solution
we know here
emf = -n A d∅ /dt
so here n = 20 and
A = 0.0015
and d∅ = B2 - B1 = 5.42 - 4.91
d∅ = 0.51 T and dt at 1 sec
so put all value
emf = -n A d∅ /dt
emf = -20 (0.0015) 0.51 / 1
emf = - 0.0153
so magnitude of the induced emf in the coil is 0.0153 V
Complete Question:
Suppose , where A has the dimensions LT, B has dimensions L²T⁻¹, and C has dimensions LT². Then the exponents n and m have the values
Answer:
The value of n = ¹/₅
The value of m = ³/₅
Explanation:
Given dimensions;
A = LT
B = L²T⁻¹
C = LT²
The values of n and m are calculated as follows;