Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
The net force is 270 N
Explanation:
We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:
where
F is the force
m is the mass
a is the acceleration
In this problem, we have
m = 90.0 kg
Substituting, we find the net force on the object:
Learn more about Newton's second law:
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Answer:
Explanation:
Current, I = 6 A
diameter of wire, d = 2.05 mm
number of electrons per unit volume, n = 8.5 x 10^28
If the diameter is doubled,
The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
