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Newton's law of gravity and Coulomb's law are both inverse-
square laws. Consequently, there should be a "Gauss's law for
gravity."
a. The electric field was defined as
to find the electric field of a point charge. Using analogous
reasoning, what is the gravitational field
Write your answer using the unit vector
signs; the gravitational force between two "like masses" is
attractive, not repulsive.
b. What is Gauss's law for gravity, the gravitational equivalent
of Equation 24.18? Use
gravitational field, and Min for the enclosed mass.
c. A spherical planet is discovered with mass M, radius R, and
a mass density that varies with radius as
where
of M and R.
Hint: Divide the planet into infinitesimal shells of thickness dr,
then sum (i.e., integrate) their masses.
d. Find an expression for the gravitational field strength inside
the planet at distance r < R.
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Chapter 24 Solutions
Student Workbook for Physics for Scientists and Engineers: A Strategic Approach, Vol 1. (Chs 1-21)
- 5. A potential divider circuit is made by stretching a 1 m long wire with a resistance of 0.1 per cm from A to B as shown. 8V A 100cm B sliding contact 5Ω A varying PD is achieved across the 5 Q resistor by moving the slider along the resistance wire. Calculate the distance from A when the PD across the 5 Q resistor is 6 V.arrow_forward4. A voltmeter with resistance 10 kQ is used to measure the pd across the 1 kQ resistor in the circuit below. 6V 5ΚΩ 1ΚΩ V Calculate the percentage difference between the value with and without the voltmeter.arrow_forward1. A 9V battery with internal resistance 5 2 is connected to a 100 2 resistor. Calculate: a. the Power dissipated in the 100 2 resistor b. The heat generated per second inside the battery. C. The rate of converting chemical to electrical energy by the battery. 2. A 230 V kettle is rated at 1800 W. Calculate the resistance of the heating element.arrow_forward
- 2. If each of the resistors in the circuit below has resistance R show that the total resistance between A and B is 5R/11 A Barrow_forward1. At 0°C a steel cable is 1km long and 1cm diameter when it is heated it expands and its resistivity increases. Calculate the change in resistance of the cable as it is heated from 0-20°C The temperature coefficient of resistance a, gives the fractional increase in resistance per °C. So increase in resistance AR = Ra.AT Where R, is the resistance at 0°C For steel a, 0.003 °C The coefficient of linear expansion a- gives the fractional increase in length per °C temperature rise. So increase in Length AL La-AT Where L, is the length at 0°C For steel a₁ = 12 x 10 °C-1 The resistivity of steel at 0°C = 1.2 x 10 Qmarrow_forward1. F E 6V 10 1.1. B a 6V b C C Apply Kirchoff's 1st law to point C for the circuit above Apply Kirchoff's 2nd Law to loops: a. ABCFA b. ABDEA C. FCDEF d. Find values for currents a,b and c Darrow_forward
- 2. The results of the Rutherford experiment can be categorized in 3 statements. Fill in the missing words Most 11. Some III. A few State which result gives evidence that the nucleus is a. heavier than an alpha particle b. very small compared to the size of the atom c. positively charged 3. Using values in the diagram derive an expression for r .0 e marrow_forward3. A 100 W light bulb is connected to 230 V mains supply by a cable with resistance 0.12. Determine the heat loss per second by the cable.arrow_forward1. The image shows electrons flowing in a conductor with cross sectional area 1mm². A electron flow • Add an arrow showing the direction of current. B • Which end has the highest potential? • Calculate the current when 1019 electrons flow through the wire in 10 s. If there are 1026 electrons per unit volume what is the drift velocity of the electrons?arrow_forward
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