Answer:
a) k = 891.82 N/m
b) e = 0.0143 m = 1.43 cm
c) W = 5.02 J
Explanation:
Step 1: Data given
Mass = 2.60 kg
the spring stretches 2.86 cm = 0.0286
Step 2: What is the force constant of the spring?
Force constant, k = force applied / extension produced
k = (2.60kg * 9.81N/kg) / 0.0326 m
k = 891.82 N/m
b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it
Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm
Half the mass means half the extension
c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
W = average force used * distance
W = 1/2 * k*e * e = 1/2 k*e²
W = 1/2 * 891.82 * (0.075)² = W = 5.02 J
The kinetic energy of a 750 kg car moving at 50.0 km h is 72300 joules.
Given,
mass of car (m)= 750 kg
speed of car (v)= 50 km/h
= 50 x 5/18
= 13.9 m/s
we know that kinetic energy= 1/2 x m x v*v
= 1/2 x 750 x 13.9 *13.9
= 72300 joules
so the kinetic energy is 72300 joules.
Energy is the capacity of an object to do work, and like work, energy's unit is the joule (J). Energy exists in many different forms, but the one we think of most often when we think of energy is kinetic energy. Kinetic energy is often thought of as the energy of motion because it is used to describe objects that are moving.
Learn more about kinetic energy here :-
brainly.com/question/999862
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Answer:
C. Both A and B
Explanation:
Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.
Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.
Answer:
19.2 mA
Explanation:
C = 9 micro farad = 9 x 10^-6 F
L = 16 H
f = 60 Hz
Vrms = 116 V


XL = 6028.8 ohm


Xc = 294.88 ohm
Let Z be the impedance of the circuit.


Z = 6036 ohm


Irms = 0.0192 A
Irms = 19.2 mA