Answer:
Explanation:
First we need to determine the distance covered during deceleration. According to the equation of motion.
S = ut+1/2at²
Given:
u = 20m/s
t = 0.50s
a = -10m/s (deceleration is negative acceleration)
S = 20²+1/2(-10)(0.5)²
S = 400-5(0.5)²
S = 400-5(0.25)
S = 400-1.25
S = 398.75m
If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m
I use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop
Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.
The force is repulsive.
Explanation: Please see the attachments below
"<span>During radioactive decay, atoms break down, releasing, particles or energy" is the one statement about radioactive decay among the following choices given in the question that is true. The correct option is option "b".
"H</span>alf-life" is the term among the following that <span>refers to the time it takes for one-half of the radioactive atoms in a sample of a radioactive element to decay. The correct option is option "d".</span>
Answer:
Explanation:
Let c be the circumference and r be the radius
c = 2πr , r = c / 2π , area A = π r² = π (c/2π )² = (1/4π) x c²
flux (ψ) = BA = 1 X 1/4π X c²
dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt
at t = 8 s
c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s
e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.
