The llamas average acceleration from 15 to 20 seconds is 5
An object with more mass has more kinetic energy than an object with less mass, if both objects are moving at the same speed. <em>(c)</em>
Answer:
The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm
Explanation:
Given that,
Emf of battery = 15.0 V
Voltage = 12.2 V
Power = 14.0 W
(a). We need to calculate the value of R
Using formula of power
Where, R = resistance
P = power
V = voltage
Put the value into the formula
(b). We need to calculate the internal resistance of the battery
Firstly we calculate the current
Using formula of current
Put the value of P and V into the formula
We calculate the internal resistance
Using formula of emf
Put the value into the formula
Hence, The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm
The lungs art part of The excretory<span> system....
</span><span>somatic nervous system is ..... </span><span>autonomic nervous system<span>....
</span></span>
The process in which organ systems work to maintain a stable internal environment is called homeostasis. Keeping a stable internal environment requires constant adjustments. Here are just three of the many ways that human organ systems help the body maintain homeostasis:
Respiratory system: A high concentration of carbon dioxide in the blood triggers faster breathing. The lungs exhale more frequently, which removes carbon dioxide from the body more quickly.
Excretory system: A low level of water in the blood triggers retention of water by the kidneys. The kidneys produce more concentrated urine, so less water is lost from the body.
Endocrine system: A high concentration of sugar in the blood triggers secretion of insulin by an endocrine gland called the pancreas. Insulin is a hormone that helps cells absorb sugar from the blood.
Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules