Answer:
In which direction does the current in circuit A flow?
counterclockwise
<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms?
</h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 =
1
</h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?
</h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2>
</h2>
Explanation:
Answer:
Independent variable: Surrounding color
Dependent Variable: Eating Habits
Explanation:
Dependent variables are dependent on Independent variables which means that in this question Eating habits are affected by the Surrounding color.
Hope it Helps!!
Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )
This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.
Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.
I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.
