Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
Answer:
See the explanation below
Explanation:
Density is defined as the relationship between mass and volume, i.e. the following equation can be used:
density = m/v
where:
density [kg/m^3]
m = mass [kg]
v = volume [m^3]
If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.
m = density*v
To understand this concept more clearly, let's use the following example:
We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.
1000 = m /1
m = 1000*1 = 1000 [kg]
Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.
1000 = 500/v
v = 500/1000
v = 0.5 [m^3]
We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
1.) The object's Velocity
Faster it goes, more kinetic energy it has
