Speed. units m/s
Because you want to know how fast he can run without involving any direction thus the best is speed
speed = distance / time
To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to
The ideal gas equation said us that,
PV = nRT
Here,
P = pressure
V = Volume
R = Gas ideal constant
T = Temperature
n = Amount of substance (at this case the mass)
Then
The amount of substance per volume is the density, then
Replacing with our values,
Finally the specific volume would be
Answer:
Both are moving at 30 km/h, so their speed is the same. ... enough fuel for the trip/how long it will take. 4 Weight is a force, and so is a vector. ... c At 10 seconds David's displacement is.
Answer:
(a) μ ≈ 0.57
(b) not enough information
Explanation:
Draw a free body diagram.
There are four forces acting on block B:
- Weight force Mb g pulling down
- Normal force N pushing up
- Friction force Nμ pulling left
- Tension force T₁ pulling right
There are three forces acting on the node:
- Tension force T₁ pulling left
- Tension force T₂ pulling 35° above the horizontal
- Weight force Ma g pulling down
(a)
Sum of forces on block B in the y direction:
∑F = ma
N − Mb g = 0
N = Mb g
Sum of forces on block B in the x direction:
∑F = ma
T₁ − Nμ = 0
T₁ = Nμ
T₁ = Mb g μ
Sum of forces on the node in the y direction:
∑F = ma
T₂ sin 35° − Ma g = 0
T₂ sin 35° = Ma g
Sum of forces on the node in the x direction:
∑F = ma
T₂ cos 35° − T₁ = 0
T₂ cos 35° = T₁
T₂ cos 35° = Mb g μ
Divide the previous equation by this equation, eliminating T₂.
tan 35° = Ma g / (Mb g μ)
μ = Ma / (Mb tan 35°)
μ = 0.4 / tan 35°
μ ≈ 0.57
(b) T₂ sin 35° = Ma g = 0.4 Mb g
Without knowing the value of Mb, we cannot find the value of the tension force T₂.
Answer:
increases by a factor of 4.
Explanation:
The power dissipated through a resistor is I2RI2R
Current=IResistance=RPower=(current)2×(resistance)=I2RIf we double the current, P=(2I)2R=4I2RCurrent=IResistance=RPower=(current)2×(resistance)=I2RIf we double the current, P=(2I)2R=4I2R
Thus, doubling the current increases the power dissipated through a resistor by a factor of four.