Answer:
Explanation:
Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s
An impulse results in a change of momentum
The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s
The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s
The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s
The tug applies 0.012e6 N•s of impulse each second.
The initial barge momentum will be zero in
t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds
To stop the barge in one minute(60 s), the tug would have to apply
4.5e6 / 60 = 75000 N•s /s or 75 000 N
Answer:
The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.
Explanation:
Answer:
Explanation:
When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.
This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ = 
Hence for a given w , θ depends upon g or weight .
Answer:
The the speed of the car is 26.91 m/s.
Explanation:
Given that,
distance d = 88 m
Kinetic friction = 0.42
We need to calculate the the speed of the car
Using the work-energy principle
work done = change in kinetic energy
Put the value into the formula
Hence, The the speed of the car is 26.91 m/s.
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
