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Ksenya-84 [330]
4 years ago
12

Bobo, the clown, can swim at 2.0 m/s. he must make a landing directly across to the north side of the styx river, which is 100.

m wide. the river flows at 6.0 m/s due east at this point. bobo’s biggest problem is that he can only swim while facing due north. how can he possibly make a landing at the desired location?
Physics
1 answer:
victus00 [196]4 years ago
6 0
We are given the following:

Bobo's swimming speed = 2.0 m/s
Width of the river = 100 m
Flowrate of the river = 6.0 m/s due east

First, we need to illustrate the problem. Draw the river with a width of 100 meters. Then, the flow of the river, east at 6 meters per second. Lastly, draw Bobo at one side of the river facing north and an arrow representing swimming speed at 2 meters per second.

Now, we can use the Pythagorean theorem to solve this rate problem.

c^2 = a^2 + b^2 

c = speed of Bobo needed
a = speed of Bobo facing north
b = flow rate of the river going east

c^2 = 2^2 + 6^2 

c = 6.32 m / s should be his speed to overcome the current and make a landing at the desired location. 
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth
Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

W=mg

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center


The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

800 N= W=mg=m \frac{GM}{R^2} (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

200 N = W' = mg' = m \frac{GM}{(R+h)^2} (2)

If we divide eq.(1) by eq.(2), we get

\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}

4=\frac{(R+h)^2}{R^2}

By solving the equation, we find:

4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0

which has two solutions:

h=-3R --> negative solution, we can ignore it

h=R --> this is our solution

Since the Earth's radius is R=6.4\cdot 10^6 m, the person should be at h=R=6.4\cdot 10^6 m=6400 km above Earth's surface.

5 0
4 years ago
Two planets with the same mass and atmospheric conditions orbit a single star. Planet A is closer to the star than Planet B. Whi
8_murik_8 [283]

Answer:

is not this one Planet B is expected to be hotter than Planet A.

Explanation:

3 0
2 years ago
A hydraulic press must be used to press-fit two components together in an assembly process. the system must generate at least 77
Kobotan [32]
The pressure generated by the piston at the input cylinder must be:

p = F / A = 250 lb-f / 30 in^2


The pressure generated by the piston at the output cylinders is p = F / A = 775 lb-f / A.


Pascal principle rules that both pressures are equal, so:


250 lb-f / 30 in^2 = 775 lb-f / A => A = 775 lb-f * 30 in^2 / 250 lb-f = 93 in^2


Gvien that each output cylinder are 30in^2 you need 93 / 30 = 3.1 cylinders.


Which means that at least you need 4 cylinders to generate a force at least of 775 lb-f.


Answer: 4
8 0
4 years ago
A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th
zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

        v = ?

        u = 0 m/s

        a = 4.0 m/s²

         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

        a = -2.0 m/s²

         t = 4 s

         v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0
3 years ago
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