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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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Answer:

Explanation:

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mass of block = 3 kg

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F = 17.64 N

using energy conservation

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2+Fx

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250 x² + 17.64 x - 37.5 = 0

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4 years ago
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.
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Answer:

D=99.4665307m \approx 99.5m

Explanation:

From the question we are told that

Mass  m=6kg

Velocity of mass  V_m=16

Force of Tunnel  F_t=8N

Length of Tunnel L_t=1.6

Height of frictional incline H_i=2.9

Angle of inclination  \angle =16 \textdegree

Acceleration due to gravity  g=9.8m/s^2

First Frictional surface has a coefficient  \alpha_1 =0.21\ for\ d_c=1

Second Frictional surface has a coefficient \alpha _2=0.1

Generally the initial Kinetic energy is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}(6)(16)^2

K.E=768

Generally the work done by the Tunnel is mathematically given as

w_t=F_t*d_t

w_t=8*1.6

w_t=12.8J

Therefore

Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t

E_t=K.E+E_t\\E_t=768J+12.8J

E_t=780.8J

Generally the energy lost while climbing is mathematically given as

E_c=mgh

E_c=(6)(9.8)(2.9)

E_c=170.52J

Generally the energy lost to friction is mathematically given as

E_f=\alpha *m*g*cos\textdegree*d_c

E_f=0.21*6*9.8*cos16*1

E_f=11.86965942 \approx 12J

Generally the energy left in the form of mass Em is mathematically given as

E_m=E_t+E_c+E_f

E_m=(768J)-(170.52)-(12)

E_m=585.48J

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E_m=\alpha_2*g*m*d

Therefore

It slide along the second frictional region

D=\frac{585.46}{0.1*9.81*6}

D=99.4665307m \approx 99.5m

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