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Alex_Xolod [135]

2 years ago

6

The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs

erved a sea lion making a circular turn with a radius of 0.37 m while swimming at 4.0 m/s.
What is the sea lion's centripetal acceleration, in units of g?

What percentage is this acceleration of that of an F-15 fighter jet's maximum centripetal acceleration of 9g?

1 answer:

anygoal [31]2 years ago

6 0

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

$v = 4 m/s$

Now the centripetal acceleration of the sea lion is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{0.37}$

$a_c = 43.2 m/s^2$

as we know that

$g = 9.8 m/s^2$

so we have

$a = 4.41 g$

Now Percentage of this acceleration wrt maximum jet acceleration is given as

$P = \frac{4.41 g}{9g} \times 100$

$P = 49$ %

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)

nikdorinn [45]

Answer:

$E = 55.9583\ Volts/meter$

Explanation:

First let's find the electric potential using y = 22.5:

$V(y) = 1.69y^2 +15.6y+52.5$

$V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5$

$V(22.5) = 1259.0625\ Volts$

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

$E = V/d$

$E = 1259.0625/22.5$

$E = 55.9583\ Volts/meter$

3 0

2 years ago

A stock person at the local grocery store has a job consisting of the following five segments:

vaieri [72.5K]

Answer:

B

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

6 0

2 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

2 years ago

According to Dr. paul Narguizian professor of Biology and Science Education at California State University, ______ are generaliz

Mazyrski [523]

Answer:

I believe the correct answer would be A :)

Explanation:

3 0

1 year ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w

OLga [1]

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\ Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

8 0

2 years ago