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2 years ago

8

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

r speed is decreased by the water's speed. The current in a river flows at 0.52 m/s. In still water you can swim at 1.73 m/s. If you swim downstream a certain distance, then back again upstream, how much longer, in percent, does it take compared to the same trip in still water?

1 answer:

Blababa [14]2 years ago

5 0

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

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DIA [1.3K]

Answer:

1st law--Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net forceacting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.

2nd— The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum(mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the "changes" expressed in the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass m, the second law states that the force F is the product of an object's mass and its acceleration a:

F = m * a

For an external applied force, the change in velocity depends on the mass of the object. A force will cause a change in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.

3rd law-- The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine.

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1 year ago

The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.

Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

$\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3$

$1\ kg=2.20462\ lb$

$1\ m=39.3701\ in$

$\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3$

So,

$10.1\ kN/m^3=0.03719\ lbs/in^3$

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2 years ago

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

Brut [27]

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

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3 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

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2 years ago

A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the

nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

7 0

2 years ago