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2 years ago

5

A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the

speed of the boat is 10 mph and the river is flowing east at 5mph, in what direction should she head the boat in order to arrive at the desired landing point

1 answer:

I am Lyosha [343]2 years ago

4 0

Answer:

If she stands on the North side of a river flowing to the East at 5 mph,

she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have

x^2 + 5^2 = 10^2 where x is her speed directly across the river

x = (75)^1/2 = 8.66 mph towards the South

sin theta = 5 / 10 = 1/2

She must angle the boat at 30 deg from straight South

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Two rods are identical, except that one is brass (Y = 9.0 × 1010 N/m2) and one is tungsten (Y = 3.6 × 1011 N/m2). A force causes

PIT_PIT [208]

To solve this problem it is necessary to apply the concepts related to Young's Module, and find the radius that gives the ratio between the two given materials. Young's module can be defined as,

$Y=\frac{FL}{A \Delta L}$

Where,

F= Force

L = Initial Length

A = Cross-sectional Area

$\Delta L =$ Change in Length

Re-arrange the equation to find the change in Length we have,

$\Delta L = \frac{FL}{AY}$

If both the Force, as the Area and the initial length are considered constant, we can realize directly that the change in length is inversely proportional to Young's Module, therefore

$\Delta L \propto \frac{1}{Y}$

Applying this concept to that of the two materials (Brass and Tungsten),

$\frac{\Delta L_T}{\Delta L_B} = \frac{Y_B}{Y_T}$

$\frac{\Delta L_T}{\Delta L_B} = \frac{9*10^{10}}{3.6*10^{11}}$

$\frac{\Delta L_T}{\Delta L_B} = 0.25$

If the force caused $3 * 10^ {- 6}m$ to be stretched, the tungsten will stretch 0.25 of that ratio

$L_T = 3*10^{-6}*0.25$

$L_T = 7.5*10^{-7}m$

Therefore the amount of stretch of Tungsten is 7.5*10^{-7}m

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3 years ago

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2 years ago

Science assessment help pls

DENIUS [597]

Answer:

Work Done = W

force = F

Distance = d

W = Fd

or W = F*d

W (in joules) = 3.5*4 = 14 Nm (or J)

1Nm = 1J

so newton meters and joules are the same

Power = Work (in joules) /time (in seconds)

i don’t know the time so i can’t solve it

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2 years ago

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A 1700kg rhino charges at a speed of 50.0km/h. what average force is needed to bring the rhino to a stop in 0.50s?

uranmaximum [27]

From 50km/h to 0km/h in 0.5s we need next acceleration:
First we convert km/h in m/s:
50km/h = 50*1000/3600=13.8888 m/s
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Now we use Newton's law:

F=m*a

F=1700*27.7777 = 47222N

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3 years ago

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

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3 years ago