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2 years ago

9

A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t

he buoyant force acting on the submerged log ?

1 answer:

Sedbober [7]2 years ago

3 0

Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

where:

m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

V = 0.22 [m^3]

The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

F = 1618.65[N]

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0

2 years ago

Seven seconds after a brilliant flash of lightning, thunder shakes the house. approximately how far was the lightning strike fro

tangare [24]

Very roughly 7,700 feet ... about 1.5 miles.

8 0

2 years ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago

A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0

Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

$F_{lift}=F_{up} - F_{dw}=0.5*p*V^2$

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

$F_{lift}=0.5*p*V^2=W$

By replacing the known values it is possible to find the plate's weight:

$F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W$

$W=72.6 N$

When the plate kept to 30° from the vertical the moment equation balance is written as:

$F_{lift}=0.5*p*V^2=W*sen(30\°)$

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

$V=\sqrt(W*sen(30\°)/0.5p)$

$V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})$

$V=\sqrt(60.5 \frac{N}{kg/m^3})$

$V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})$

$V=\sqrt(60.5 \frac{m^2}{s^2})$

$V= 7.78 m/s$

7 0

2 years ago

Your annoying little brother is dropping rocks out of his bedroom window on the 2nd floor. You are on the ground floor and watch

Valentin [98]

Answer:

Too little information, please elaborate

5 0

2 years ago