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3 years ago

What is the maximum weight a boat can hold if a boat can displace 60.5ml?

Physics

1 answer:

Valentin [98]3 years ago

6 0

Answer:

a. 60.5 kg

Explanation:

Given data,

The maximum water a boat can displace is, 60.5 ml

According to the principle of buoyancy, the weight of the floating body is equal to the weight of the liquid displaced.

Under standard temperature and pressure, a unit mass of water equals one liter.

If a boat can displace a maximum of 60.5 ml of water, then it can hold a mass of a maximum of 60.5 kg of mass.

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A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass

saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

$W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)$ (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

$d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m$

Furthermore, the angle α is:

$\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°$

Then, you solve the equation (1) for W and replace the values of all parameters:

$W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J$

The work done by the student is 609.97J

6 0

4 years ago

The total circuit resistance is 6½ ohms. If a second 6½-ohm resistor is placed in parallel with the first, the total circuit res

mamaluj [8]

Answer:

Explanation:

It will increase cause the more branches you add the total current will increase

5 0

3 years ago

The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 k

Andrew [12]

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

$\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2$

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

$\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}$

a) The velocity (V2) is:

$\frac{P1}{pg}+Z1=\frac{V2^2}{2g}$

$(\frac{P1}{pg}+Z1)(2g)=V2^2$

$V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}$

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

$V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}$

$V2=24.692 m/s$

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

$\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3$

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

$\frac{V2^2}{2g}=\frac{P3}{pg}+Z3$