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3 years ago

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The motor of a ski boat generates an average power of 2.74 × 10^4 W when the boat is moving at a constant speed of 14.3 m/s. Whe

n the boat is pulling a skier at the same speed, the engine must generate an average power of 9.20 × 10^4 W. What is the tension in the tow rope that is pulling the skier?

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

T = 4517.48 N

Explanation:

It is given that,

Average power of the motor, $P_1=2.74\times 10^4\ W$

Speed of the boat, $v_1=14.3\ m/s$

Force or tension acting on the motor at this point is given by :

$P=T\times v$

$T=\dfrac{P_1}{v_1}$

$T=\dfrac{2.74\times 10^4}{14.3}$

T = 1916.08 N

Average power generated by the engine, $P_2=9.2\times 10^4\ W$

Force or tension acting on the motor at this point is given by :

$P=T\times v$

$T=\dfrac{P_2}{v_2}$

$T=\dfrac{9.2\times 10^4}{14.3}$

T = 6433.56 N

So, the tension in the tow rope that is pulling the skier is, T = 6433.56 N - 1916.08 N = 4517.48 N

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A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th

blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

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Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

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3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

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