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3 years ago

Your boat runs aground at high speed. What should you do FIRST?

1 answer:

7 0

In the event that you run on solid land, ensure nobody is harmed and after that check for spills. On the off chance that the effect did not cause a break, take after these means to attempt to get free. Try not to put the boat in turn around. Rather, stop the motor and lift the outdrive.

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A rod of length L is pivoted about its left end and has a force F applied perpendicular to the other end. The force F is now rem

timurjin [86]

Answer:

$F' = 4F$

Explanation:

When force applied to the end of the rod in perpendicular direction then net torque on the rod is given as

$\tau = L F sin90$

$\tau = LF$

Now another force is applied at mid point of the rod at an angle of 30 degree with the rod

so new value of torque is given as

$\tau = \frac{L}{2}F' sin\theta$

$LF = \frac{L}{2}F' sin30$

$LF = \frac{F'L}{4}$

so we have

$F' = 4F$

3 0

3 years ago

A spinning flywheel has rotational inertia I = 474.0 kg·m2. Its angular velocity decreases from 26.2 rad/s to zero in 204.0 s du

Flura [38]

Answer:

$\tau = 60.88 N.m$

Explanation:

given,

rotational inertia, I = 474.0 kg·m²

decrease in angular velocity

ω₁ = 26.2 rad/s ω₂ = 0 rad/s

time = 204 s

torque = ?

$\tau = I \alpha$

$\alpha = \dfrac{\omega_1-\omega_2}{t}$

$\alpha = \dfrac{26.2-0}{204}$

α = 0.128 rad/s²

$\tau = I \alpha$

$\tau = 474\times 0.128$

$\tau = 60.88 N.m$

frictional torque acting by the flywheel is equal to 60.88 N.m

7 0

3 years ago

Which takes more energy, going from 0 to 30, or from 30 to 60?

AfilCa [17]

0 to 30 takes more energy

3 0

3 years ago

As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the dens

azamat

Answer:

1.2825 * 10^3 kg/m³

Explanation:

Given that :

Mass of aluminum ball (m1) = 4kg

Apparent mass of ball (m2) = 2.10 kg

Density of aluminum (d1) = 2.7 * 10^3 kg/m³

Density of liquid (d2) =?

Using the relation :

d1 / d2 = m1 / (m2 - m1)

(2.7 * 10^3) / d2 = 4 / (4 - 2.10)

2700 / d2 = 4 / 1.9

4 * d2 = 2700 * 1.9

4 * d2 = 5130

d2 = 5130 / 4

d2 = 1282.5 kg/m³

Hence, density of liquid = 1.2825 * 10^3

6 0

3 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

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