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3 years ago

8

In general, the ________ of a simple machine is the ratio of the distance over which the force is applied to the distance over w

hich the load is moved.

2 answers:

aksik [14]3 years ago

3 0

Mechanical advantage, i hope i helped you!

andrey2020 [161]3 years ago

3 0

Mechanical advantage

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What is it? Please help me.

Fed [463]

I believe that is a

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3 years ago

The wetter the surface, the friction

allochka39001 [22]

Answer: Water can either increase or decrease the friction between surfaces.

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3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$

For the horse: $s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2$

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

$s_c = s_h$

$0.233t = 1.57 + 0.0068t^2$

$0.0068t^2 - 0.233t + 1.57 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}$

$t= \frac{0.233\pm0.11}{0.0136}$

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0

3 years ago

Explain how surface and subsurface events are integral parts of the rock cycle.

vagabundo [1.1K]

Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc.

3 0

3 years ago