If the sound comes from the right side, the waves reach the right ear before the left ear. if the sound comes from the left side, the waves reach the left ear before the right ear. The difference between the phases of waves reaching both ears is detected by the ears and then interpreted by the brain

8 0

3 years ago

An electric current in a wire flows to the West in a magnetic field directed

vredina [299]

Answer: Current in a wire

We can use the same right-hand rule as we did for the moving charges—pointer finger in the direction the current is flowing, middle finger in the direction of the magnetic field, and thumb in the direction the wire is pushed.

Explanation:

7 0

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

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 box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago