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3 years ago

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Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in cm) be

tween two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 343 m/s?

1 answer:

kenny6666 [7]3 years ago

8 0

Answer:

34.3cm

Explanation:

Attached is the full solution

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Consider a system made of a ramp that has a run of 8 m and a rise of 6 m. Two equal masses of 4-kg are connected to each other b

djyliett [7]

Answer:

The force of friction is 15.68 N.

Explanation:

Given:

Rise is 6 m and run is 8 m.

Mass of each body (m) = 4 kg

The whole system is in equilibrium.

Now, consider the diagram below representing the scenario given in the question.

The forces acting on the mass that is hanging are tension force in the upward direction and weight of the body acting vertically downward.

As the mass is in equilibrium, the total upward force equals total downward force. So,

$T= mg=4\times 9.8=39.2\ N$ -------- (1)

Now, the forces acting on the other mass along the ramp are:

1. Tension (T) up the ramp

2. $mg sin\theta$ and frictional force (f) down the ramp

Now, as per question:

Rise = 6 m and run = 8 m

So, from figure,

$\tan\theta=\frac{6}{8}=0.75\\\\\theta=\tan^{-1}(0.75)=37\°$

Now, $\sin\theta=\sin(37\°)=0.6$

Now, as the other mass is also at rest, net force acting on it is also 0. So,

$F_{net}=0\\\\T-(mg\sin\theta+f)=0\\\\mg\sin\theta+f=T\\\\f=T-mg\sin\theta$

Now, plug in the given values and solve for 'f'. This gives,

$f=39.2-(4\times 9.8\times 0.6)\\\\f=39.2-23.52=15.68\ N$

Therefore, the force of friction is 15.68 N

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4 years ago

According to the picture, what is the truck doing?

Slav-nsk [51]

Answer:

but were is the picture

how can we see

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3 years ago

A 14-kg object moving with a constant velocity

natta225 [31]

The final velocity of the 14 kg object is 1.6 m/s in the same direction

Explanation:

We can solve this problem by using the law of conservation of momentum: the total momentum of the system must be conserved before and after the collision. Therefore, we can write

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 14 kg$ is the mass of the first object

$u_1 = 5.0 m/s$ is the initial velocity of the first object

$v_1$ is the final velocity of the first object

$m_2 = 8.0 kg$ is the mass of the second object

$u_2 = 3.0 m/s$ is the initial velocity of the second object

$v_2 = 9.0 m/s$ is the final velocity of the second object

Re-arranging the equation and substituting the values, we find:

$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 v_2}{m_1}=\frac{(14)(5.0)+(8.0)(3.0)-(8.0)(9.0)}{14}=1.6 m/s$

And the direction is the same as the initial direction, since it has the same sign.

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3 years ago

The speed of a rocket just after being launched is 12 m / s.

drek231 [11]

This question involves the concepts of the equations of motion, kinetic energy, and potential energy.

a. The kinetic energy of the rocket at launch is "3.6 J".

b. maximum gravitational potential energy of the rocket is "3.6 J".

<h3>a. KINETIC ENERGY AT LAUNCH</h3>

The kinetic energy of the rocket at launch is given by the following formula:

$K.E=\frac{1}{2} m v_i^2$

where,

K.E = initial kinetic energy = ?
m = mass of rocket = 0.05 kg
$v_i$ = initial speed = 12 m/s

Therefore,

$K.E=\frac{1}{2}(0.05\ kg)(12\ m/s)^2$

K.E = 3.6 J

<h3>b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>

First, we will use the third equation of motion to find the maximum height reached by rocket:

$2gh=v_f^2-v_i^2$

where,

g = -9.81 m/s²
h = maximum height = ?
vf = final speed = 0 m/s

Therefore,

2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²

h = 7.34 m

Hence, the maximum gravitational potential energy will be:

P.E = mgh

P.E = (0.05 kg)(9.81 m/s²)(7.34 m)

P.E = 3.6 J

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2 years ago

Which statements about the two fossil images are correct?

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What fossil images, there are no pictures attached.

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4 years ago

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