Someone please help me for 7 points.... I don't know how to do this...

The vibrations produced by a jackhammer are used to break up pavement. What type of waves did/does the jackhammer produce into t

klio [65]

<span>P waves

P-waves are sometimes called pressure waves, and sometimes they are called longitudinal waves. P-waves are transmitted through the Earth’s interior with a backwards and forwards motion along the line of travel, by alternating compression and dilatation. A jack hammer creates P-waves. They are capable of passing through any type of material they encounter, including the liquid of the Earth’s outer core, although they
will be bent and deflected when they pass across the boundaries separating
layers of different densities.

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8 0

4 years ago

Match the following list of key words with their definitions

Korvikt [17]

Question 1 :

1. VELOCITY

2. SPEED

3. NEWTON

4. MOMENTUM

5. MASS

Question 2

1. FORCE

2. FRICTION

3. GRAVITY

4. ELECTROAGNETIC FORCE

5. ACCELERATION

6 0

2 years ago

Which method of separation would you use when working in the garden and trying to separate the dirt from the weeds?

Thepotemich [5.8K]

Sifting is the best method cuz all the dirt will be carried by wind.

3 0

3 years ago

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In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d)

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

So

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

A 500-n parachutist opens his chute and experiences an air resistance force of 800 n. the net force on the parachutist is then

Tasya [4]

Force of 500 N is acting on the parachutist.

Parachutist applies 500 N force in downward direction.

Answer:

300 N upward

Solution:

Parachutist feels air resistance of 800 N.

Thus, 800 N of force is acting in upward direction.

Total force acting on the parachutist is given by,

$F_{net}$ = air resistance force - force of parachutist

$F_{net}$ = 800-500

$F_{net}$ = 300 N

Direction of force is in upward direction because the air resistance force is more than force of parachutist.

6 0

3 years ago

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