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3 years ago

15

Can you please answer qeastion

1 answer:

notsponge [240]3 years ago

8 0

Answer:

the mantel

Explanation:

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The troughs or zero amplitude of a standing wave are called ____________.

konstantin123 [22]

Each successive graph is at a later time. You can see from these graphs how the amplitude of the total electric field changes, but the positions of the crests and troughs (called antinodes) and places of zero field (called nodes) never change.!!!!!!!!!!!!!!!!!

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3 years ago

An arrow in flight has an initial velocity of 65 meters per second, and 10 seconds later, it has a velocity of 35 meters per sec

-BARSIC- [3]

Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2

7 0

3 years ago

At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec

evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out= 3.14 inch³/sec

4 0

3 years ago

Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

$Range=v_i^2\frac{sin(2\theta)}{g}$

which for our case renders:

$Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m$

4 0

2 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

2 years ago