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3 years ago

A rock is dropped from a bridge. What happens

1 answer:

5 0

When a rock is dropped from a bridge in this way, in terms of the magnitude of the acceleration and the speed of the rock as it falls, the best option is "<span>(1) Both acceleration and speed increase."</span>

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Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

Nataliya [291]

Answer:

Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft

Explanation:

7 0

3 years ago

Define average velocity.Immersive Reader

vagabundo [1.1K]

Answer:

The slope of a graph of position vs time

4 0

3 years ago

A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that

Nana76 [90]

Answer:

m=417.24 kg

Explanation:

Given Data

Initial mass of rocket M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s

velocity of gas vg = 4300 m/s

Θ = 11° angle in degrees

To find

m = mass of gas

Solution

Let m = mass of gas

first to find Initial speed with angle given

Vi=vi×tanΘ...............tan angle

Vi= 2900m/s × tan (11°)

Vi=563.7 m/s

Now to find mass

m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

put the values as we have already solve vi ×tanΘ

m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

m=417.24 kg

7 0

3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

OverLord2011 [107]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

$\theta_i = 34 degree$

so we know that

$\theta_i = \theta_r$

so here we can say that reflection angle of light will be same as angle of incidence

$\theta_r = 34 degree$

8 0

3 years ago