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3 years ago

Help a Girl Out and Answer This you will get Brainliest, Thanks, and more points if your answer is correct (:

Physics

1 answer:

melamori03 [73]3 years ago

3 0

Answer:

this was 2 weeks ago, but im pretty sure the correct answers are:

B=c

C=d

D=b

hope this helps and is correct :)

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Andrea asked her brother to take a 4 ft floating raft out of the water near the wave-swept shore. Using this raft as a measuring

lutik1710 [3]

Answer:

176ft/s

Explanation:

Speed of a wave is a function of the frequency and wavelength of the wave. It is expressed mathematically as:

V = fλ where:

V is the speed of the wave

f is the frequency

λ is the wavelength

Given f = 16Hz, λ = 11ft

V = 16×11

V = 176ft/s

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4 years ago

What is your average speed if you walk 2km in 20 minutes

vlabodo [156]

You're average speed would be 6km/hour or 100m per minute.

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3 years ago

What is light intensity?

Stels [109]

Answer:

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3 years ago

A ball rolls off a desk at a speed of 2 m/s and lands 0.50 seconds later.

Hoochie [10]

Answer:

(a) The ball lands <u>1 m</u> ahead of the desk's base.

(b) The height of the desk is <u>1.225 m.</u>

Explanation:

Given:

Initial velocity of the ball is, $u_{0}=2\ m/s$

Time of flight of the ball is, $t=0.50\ s$

(a)

Let the ball fall at a distance of $x$ from the base of the desk and let the height of the desk be $y$ .

The given motion of the ball is a projectile motion that can be divided into 2 mutually perpendicular directions.

Now, considering only the horizontal motion of the ball. The ball not under the influence of any force in the horizontal direction.

Therefore, the distance covered by the ball horizontally is given as:

$x=v_0\times t\\x=2\times 0.50=1\ m$

So, the ball lands 1 m ahead of the desk's base.

(b)

Now, consider only the vertical motion of the ball. The ball is under the influence of gravity.

So, vertical distance can be obtained using Newton's equations of motion.

We use the following equation of motion:

$y-y_0=u_{oy}t+\frac{1}{2}gt^2$

Where,

$y_{0}\rightarrow \textrm{initial vertical position of ball}\\u_{oy}\rightarrow \textrm{initial vertical velocity}\\g\rightarrow \textrm{acceleration due to gravity}\\y\rightarrow \textrm{final position of ball vertically}$

Here, as the ball leaves the desk horizontally, initial velocity is only in the horizontal direction and doesn't have any component in the vertical direction. So, $u_{oy}=0\ m/s$