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valkas [14]
3 years ago
6

A horse ran 600 m in a time of 33 s. What was the average speed of the horse?

Physics
1 answer:
GalinKa [24]3 years ago
5 0
<h3>Average Speed ≈ 18m/s</h3>

Explanation:

Distance : 600 m

Time : 33 seconds

Average Speed : ?

Av. Speed  =  \frac{Distance }{Time}   \\  =  \frac{600}{33}

Simplify

Av. Speed  = 18.18181 \\  = 18 \: m {s}^{ - 1}

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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
3 years ago
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Ax = 4.3 m
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Answer:

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3 years ago
A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due
Amiraneli [1.4K]

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

6 0
3 years ago
A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?
iris [78.8K]

Answer:

\Delta x=22.67786838m

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2 (1)

Where:

y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis

y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis

v_o=initial\hspace{3}velocity

t=travel\hspace{3}time

g=gravity\hspace{3}constant

\theta=Initial\hspace{3}launch\hspace{3}angle

In this case:

\theta=0

Because the dog jumps horizontally

Let's asume the gravity constant as:

g=9.8

y=0

Because when the dog reach the base the height is 0

y_o=70

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Now let's replace the data in (1)

y_o-\frac{1}{2} *(9.8)*t^2+70

Isolating t:

t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473

Finally let's find the horizontal displacement using this equation:

\Delta x=v_o*cos(\theta)*t

Replacing the data:

\Delta x=6*1*3.77964473=22.67786838m

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