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garri49 [273]
3 years ago
10

I need help with this answer

Physics
1 answer:
iren2701 [21]3 years ago
7 0
It’s oxygen i’m pretty sure
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A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
ANEK [815]

Answer:\dfrac{\pi}{2} ms^{-1}

Explanation:

Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So,d=2\pi r

Given,d=0.5m\\T=2sec

d=2\times \pi \times 0.5=\pi m

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

5 0
3 years ago
What kind of graph uses points on a grid connected by line segments to represent data and is usually used to show change over ti
Brrunno [24]
Line Graph

Hope this is what you are looking for! 
5 0
3 years ago
~~~~~NEED HELP ASAP~~~~~
Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0
2 years ago
Your tired 10 lb. baby cousin is cradled in your arms as you're walking back and forth across a 15m room to get her to fall asle
Mashcka [7]

Work done = Force x Distance

Force = 10 lb = 44.5 N

Work Done = 44.5 N x 15 m

= 667.5 N-m

6 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

while the vertical distance covered must be

d_y = 6 m

The rock is thrown horizontally with velocity v_x, so we can rewrite the horizontal distance as

d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

The vertical distance covered instead is

d_y = \frac{1}{2}gt^2

where we omit the term ut since the initial vertical velocity is zero. From this equation,

t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
3 years ago
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