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madam [21]
3 years ago
11

A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

swings from this rope that is 6.00 m long, his tangential speed at the bottom of the swing is 9.50 m/s. What is the centripetal acceleration, in m/s2, of the child at the bottom of the swing
Physics
2 answers:
solong [7]3 years ago
4 0

Answer:

10.8n

Explanation:

For this case the centripetal force is given by:

Where,

m: mass of the object

v: tangential speed

r: rope radius

Substituting values in the equation we have:

Then, doing the corresponding calculations:

Answer:

The centripetal force exerted on the rope is:

trasher [3.6K]3 years ago
3 0

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

The centripetal acceleration is given by:

a_{c} = \frac{v^{2}}{r}

Where:

v^{2}: is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

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When working with a powered instrument, the swirling effect produced by the water stream within the confined space of a periodon
Lina20 [59]

Answer:

Acoustic microstreaming

Explanation:

Acoustic microstreaming is the swirling effect produced by water stream confined in a spaced of a periodontal pocket.

  • It is the movement of water in a particular direction as a result of mechanical pressure within the fluid body.
  • They are often used in dental procedures to remove particulates from the teeth.
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3 0
3 years ago
Can you plz help me in on 11 12 and 14?
anygoal [31]
I’m pretty sure 14 is mutations
3 0
3 years ago
a catcher "gives" with the ball when he catches a 0.196 kg baseball moving at 31 m/s. if he moves his glove a distance of 5.32 c
Aloiza [94]

Answer:

3540.5N

Explanation:

Step one:

given data

mass m= 0.196kg

speed  v= 31m/s

distance r= 5.32cm = 0.0532m

Step two

The expression relating force, mass, velocity and distance is

F= mv^2/r

substitute we have

F=0.196*31^2/0.0532

F=0.196*961/0.0532

F=188.356/0.0532

F=3540.5N

6 0
3 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
3 0
3 years ago
after the car leaves the platform , gravity causes it to accelerate downward at a rate of 9.8 m/s2. what is the gravitational fo
alexandr1967 [171]

The gravitational force on the car is the force popularly known
as the car's "weight".  Its magnitude is

            (9.8 m/s²) times (the car's mass, in kilograms) .

The unit of this quantity is [newton] .

5 0
3 years ago
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